y = sin 5 + log₁₀ x + 2 sec x
\(\therefore \cfrac { dy }{ dx } =0+\left( \cfrac { 1 }{ x } \right) { log }_{ 10 }e+2\left[ secxtanx \right] =\cfrac { { log }_{ 10 }e }{ x } +2secxtanx\) 

\(f(x)=\{ \begin{matrix} 2x+3, & x\le 0 \\ 3(x+1), & x>0 \end{matrix}\)
\(\therefore \underset { x\rightarrow { 0 }^{ - } }{ lim } f(x)=\underset { x\rightarrow { 0 }^{ - } }{ lim } (2x+3)=2\times 0+3=3\) 
and \(\underset { x\rightarrow { o }^{ + } }{ lim } f(x)=\underset { x\rightarrow { 1 }^{ - } }{ lim } 3(x+1)=3(0+1)=3\) 
So,\(\underset { x\rightarrow 0 }{ lim } f(x)\) exists and is equal to 3.
\(\underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)=\underset { x\rightarrow { 1 }^{ - } }{ lim } 2x+3=2\times 1+3=5\) 
\(\underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)=\underset { x\rightarrow { 1 }^{ - } }{ lim } 2x+3=2\times 1+3=5\) 
\(\underset { x\rightarrow { 1 }^{ + } }{ lim } f(x)=\underset { x\rightarrow { 1 }^{ + } }{ lim } 3(x+1)=3(1+1)=6\) 
\(\underset { x\rightarrow { 1 }^{ - } }{ lim } f(x)\neq \underset { x\rightarrow { 1 }^{ + } }{ lim } f(x)\) 
Hence,\(\underset { x\rightarrow 1 }{ lim } f(x)\) does not exist.

= \(\int { \sqrt { 169-\left( 3x+1 \right) ^{ 2 } } } dx\) 
= \(\cfrac { \frac { (3x+1) }{ 2 } \sqrt { 169-\left( 3x+1 \right) ^{ 2 } } +\frac { 169 }{ 2 } { sin }^{ -1 }\left( \frac { 3x+1 }{ 13 } \right) }{ 3 } +c\) 
= \(\frac { 1 }{ 6 } \left\{ 3x+1\sqrt { 169-(3x+1)^{ 2 } } +169{ sin }^{ -1 }\left( \frac { 3x+1 }{ 13 } \right) \right\} +c\)

\(\vec { p } \times \vec { q } =\left| \begin{matrix} \vec { i } & \vec { j } & \vec { k } \\ -3 & 4 & -7 \\ 6 & 2 & -3 \end{matrix} \right| \) 
= \(2\vec { i } -5\vec { j } -30\vec { k } \) 
Now, \(\vec { p } .\left( \vec { p } \times \vec { q } \right) =\left( -3\vec { i } +4\vec { j } -7\vec { k } \right) .\left( 2\vec { i } -51\vec { j } -30\vec { k } \right) \) 
= - 6 - 204 + 210 = 0
Hence \(\vec { p } \) and \(\vec { p } \times \vec { q } \) are perpendicular to each other.
Now \(\vec { q } .\left( \vec { p } \times \vec { q } \right) =\left( 6\vec { i } +2\vec { j } -3\vec { k } \right) .\left( 2\vec { i } -51\vec { j } -30\vec { k } \right) \) 
Hence \(\vec { q } \) and \(\vec { p } \times \vec { q } \) are perpendicular to each other.

\(\underset { 5x\rightarrow 0 }{ lim } \frac { { e }^{ 5x }-1 }{ x } \times5=5(1)\)

\(\frac { 2 }{ { x }^{ 2 }-1 } =\frac { 2 }{ (x+1)(x-1) } =\frac { A }{ x+1 } +\frac { B }{ x-1 } \)
\(\frac { 2 }{ { x }^{ 2 }-1 } =\frac { A(x-1)+B(x+1) }{ (x+1)(x-1) } \)
2 = A(x-1)+B(x+1) ...(1)
Putting x = 1 we get,
2 = B(1 + 1) ⇒ B(2) = 2 ⇒ B = 1
Putting x = -1 in (1) we get,
2 = A(-1-1) + B(-1 + 1)
2 = A(-2) ⇒ A = -1
\(\\ \therefore \frac { 2 }{ { x }^{ 2 }-1 } =\frac { 1 }{ x+1 } +\frac { 1 }{ x-1 } \)

f (x) = x^{2 + 1} causes the graph of the function
f(x) = x² shifts to the upward for one unit.
f (x) = (x + 1)² causes the graph of the
function f(x) = x² shifts to the left for one unit.

Let \(\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}\) and \(\vec{b}={-4\hat{i}+6\hat{j}-8\hat{k}}\)
Then \(\left| \vec { a } \right| =\sqrt { { 2 }^{ 2 }+{ (-3) }^{ 2 }+{ 4 }^{ 2 } } =\sqrt { 4+9+17 } =\sqrt { 29 } \)
and \(\left| \vec { b } \right| =\sqrt { { (-4) }^{ 2 }+6^{ 2 }+(-8)^{ 2 } } =\sqrt { 16+36+64 } =\sqrt { 116 } \)
\(\therefore \left| \vec { b } \right| =2\left| \vec { a } \right| \)
Thus, \(\vec{a}\)and \(\vec{b}\) are collinear.

Let \(\vec{a}=2\vec{i}-\vec{j}+3\vec{k}\) and \(\vec{b}=\vec{i}+2\vec{j}-3\vec{k}\) be two vectors.
Then, \(\left| \vec { a } \right| =\sqrt { { 2 }^{ 2 }+{ (-1) }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 14 } \)
and \(\left| \vec { b } \right| =\sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+(-{ 3) }^{ 2 } } =\sqrt { 14 } \)
Hence the required vectors are\(2\vec{i}-\vec{j}+3\vec{k}\) and \(\vec{i}+2\vec{j}-3\vec{k}.\)

Given P(A) = \(\frac { 1 }{ 2 } \), P(B) = \(\frac { 7 }{ 12 } \) and P(\(\bar { A } \cup \bar { B } \)) = \(\frac { 1 }{ 4 } \).
Now, \(P(\bar { A } \cup \bar { B } )=P(\overline { A\cap B } )=1-P(A\cap B)\)
\(\Rightarrow \frac { 1 }{ 4 } =1-P(A\cap B)\quad \Rightarrow P(A\cap B)=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
Now P(A) \(\times\) P(B) = \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } =\frac { 7 }{ 24 } \)
\(\therefore P(A\cap B)\neq P(A)\times P(B)\)
Thus, A and B are not independent.

The matrix A is singular if |A| = 0
\(\left| \begin{matrix} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{matrix} \right| =0\)
\(1\left| \begin{matrix} 2 & 1 \\ 2 & -3 \end{matrix} \right| +2\left| \begin{matrix} 1 & 1 \\ x & -3 \end{matrix} \right| +3\left| \begin{matrix} 1 & 2 \\ x & 2 \end{matrix} \right| =0\)
\(\Rightarrow\)(-6 - 2) + 2(-3 - x) + 3(2 - 2x) = 0
\(\Rightarrow\) 
\(\Rightarrow\) -8x = 8 => x = -1

Let S be the sample space and A be the event of getting at least two heads.
Therefore, the event Ā denotes, getting at most one head.
n(S) = 2⁹ = 512, n(Ā) = 9C₀ + 9C₁= 1 + 9 = 10
\(P(\bar{A})=\frac{10}{512}=\frac{5}{256} \)
\( P(A)=1-P(\bar{A})=1-\frac{5}{256}=\frac{251}{256}\)

\(P(\bar{A} / B) =\frac{P(\bar{A} \cap B)}{P(B)} \)
\(=\frac{P(B)-P(A \cap B)}{P(B)} \)
\(=\frac{0.5-0.2}{0.5}=\frac{0.3}{0.5}=\frac{3}{5}\)

(iii) P(queen card or 9)=\(\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}\)

P(card will be 6 or smaller)
= \(\frac{5+5+5+5}{52}=\frac{20}{52}=\frac{5}{13}\)   [∵ 5 cards which are 6 or smaller from each variety]

since the experiment has exactly the three possible mutually exclusive outcomes A, B and C, they must be exhaustive events.
\(\Rightarrow S=A\cup B\cup C\)
Therefore, by axioms of probability
\(P(A)\ge 0,P(B)\ge P(C)\ge 0\) and
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
Even though P(A) + P(B) + P(C) = 0.421 + 0.527 + 0.042 = 0.990 < 1
therefore, the assignment is not permissible

Given A and B are independent
\( \Rightarrow P(A \cap B) =P(A) P(B) \)
\(P(A \cup B) =0.6, P(A)=0.2\)
\(P(A \cup B) =P(A)+P(B)-P(A \cap B) \)
\(=P(A)+P(B)-P(A) \cdot P(B) \)
\(0 \cdot 6 =0 \cdot 2+P(B)(1-0 \cdot 2) \)
\(\frac{0 \cdot 4}{0 \cdot 8} =P(B) \)
\(\text { i.e., } P(B) =\frac{1}{2}=0.5\)

Let = \(\int { { e }^{ 2x } } sinx\quad dx\)
We know \(\int { { e }^{ ax } } sin\) bx dx = \(\frac { { e }^{ ax } }{ { a }^{ 2 }+{ b }^{ 2 } } \) (asin bx - b cos bx)
Here a = 2, b = 1
\(\int e^{2 x} \sin x d x=\frac{e^{2 x}}{5}[2 \sin x-\cos x]+c\)

\(\int e^{-5x}sin 3x \ dx\)
Using the formula
\(\int e^{ax}sin \ bx \ dx={e^{ax}\over a^2+b^2}[a \ sin \ bx +b \ cos \ bx]+c\)
for a = -5, b = 3, we get
\(\int e^{-5x}sin \ 3x \ dx=({e^{-5x}\over (-5)^2+3^2})(-5sin 3x-3cos 3x)+c\)
\(\int e^{-5x}sin \ 3x \ dx=-({e^{-5x}\over 34})(5sin 3x-3cos 3x)+c\)

Let \(I=\int \frac{x^2}{1+x^6} d x\)
put t = \(x^3 \Rightarrow d t=3 x^2 d x\)
\( \frac{d t}{3} =x^2 d x\)
\(I =\int \frac{d t}{1+t^2} \)
\(=\frac{1}{3} \int \frac{d t}{1+t^2} \)
\(=\frac{1}{3} \tan ^{-1}(t)+c \)
\(=\frac{1}{3 \tan ^{-1}}\left(x^3\right)+c \)

\( \int e^{z \log a} \times e^x d x =\int e^{\log a^2} \times e^x d x \)
\(=\int a^x \times e^x d x \quad\left(\because e^{\log u}=u\right) \)
\(=\int(a e)^x d x \)
\(=\frac{(a e)^x}{\log a e}+c \)
Aliter :
\( \int e^{x \log a} \times e^x d x =\int e^{x \log a+x} d x \)
\(=\int e^{x(\log a+1)} d x \)
\(=\frac{e^{\log a+1) x}}{(\log a+1)}+c\)

\(\int {e^{2x}-1\over e^{x}}dx=\int ({e^{2x}\over e^x}-{1\over e^x})dx\)
\(=\int (e^x-e^{-x})dx=e^x+e^{-x}+c\)

\(\int { \frac { { x }^{ 24 } }{ { x }^{ 25 } } } dx=\int { \frac { 1 }{ x } } dx\)
= \(log|x|+c\)

\( y=x^{\log x}+(\log x)^x \)
Take log on both sides
\( \log y=\log x^{\log x}+\log (\log x)^x \)
\(\log y=\log x(\log x)+x \log (\log x) \)
\(\frac{1}{y} \cdot \frac{d y}{d x}=\log x\left(\frac{1}{x}\right)+\log (x) \cdot \frac{1}{x}+\log (\log x) \) \(+x \frac{1}{\log x} \cdot \frac{1}{x}\)
\( \frac{1}{y d x} =2 \log x\left(\frac{1}{x}\right) \div \log (\log x) \div \frac{1}{\log x}\)
\( \frac{d y}{d x} =y\left[\frac{2}{x} \log x+\log (\log x) \div \frac{1}{\log x}\right] \)
\(=\left[x^{\log x} \div(\log x)^z\right] \) \( {\left[\frac{2}{x} \log x \div \log (\log x) \div \frac{1}{\log x}\right] } \)

Taking logarithm on both sides of the equation and using the rules of logarithm we have,
log y = \({3\over 4}log \ x+{1\over2}log(x^2+1)-5log(3x+2)\)
Differentiating implicitly
\({y'\over y}={3\over 4x}+{1\over 2}{2x\over (x^2+1)}-{5\times 3\over 3x+2}\)
\(={3\over 4x}+{x\over (x^2+1)}-{15\over 3x+2}\)
Therefore,\({dy\over dx}=y'={x^{3\over4}\sqrt{x^2+1}\over (3x+2)^5}[{3\over 4x}+{x\over x^2+1}-{15\over 3x+2}]\)

Let y = 2^x = e^xlog2.
Take u = (log 2)x so that
y = e^u
\({dy\over dx}={dy\over du}\times {du\over dx}=e^u \times log2=e^{xlog 2}\)
= (log2)2^x.

First we write : f(x) = (x² + x + 1)^{\({-1\over 3}\)}
Then, f '(x) = -\({1\over3}(x^2+x+1)^{{-1\over3}-1} {d\over dx} (x^2+x+1)\)
\(=-{1\over 3}(x^2+x+1)^{-4\over 3}\times (2x+1)\)
\(=-{1\over 3}(2x+1)(x^2+x+1)^{-4\over 3}\).

The limit of f(x) does not exist at x = x_o.

\(lim_{x\rightarrow \infty}(1+{1\over x})^{7x} \)\(=lim_{x\rightarrow \infty}[(1+{1\over x})^{x}]^7 \)
Put \({1\over x}=t,\)
when \(x\rightarrow \infty\) means \({1\over x}\rightarrow 0\)
\(\therefore {1\over x}\rightarrow 0\) means \(t \rightarrow 0\)
\(=[lim_{t\rightarrow 0}[(1+t)^{1\over t}]^7 =e^7\) \([\because lim_{x\rightarrow 0}(1+x)^{1\over x}=e]\)
\(\therefore lim_{x\rightarrow \infty}(1+{1\over x})^{7x}=e^7\)

\(lim_{x\rightarrow-2}(-{3\over 2}x)=-{3\over2}lim_{x\rightarrow-2}(x)=({-{3\over2}})(-2)=3.\)

Let \( f(x)=\frac{\sqrt{1-x}-2}{x+3} \)
\(\therefore \lim _{x \rightarrow-3} \frac{\sqrt{1-x}-2}{x+3}=-0.250 \)

Let \( f(x)=\frac{x-2}{x^2-4}=\frac{x-2}{(x-2)(x+2)}=\frac{1}{x+2} \)
\( \therefore \lim _{x \rightarrow 2} \frac{x-2}{x^2-4}=\lim _{x \rightarrow 2} \frac{1}{x+2}=\frac{1}{4}=0.25 \)

Let x = 3, y = -1, z = 3
\(\therefore r= \sqrt{x^2+y^2+z^2}=\sqrt{9+1+9}=\sqrt{19}\)
Hence, the direction consines are \({3\over \sqrt{19}},{-1\over \sqrt{19}},{3\over \sqrt{19}}\)

Given B is singular
\(\therefore|B|=0\)
\(\left|\begin{array}{ccc} b-1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{array}\right|=0\)
\((b-1)(4+4)-2(12-2)+3(-6-1)=0\)
\((b-1)(8)-2(10)+3(-7)=0\)
\(8 b-8-20-21=0\)
\(8 b-49=0\)
\(8 b=49\)
\(b=\frac{49}{8}\)

Given A is singular
\(\therefore|A|=0\)
\(\left|\begin{array}{cc} 7 & 3 \\ -2 & a \end{array}\right|=0\)
7a + 6 = 0
7a = -6
\(a=\frac{-6}{7} .\)

Let A = \(\begin{vmatrix} s & a^2 & b^2+c^2 \\ s & b^2 &c^2+a^2 \\ s & c^2 & a^2+b^2 \end{vmatrix}\)
Applying C₂ ⟶ C₂ + C₃ we get,
A = \(\left| \begin{matrix} s & { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } & { b }^{ 2 }+{ c }^{ 2 } \\ s & { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } & { c }^{ 2 }+{ a }^{ 2 } \\ s & { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } & { a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right| \)
A = s(a² + b² + c²)\(\left| \begin{matrix} 1 & 1 & { b }^{ 2 }+{ c }^{ 2 } \\ 1 & 1 & { c }^{ 2 }+{ a }^{ 2 } \\ 1 & 1 & { a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right| \) = s(a² + b² + c²)(0) = 0 [\(\therefore\) C₁ ≡ C₂]
Hence, \(\left| \begin{matrix} s & { a }^{ 2 } & { b }^{ 2 }+{ c }^{ 2 } \\ s & { b }^{ 2 } & { c }^{ 2 }+{ a }^{ 2 } \\ s & { c }^{ 2 } & { a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right| \) = 0
= RHS Hence Proved.

\(\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix}\) = (2 \(\times\) 2) - (- 1 \(\times\) 4) = 4 + 4 = 8.

The number of elements is the product of number of rows and number of columns.
Therefore, we will find all ordered pairs of natural numbers whose product is 12.
Thus, all the possible orders of the matrix are 1 \(\times\) 12, 12 \(\times\) 1, 2 \(\times\)6, 6 \(\times\)2, 3 \(\times\) 4 and 4 \(\times\) 3.
Since 7 is prime, the only possible orders of the matrix are 1 \(\times\) 7 and 7 \(\times\) 1.

When at least one by and one girl are to be selected, then
Number of ways=⁴C₁ \(​​\times\) ⁷C₄ +⁴C₇\(​​\times\) ⁷C₃ +⁴C₃\(​​\times\)⁷C₂ +⁴C₄+ ⁷C₁
\(=4\times{7\times6\times5\times4\over4\times3\times2\times1}+{4\times3\over2\times1}\times{7\times6\times5\over3\times2\times1}+4\times{7\times6\over2\times1}+1\times7\)
= (4 \(​​\times\) 35) + (6 \(​​\times\) 35) + (4 \(​​\times\)21) + 7
= 140 + 120 + 84 + 7 = 441 ways
Hence the required number of ways are 441 ways​​​​​​​

Let m be the slope of the line joining A (a, 2a) and B (-2,3).Then m₁ = \(\frac { 2a-3 }{ a+2 } \)
Let m₂ be the slope of the line 4 x + 3y + 5=0. Then m₂=-\(\frac { 4 }{ 3 } \)
Since given lines are perpendicular. Therefore,
m₁m₂=-1 ⇒ \(\frac { 2a-3 }{ a+2 } \times -\frac { 4 }{ 3 } \)=-1 ⇒ 3a+6 ⇒ a=18/5.

cos(300°) = cos(270° + 30°)
= sin 30° = \(\frac{1}{2}\)

(3x +y) (x - 3y) = 0 \(\Rightarrow\) 3x²- 8xy - 3y² = 0

3x - 8y + 13= 0

0

Let cosec^-1\(({2\over\sqrt{3}})\) = y, where - \({\pi\over 2}\le y \le {\pi\over 2}\)
\(\Rightarrow cosec\quad y={2\over\sqrt{3}}=sin \ y={\sqrt{3}\over2}\)
Thus, the principal value of cosec^-1\(({2\over\sqrt{3}})\) = \({\pi\over 3}\) 

Let sin^-1\(({\sqrt{3}\over2})\) = y, where -\({\pi\over 2}\le y \le {\pi\over 2}\)
\(\Rightarrow \) sin y = \({\sqrt{3}\over2}\) = sin \({\pi\over3}\Rightarrow y ={\pi\over3}\)
Thus, the principal value of sin^-I \(({\sqrt{3}\over2})\) = \({\pi\over 3}\)

Here n = 6, which is even.
Thus the middle term in the expansion of (x +y)⁶ is the term containing \({x}^{{6\over 2}}{y}^{{6\over 2}},\) that is the term ⁶C₃ x³y³ which is equal to 20x³y³.

⁵P₃ = 5 \(\times\)4 \(\times\) 3 = 60

Now, 180° =π radians gives \(1^0=\frac{\pi}{180}\) radians
\(18^0=\frac{\pi}{180}\times\ 18\ radians\ =\frac{\pi}{10}\ radians\).

\(\frac { 6! }{ n! } =\frac { 1.2.3.4.5.6 }{ 1.2.3...n } \) = 6. As n < 6 we get, n = 5

Given a, b, c are in A.P
\(\Rightarrow b=\frac { a+c }{ 2 } \)
RHS = 4[b² - ac]
\(=4\left[ { \left( \frac { a+c }{ 2 } \right) }^{ 2 }-ac \right] =4\left[ { \left( \frac { a+c }{ 4 } \right) }^{ 2 }-ac \right] \)
\(=4\left[ \frac { { \left( a+c \right) }^{ 2 }-4ac }{ 4 } \right] ={ a }^{ 2 }+{ c }^{ 2 }+2ac-4ac\)
= a² + c² - 2ac
= (a - c)² = LHS
Hence proved.

\({ a }_{ n }=\begin{cases} n+1\quad if\quad n\quad is\quad odd \\ n\quad \quad if\quad n\quad is\quad even \end{cases}\)
a₁ = 1 + 1 = 2, a2 = 2, a₃ = 3 + 1 = 4
a₄ = 4, a₅ = 5 +1 = 6, a₆ = 6
hence the first 6 terms are 4, 2, 2, 4, 6, 6...

Given equation is \(\frac{x}{a}+\frac{y}{b}=1\)
\(\Rightarrow \frac{bx+ay}{ab}=1\)
\(\Rightarrow\)bx + ay - ab = 0....(1)
Given that p= Length of the perpendicular from the origin to the line (1)
\(\Rightarrow p=\left|\frac{b(0)+a(0)-ab}{\sqrt{b^2+a^2}}\right|\)
\(\Rightarrow p^2=\left(\frac{ab}{\sqrt{a^2+b^2}}\right)^2\Rightarrow P^2=\frac{a^2b^2}{a^2+b^2}\)
\(\Rightarrow\frac{1}{p^2}=\frac{a^2+b^2}{a^2b^2}\)
\(\Rightarrow\frac{1}{p^2}=\frac{a^2}{a^2b^2}+\frac{b^2}{a^2b^2}\)
\(\Rightarrow\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{a^2}\)
Hence proved.

Since the given line, is perpendicular to y-axis, it will be parallel to y-axis.
\(\therefore\) Equation of the line is x = -2

T_n = a + (n - 1)d
T_m+n = a + (m + n - 1)d
& T_m-n = a + (m - n - 1)d
T_m+n + T_m-n = a + (m + n - 1)d + a + (m - n - 1)d
= 2a + d(m + n - 1 + m - n - 1)
= 2a + d(2m - 2)
= 2[a + (m - 1)d]
T_m+n + T_m-n = 2. T_m

Given ^(n-1)P₃ :ⁿ P₄ = 1 : 10 
⇒ \(\frac { (n-1){ P }_{ 3 } }{ n{ P }_{ 4 } } =\frac { 1 }{ 10 } \)
⇒ 10.(n-1)P₃ = 1.nP₄  \(\left[ \because npr\quad =\frac { n! }{ (n-r)! } \right] \)
⇒ \(10\times \frac { (n-1)! }{ (n-1-3)! } =\frac { n! }{ (n-4)! } \)
⇒   \(\frac { 10\times (n-1)! }{ (n-4)! } =\) \(\frac { 10\times (n-1)! }{ (n-4)! }\)⇒\(\frac { n(n-1)! }{ (n-4)! } \)
⇒10 = n
∴ n = 10.

For any n with r = 2
\(\therefore \frac { n! }{ r!(n-r)! } =\frac { n! }{ 2!(n-2)! } =\frac { n(n-1)(n-2)! }{ 2\times 1(n-2)! } =\frac { { n }^{ 2 }-n }{ 2 } \)

Each I prize can be distributed to any one of the 10 students.
∴ By the rule of product, the number of ways of distributing 12 distinct prizes to 10 students are 
\(10 \times 10 \times 10 \times 10 \times 10 \times 10 \times\) \(10 \times 10 \times 10 \times 10 \)  = 10¹²

cos 65^o + cos 15^o = \(2\cos { \left( \frac { 65+15 }{ 2 } \right) } .\cos { \left( \frac { 65-15 }{ 2 } \right) } \)
= 2 cos 40^o cos 25^o

sin 75^o - sin 35^o = \(2\cos { \left( \frac { 75+35 }{ 2 } \right) } .\sin { \left( \frac { 75-35 }{ 2 } \right) } \)
= 2 cos(55^o) sin 20^o

sin\(\left( \frac { -11\pi }{ 3 } \right) =-sin\frac { 11\pi }{ 3 } \)
= \(-sin\left( \frac { 11\times 180 }{ 3 } \right) \) = -sin(660⁰)
= -sin(2 \(\times\) 360⁰ - 60⁰)
= -(-sin(60⁰))   [Angle is in the IV quadrant and sine is negative
= sin 60⁰ = \(\frac { \sqrt { 3 } }{ 2 } \).

-x² + 3x + 1 = 0
Given equation is -x² + 3x + 1 = 0
Here a = -1, b = 3, c = 1
\(\therefore\) D = b² - 4ac = 3² - 4 (-1) (1)
= 9 + 4 = 13
Since D > 0, the two roots are real and distinct.

Given equation is x⁴ = 16
\(\Rightarrow\)   x⁴ = 16 = 0
\(\Rightarrow\) (x²)²- (4)² = 0
\(\Rightarrow\) (x² + 4)(x²- 4) = 0   [ \(\because\) a²- b² = (a + b)(a - b) ]
\(\Rightarrow\) x = -4,
\(\Rightarrow\) x² = 4 When
\(\Rightarrow\) x² = -4,
x = \(\pm\sqrt{-4}=\pm2\) i where \(i=\sqrt{-1}\) when
\(\Rightarrow\)  x² = 4,
x = \(\pm\sqrt{4}=\pm2\)
Hence the four roots of the given equation are x = 2, -2, 2i, -2i 

\(\frac{1}{cos\theta}=\frac{13}{5} ⇒ cos\theta=\frac{5}{13}\)

AB =\(\sqrt{13^2-5^2}\)
= \(\sqrt{169-25}\)
= \(\sqrt{144}=12\)
Since θ lies in the IV quadrant, only cos θ-. and sec θ are positive
sin θ = \(\frac{-12}{13}, cos θ=\frac{5}{13}, tanθ=\frac{-12}{5}, cosecθ=\frac{-13}{12} and\quad cotθ=\frac{-5}{12}\)

\(cot\theta =\sqrt { 3 } \Rightarrow tan\theta =\frac { 1 }{ \sqrt { 3 } } \)
Since tan\(\theta\) = \(\frac { 1 }{ \sqrt { 3 } } \)>0, the principal value lies in the I quadrant
tan \(\theta\) = \(\frac { 1 }{ \sqrt { 3 } } \) = tan\(\left( \frac { \pi }{ 6 } \right) \)
\(\Rightarrow\) \(\theta\) = \(\left( \frac { \pi }{ 6 } \right) \) is the principal

The relation R is defined by xRy if x + 2y = 1 for x, y \(\in \)  N.
Reflexivity : Let x, y \(\in \)  N
xRx \(\Rightarrow\)  x + 2x = 1 \(\Rightarrow\) 3x = 1 \(\Rightarrow\) x = \(\frac { 1 }{ 3 } \notin N\)
\(\therefore\)  R is reflexive.
Symmetricity: xRy \(\Rightarrow\) yRx for x, y \(\in \)  N
xRy \(\Rightarrow\)  x + 2y = 1 which is not possible for any values of x, Y \(\in \) N
\(\therefore\) R is not symmetric
Transitivity: xRy and yRz \(\Rightarrow\) xRz.
xRy and yRz are not possible for any values of x, y, z \(\in \) N
\(\therefore\)  R is not transitive.
\(\therefore\) R is neither reflexive, nor symmetric and not transitive.

Given relation is "aRb if a is not a sister of b". and a, b, c \(\in \) A.
Reflexivity: aRa \(\Rightarrow\) a is not a sister of a
\(\therefore\)  R is reflexive.         
Symmetric : aRb \(\Rightarrow\) bRa
a is not a sister of b \(\Rightarrow\) b is not a sister of a
\(\therefore\) R is not symmetric.
Transitivity: aRb and bRC \(\Rightarrow\) aRC
a is not a sister of b and b is not a sister of C. [Eg : Mother is not a sister of daughter, daughter is not a sister of chithi, but mother is a sister of chithi.]
\(\Rightarrow\) a is not a sister of C.
\(\therefore\)  R not is transitive.

tan (45^o + A) = \(\frac { 1+\tan { A } }{ 1-\tan { A } } \)
LHS = tan (45^o + A) = \(\frac { \tan { { 45 }^{ o } } +\tan { A } }{ 1+\tan { { 45 }^{ o } } .A\tan { } } =\frac { 1+\tan { A } }{ 1-\tan { A } } \) = RHS
Hence proved.

\(\frac { 7\pi }{ 3 } \)
\(\frac { 7\pi }{ 3 } \) = \(\frac { 7\pi }{ 3 } \) \(\times\) \(\frac { 180 }{ \pi } \) = 420⁰

\(\frac { 2\pi }{ 5 } \)
\(\frac { 2\pi }{ 5 } \) \(\times\) \(\frac { 180 }{ \pi } \) = 2 \(\times\) 36⁰ = 72⁰

-2x > 0 or 3x - 4 < 11

⇒ -x > 0 or 3x < 11 + 4
⇒ x < 0 or 3x < 15 [a > b ⇒ -a < -b]
⇒ x < 0 or x <  \(\frac{15}{3}\)
⇒ x < 0 or x < 5
⇒ x ∈ (-\(\infty \), 5)

B-A = {4, 5, 6, 7}
LHS = C-(B-A) = {3, 9}...(1)
C\(\cap \)A = {3}
B' = {1,2,3,8,9}
C\(\cap \)B' = {3, 9}
RHS = (C\(\cap \)A)\(\cup \) (C\(\cap \)B') = {3, 9}...(2)
From (1) and (2), LHS = RHS

Let cosec-1(-1) = y, where \(-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 } \)
⇒ -1 = cosec y
⇒ cosec y = -cosec \(\left( \frac { \pi }{ 2 } \right) \)
⇒ cosec y = cosec -\(\left( \frac { \pi }{ 2 } \right) \)    [∵ cosec (-ፀ) = -cosec ፀ]
⇒ y = -\(\left( \frac { \pi }{ 2 } \right) \)
Thus, the principal value of cosec-1 is -\(\left( \frac { \pi }{ 2 } \right) \).

328⁰
328⁰ = 270⁰ + 58⁰
∴ 328⁰ lies in the IV quadrant

(i) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) 
= 0.3 + 0.6 - 0.25
= \(0.9-0.25=0.65=\cfrac { 65 }{ 100 } =\cfrac { 13 }{ 20 } \) 
(ii) \(P(A/B)=\cfrac { P(A\cap B) }{ P(B) } =\cfrac { 0.25 }{ 0.6 } =\cfrac { 25 }{ 60 } =\cfrac { 5 }{ 12 } \) 
(iii) \(P(B/\bar { A } )=\cfrac { P\left( B\cap \bar { A } \right) }{ P\left( \bar { A } \right) } \) 
= 0.35
\(P(\bar { A } )=1-P(A)=1-0.3=0.7\)
\(\therefore P(B\cap \bar { A } )=\cfrac { 0.35 }{ 0.7 } =\cfrac { 1 }{ 2 } \)
(iv)  \(P(\bar { A } /B)=P\left( \bar { A } \cap B \right) /P(B)\) 
\(P(\bar { A } \cap B)=0.35\) (from iii)
\(\therefore P(\bar { A } /B)=\cfrac { P\left( \bar { A } \cap B \right) }{ P(B) } =\cfrac { 0.35 }{ 0.6 } =\cfrac { 35 }{ 60 } =\cfrac { 7 }{ 12 } \)
(v) \(P(\bar { A } /\bar { B } )=\cfrac { P\left( \bar { A } \cap \bar { B } \right) }{ P(\bar { B } ) } \) 
 = \(1-\left( \bar { A } \cap \bar { B } \right) =P(\bar { A\cup B } )\) 
=1 - 0.65 = 0.35
\(\therefore P(\bar { A } /\bar { B } )=\cfrac { P\left( \bar { A } \cap \bar { B } \right) }{ P(\bar { B } ) } =\cfrac { 0.35 }{ 0.4 } =\cfrac { 35 }{ 40 } =\cfrac { 7 }{ 8 } \)

Intercept form √3x - y + 4 = 0 ⇒ √3x - y = -4
\(\frac{-\sqrt{3}}{4}x+\frac{y}{4}=1\)
That is \(\frac{x}{(-\frac{4}{\sqrt{3}})}+\frac{y}{4}=1\)
Comparing the above equation with the equation \(\frac{x}{a}+\frac{y}{b}=1\)
We get, x-intercept = -\(\frac{4}{\sqrt{3}}\) and y-intercept = 4

(i) For a relation on {0, 1, 2, 3} to be reflexive, it must have the pairs (0, 0), (1, 1), (2, 2), (3, 3). Fortunately, it becomes symmetric and transitive. Therefore, as in (i) if we insert (1, 2) and (2, 3) we get the required one. Thus {(0, 0), (1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive; it is not symmetric and it is not transitive. 
(ii) Proceeding like this we get the relation {(0, 0), (1, 1), (2, 2), (3, 3), (1, 2)} that is reflexive, transitive and not symmetric.

\(I=\int { { e }^{ 2x } } sin3xdx\) put u = sin 3x; du = 3 cos 3x dx
e^2xdx = dv \(\therefore v=\cfrac { { e }^{ 2x } }{ 2 } \) 
= \(\left( sin3x \right) \left( \cfrac { { e }^{ 2x } }{ 2 } \right) -\int { \cfrac { { e }^{ 2x } }{ 2 } \left( 3cos3x \right) dx } \) 
= \(\cfrac { { e }^{ 2x } }{ 2 } sin3x-\cfrac { 3 }{ 2 } \int { { e }^{ 2x }cos3xdx } =\cfrac { { e }^{ 2x } }{ 2 } sin3x-\cfrac { 3 }{ 2 } { I }_{ 1 }\) 
\(I=\int { { e }^{ 2x }sin3xdx } \) 
= \(\cfrac { { e }^{ ax } }{ { a }^{ 2 }+{ b }^{ 2 } } \left( asinbx-bcosbx \right) =\cfrac { { e }^{ ax } }{ 13 } \left[ 2sin3x-3cos3x \right] +c\) 
where \({ I }_{ 1 }=\int { { e }^{ 2x }cosx3xdx } \) 
Let u = cos 3x; du = -3 sin 3x dx
e^2xdx=dv,\(\therefore v=\cfrac { { e }^{ 2x } }{ 2 } \) 
\({ I }_{ 1 }=\left( cos3x \right) \cfrac { { e }^{ 2x } }{ 2 } -\int { \cfrac { { e }^{ 2x } }{ 2 } -\left( -3sin3x \right) } dx\) 
(i.e.,) \({ I }_{ 1 }=\cfrac { { e }^{ 2x } }{ 2 } cos3x+\cfrac { 3 }{ 2 } \int { { e }^{ 2x }sin3x } dx\) 
= \(\cfrac { { e }^{ 2x } }{ 2 } cos3x+\cfrac { 3 }{ 2 } I\) 
Substituting (2) in (1) we get
\(I=\cfrac { { e }^{ 2x } }{ 2 } sin3x-\cfrac { 3 }{ 2 } \left\{ \cfrac { { e }^{ 2x } }{ 2 } cos3x+\cfrac { 3 }{ 2 } I \right\} \) 
(i.e.,) \(I=\cfrac { e^{ 2x } }{ 2 } \left[ sin3x-\cfrac { 3 }{ 2 } cos3x \right] -\cfrac { 9 }{ 4 } I\) 
\(I\left( 1+\cfrac { 9 }{ 4 } \right) =\cfrac { { e }^{ 2x } }{ 2 } \left[ 2sin3x-3xos3x \right] \) 
 (i.e.,)  \(I\left( \cfrac { 13 }{ 14 } \right) =\cfrac { { e }^{ 2x } }{ 2 } \left[ 2sin3x-3cos3x \right] \)
So,\(I=\cfrac { { e }^{ 2x } }{ 13 } \cfrac { { e }^{ 2x } }{ 4 } \left[ 2sin3x-3cos3x \right] \)
(i.e)\(I=\cfrac { { e }^{ 2x } }{ 13 } \left[ 2sin3x-3cos3x \right] +c\)

\(A=\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] \) 

= \(\left[ \begin{matrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix} \right] \) 
\(-4A=-4\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] =\left[ \begin{matrix} -4 & -8 & -8 \\ -8 & -4 & -8 \\ -8 & -8 & -4 \end{matrix} \right] \) 
\(-5I=-5\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} -5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{matrix} \right] \) 
LHS :\(A^{ 2 }-4A-51=\left[ \begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix} \right] +\left[ \begin{matrix} -4 & -8 & -8 \\ -8 & -4 & -8 \\ -8 & -8 & -4 \end{matrix} \right] +\left[ \begin{matrix} -5 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -5 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 9-4-5 & 8-8+0 & 8-8+0 \\ 8-8+0 & 9-4-5 & 8-8+0 \\ 8-8+0 & 8-8+0 & 9-4-5 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)

Given \(\vec{a}=2\hat{i}+\hat{j}-2\hat{k}\)
\(\left| \vec { a } \right| =\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+({ -2) }^{ 2 } } =\sqrt { 4+1+4 } =\sqrt { 9 } \)=3
\(\vec { b } =\hat { i } +\hat { j } \)
\(\left| \vec { b } \right| =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 2 } \)
\(\left| \vec { c } -\vec { a } \right| =2\sqrt { 2 } \)
\({ \left| \vec { c } -\vec { a } \right| }^{ 2 }={ (2\sqrt { 2 } ) }^{ 2 }\)
\(\therefore { \left| \vec { c } \right| }^{ 2 }+{ \left| \vec { a } \right| }^{ 2 }-2(\vec { c } .\vec { a } )=8\)
\({ \left| \vec { c } \right| }^{ 2 }+9-2\left| \vec { c } \right| =8\quad [\therefore \left| \vec { a } \right| =3,\vec { c } .\vec { a } =\left| \vec { c } \right| ]\)
\({ \left| \vec { c } \right| }^{ 2 }+2\left| \vec { c } \right| +1=0\)
\({ [\left| \vec { c } \right| -1] }^{ 2 }=0\)
\({ \left| \vec { c } \right| -1 }=0\)  \(\Rightarrow { \left| \vec { c } \right| }=0\)
Also \(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{matrix} \right| \begin{matrix} =\hat { i } (0+2)-\hat { j } (0+2)+\hat { k } (2-1) \\ =2\hat { i } -2\hat { j } +\hat { k } \end{matrix}\)
\(\left| \vec { a } \times \vec { b } \right| =\sqrt { { 2 }^{ 2 }+{ (-2) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 4+4+1 } =\sqrt { 9 } =3\)
\(\therefore \left| (\vec { a } \times \vec { b } ).\vec { c } \right| =\left| \vec { a } \times \vec { b } \right| \left| \vec { c } \right| sin30\) [angle between \(\vec{a} \times \vec{b}\) and \(\vec{c}\) is 30^o]
=3(1)\(\left( \frac { 1 }{ 2 } \right) =\left( \frac { 3 }{ 2 } \right) \)

Since \(\overrightarrow { c } \) is co-planar with \(\overrightarrow { a } \) and \(\overrightarrow { b } \)
Let \(\overrightarrow { c } =x\overrightarrow { a } +y\overrightarrow { b } \)  where x, y are sclars.
\(\Rightarrow \overrightarrow { c } =x(\hat { i } +\hat { j } +2\hat { k } )+y(\hat { i } +2\hat { j } +\hat { k } )\)
\(\Rightarrow \overrightarrow { c } =\hat { i } (x+y)+\hat { j } (x+2y)+\hat { k } (2x+y)\)
\(\therefore |\overrightarrow { c } |=1\Rightarrow \sqrt { { (x+y) }^{ 2 }+{ (x+2y) }^{ 2 }+{ (2x+y) }^{ 2 } } =1\) ....(1)
Also, \(\overrightarrow { c } \) is perpendicular to \(\hat { i } +\hat { j } +\hat { k } \)
\(\Rightarrow \overrightarrow { c } .(\hat { i } +\hat { j } +\hat { k } )=0\)
\(\\ \\ \\ \Rightarrow \) (x+y)(1)+(x+2y)1+(2x+y)1 = 0
\(\Rightarrow \) 4x+4y = 0 \(\Rightarrow \) x = -y
Substituting y = -x in (1) we get,
\(\sqrt { 0+{ (-x) }^{ 2 }+{ (x) }^{ 2 } } =1\Rightarrow \sqrt { { 2x }^{ 2 } } =1\)
\(\Rightarrow{ 2x }^{ 2 }=1\Rightarrow { x }^{ 2 }=\frac { 1 }{ 2 } \Rightarrow x=\pm \frac { 1 }{ \sqrt { 2 } } \quad \)
\(\therefore\) when \(x=\frac { 1 }{ \sqrt { 2 } } ,y=-\frac { 1 }{ \sqrt { 2 } } \) and when \(x=-\frac { 1 }{ \sqrt { 2 } } ,y=\frac { 1 }{ \sqrt { 2 } } \)
\(\therefore x=\frac { 1 }{ \sqrt { 2 } } \) and \(y=-\frac { 1 }{ \sqrt { 2 } } \Rightarrow \overrightarrow { c } =-\frac { \hat { j } }{ \sqrt { 2 } } +\frac { \hat { k } }{ \sqrt { 2 } } \)
when \(x=-\frac { 1 }{ \sqrt { 2 } } \)  , and \(y=\frac { 1 }{ \sqrt { 2 } } \Rightarrow \overrightarrow { c } =\frac { \hat { j } }{ \sqrt { 2 } } -\frac { \hat { k } }{ \sqrt { 2 } } \)
\(\therefore\) \(\overrightarrow { a } .\overrightarrow { c } =0+\frac { 1 }{ \sqrt { 2 } } -\frac { 2 }{ \sqrt { 2 } } =\frac { -1 }{ \sqrt { 2 } } <0\)
\(\overrightarrow { c } \) makes obtuse angle with \(\overrightarrow { a } \) means.
\(\overrightarrow { c } =\frac { \hat { j } -\hat { k } }{ \sqrt { 2 } } \)

\(Let\quad \sin ^{ -1 }{ x } =t\Rightarrow x =\sin { t } \)
\(Also\quad x\rightarrow \frac { 1 }{ \sqrt { 2 } } \Rightarrow \sin { t } \rightarrow \frac { 1 }{ \sqrt { 2 } } \Rightarrow t\rightarrow \frac { \pi }{ 4 } \)
\(\therefore \lim _{ x\rightarrow \frac { 1 }{ \sqrt { 2 } } }{ \frac { x-\cos { (\sin ^{ -1 }{ (x) } ) } }{ 1-\tan { (\sin ^{ -1 }{ x } ) } } } =\lim _{ t\rightarrow \frac { \pi }{ 4 } }{ \frac { \sin { t } -\cos { t } }{ 1-\tan { t } } } =\lim _{ t\rightarrow \frac { \pi }{ 4 } }{ \frac { \sin { t } -\cos { t } }{ 1-\frac { \sin { t } }{ \cos { t } } } } =\lim _{ t\rightarrow \frac { \pi }{ 4 } }{ \frac { (\sin { t } -\cos { t) } \cos { t } }{ (\cos { t } -\sin { t } ) } }\)
\( =\lim _{ t\rightarrow \frac { \pi }{ 4 } }{ -\cos { t } } =-\cos { \left( \frac { \pi }{ 4 } \right) } =-\frac { 1 }{ \sqrt { 2 } } \)

Given \(\log { ({ x }^{ 2 }+{ y }^{ 2 }) } =2\tan ^{ -1 }{ \frac { y }{ x } , } \)
Differentiating both sides, with respect to 'x' we get
\(\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } .\frac { d }{ dx } ({ x }^{ 2 }+{ y }^{ 2 })=2.\frac { 1 }{ 1+{ \left( \frac { y }{ x } \right) }^{ 2 } } .\frac { d }{ dx } { \left( \frac { y }{ x } \right) }\)

\(\Rightarrow 2x+2y\frac { dy }{ dx } =2x\frac { dy }{ dx } -2y\\ \Rightarrow x+y\frac { dy }{ dx } =x\frac { dy }{ dx } -y\\ \Rightarrow \frac { dy }{ dx } (y-x)=-x-y\\ \Rightarrow \frac { dy }{ dx } =\frac { -(x+y) }{ y-x } =\frac { x+y }{ x-y } \quad\)
Hence Proved

Let S be the sample space and A be the event that the smallest block is to be painted in red.
n(S) = 6 P₃ = 6 \(\times\) 5 \(\times\) 4 = 120
n(A) = 5 \(\times\) 4 = 20
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 20 }{ 120 } =\frac { 1 }{ 6 } \)

Let A be the event of getting award for design of railway bridge and B be the event of getting award for the efficient use of materials
Then, \(P(A)=0.48, P(B)=0.36, P(A \cap B)=0.2\)
(i) P (atleast one of the two awards)
\(=P(A \cup B) \)
\(=P(A)+P(B)-P(A \cap B) \)
\(=0.18+0.36-0.2=0.64\)
(ii) P (will get only one of the award)
\(=P(A \cap \bar{B})+P(\bar{A} \cap B) \)
\(=P(A)-P(A \cap B)+P(B)-P(A \cap B) \)
\(=0.48-0.20+0.36-0.20 \)
\(=0.44\)

Let A'and B be the availability of first and second fire engine respectively, then A and B are independent.
Then \(P(A)=P(B)=0.96 \)
\(P(\bar{A})=P(\bar{B})=1-0.96=0.04\)
(i) P(a fire engine is available when needed)
\(=P(A \cap \bar{B})+P(\bar{A} \cap B)+P(A \cap B)\)
\(=P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B)+P(A) \cdot P(B)\)
\(=0.96 \times 0.04+0.04 \times 0.96+0.96 \times 0.96\)
\(=0.96(0.04+0.04+0.96) \)
\(=0.96 \times 1.04 \)
\(=0.9984\)
(ii) P (Neither is available when needed)
\(=P(\bar{A} \cap \bar{B}) \)
\(=P(\bar{A}) \cdot P(\bar{B})\)
\(=0.04 \times 0.04\)
\(=0.0016\)

If the given matrix M is singular, then
\(\left| \begin{matrix} x & y \\ z & 1 \end{matrix} \right| \) = 0.
That is , x - yz = 0
Hence the possible ways of selecting (x, y, z) are
{(1,1,1), (2,1,2), (,2,2,1), (3,1,3), (3,3,1)} = A(say)
The number of favourable cases  n(A) = 5
The total number of cases are n(S) = 3³ = 27
The probability of the given matrix is a singular matrix is
P = \(\frac { n(A) }{ n(S) } =\frac { 5 }{ 27 } \)

Let A_i be the event that the traffic light opens at i th cross, for i = 1, 2, 3, 4.
Let B_i be the event that the traffic light closes at i th cross, for i = 1, 2, 3, 4.
The traffic lights are all independent.
Therefore Ai and B_i are all independent events, for i = 1, 2, 3, 4.
Given that P(A_i) = 0.4, i = 1, 2, 3, 4
P(B_i) = 0.6, i =1, 2, 3, 4
(i) Probability of car crossing the first crossroad without stopping,
P(A₁) = 0.4
(ii) Probability of car crossing first two crossroads without stopping
\(P({ A }_{ 1 }\cap { A }_{ 2 })\)= P(A₁A₂) = (0.4)(0.4) = 0.16
(iii) Probability of car crossing all the crossroads, stopping at third cross
P\(({ A }_{ 1 }\cap { A }_{ 2 }{ B }_{ 3 }\cap { B }_{ 3 }\cap { A }_{ 4 })\) = P(A₁A₂B₃A₄) = (0.4)(0.4)(0.6)(0.4) = 0.0384
(iv) Probability of car crossing all the crossroads, stopping at exactly one of the crossroads is
P(B₁A₂A₃A₄ \(\cup \) A₁B₂A₃A₄ \(\cup \) A₁A₂B₃A₄ \(\cup \) A₂A₃A₃B₄)
= P(B₁A₂A₃A₄)+P(A₁B₂A₃A₄)+P(A₁A₂B₃A₄)+P(A₁A₂A₃B₄)
= 4(0.4)(0.4)(0.6)(0.4) = 4(0.0384) = 0.1536.

Let be the event of speaks the truth, be the event of speaks the truth
∴ \(\bar { A } \) is the event of X not speaking the truth and \(\bar { B } \) is the event of Y not speaking the truth.
Let C be the event that they will contradict each other.
Given that
P(A) = 0.70 ⇒ P(\(\bar { A } \)) = 1 - P(A) = 0.30
P(B) = 0.90 ⇒ P(\(\bar { B } \)) = 1 - P(B) = 0.10
C = (A speaks truth and B does not speak truth or B speaks truth and A does not speak truth)
C =\(\left[ (A\cap \bar { B } )\cup (\bar { A } \cap B) \right] \) (see figure)
since \((A\cap \bar { B } )\) and \((\bar { A } \cap B)\) are mutually exclusively,
P(C) = \((A\cap \bar { B } )+(\bar { A } \cap B)\)
= P(A)P(\(\bar { B } \)) + P(\(\bar { A } \))P(B)
( Since A, B are independent event A, \(\bar { B } \) are also independent events
= (0.70) (0.10) + (0.30) (0.90)
= 0.070 + 0.270 = 0.34
P(C) = 0.34
 

Let A be the event of drawing a Jack in the first draw,
B be the event of drawing a Jack in the second draw.
Case (i)
Card is replaced
n(A) = 4  (Jack)
n(B) = 4  (Jack)
and n(S) = 52 (Total)
Clearly the event A will not affect the probability of the occurrence of event B and therefore A and B are independent.
P(A\(\cap \) B) = P(A). P(B)
P(A) = \(\frac{4}{52}\) , P(B) = \(\frac{4}{52}\)
P(A\(\cap \) B) = P(A).P(B)
=\(\frac{4}{52}\).\(\frac{4}{52}\)
=\(\frac{1}{169}\) .
Case (ii)
Card is not replaced
In the first draw, there are 4 Jacks and 52 cards in total. Since the Jack, drawn at the first draw is not replaced, in the second draw there are only 3 Jacks and 51 cards in total. Therefore the first event A affects the probability of the occurrence of the second event B.
Thus A and B are not independent. That is, they are dependent events.
Therefore, P( A \(\cap \) B ) = P(A). P(B)
P(A) = \(\frac{4}{52}\)
P(B/A) = \(\frac{3}{51}\)
P( A \(\cap \) B ) = P(A).P(B/A)
=\(\frac { 4 }{ 52 } .\frac { 3 }{ 51 } \)
=\(\frac { 1 }{ 121 } \) .

Let A, B, C be the events that the problems solved by 3 students. Then,
\(P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(C)=\frac{1}{5}\)
(i) P (Problem is solved) \(=P(A \cup B \cup C)\)
\(=1-P(\overline{A \cup B \cup C})\)
\(=1-P(\bar{A} \cap \bar{B} \cap \bar{C})\)
\(=1-P(\bar{A}) P(\bar{B}) P(\bar{C})\)
\(=1-\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \)
\(=1-\frac{2}{5}=\frac{3}{5}\)
(ii) P (exactly one of them will solve)
\(=P(A \bar{B} \bar{C} \cup \bar{A} B \bar{C} \cup \bar{A} \bar{B} C)\)
\(=P(A) \cdot P(\bar{B}) \cdot P(\bar{C})+P(\bar{A}) \cdot P(B) \cdot P(\bar{C}) +P(\bar{A}) \cdot P(\bar{B}) \cdot P(C)\)

\(=\frac{1}{5}+\frac{2}{15}+\frac{1}{10}=\frac{1}{5}\left(1+\frac{2}{3}+\frac{1}{2}\right) \)
\(=\frac{6+4+3}{30}=\frac{13}{30}\)

\(\int \sqrt{81+(2 x+1)^2} d x =\frac{\frac{2 x+1}{2} \sqrt{51+(2 x+1)^2}+\frac{81}{2} \log \mid 2 x+1+\sqrt{31+(2 x+1)^2} \mid+6}{\frac{d}{dx}(2 x+1)}\)
\( =\frac{1}{4}\left[(2 x+1) \sqrt{81+(2 x+1)^2}\right. \left.+81 \log \left|(2 x+1)+\sqrt{81+(2 x+1)^2}\right|\right]+c \)

Let \(2 x-3=A \frac{d}{d x}\left[x^2+4 x-12\right]+B\)
Then 2x - 3 = A(2x + 4) + B
Equating corresponding terms'on both sides,
2A = 2 ........(1)
4A + B = -3 .........(2)
from (1), A = 1
Putting this in (2)
4(1) + B = -3
\(\therefore\) B = -7
\(\therefore\) 2x - 3 = 1(2x + 4) - 7
\(\therefore \frac{2 x-3}{x^2+4 x-12}=\frac{2 x+4}{x^2+4 x-12}-\frac{7}{x^2+4 x-12}\)
\(=\frac{2 x+4}{x^2+4 x-12}-\frac{7}{\left(x^2+4 x+2^2\right)-2^2-12} \)
\(=\frac{2 x+4}{x^2+4 x-12}-\frac{7}{(x+2)^2-16}\)
\(=\frac{2 x+4}{x^2+4 x-12}-\frac{7}{(x+2)^2-4^2}\)
\(\therefore \int \frac{2 x-3}{x^2+4 x-12} d x=\int \frac{2 x+4}{x^2+4 x-12} d x -7 \int \frac{1}{(x+2)^2-4^2} d x\)
\(=\log \left|x^2+4 x-12\right|-7 \times \frac{1}{2(4)} \log \left|\frac{(x+2)-4}{(x+2)+4}\right|+c \)
\(=\log \left|x^2+4 x-12\right|-\frac{7}{8} \log \left|\frac{x-2}{x+6}\right|+c\)

Let I = \(\int {x+1\over x^2-3x+1}dx\)
\(x+1=A{d\over dx}(x^2-3x+1)+B\)
\(x+1=A(2x-3)+B\)
Comparing the coefficients of like terms, we get
\(2A=1 \Rightarrow A={1\over2};-3A+B=1\Rightarrow B={5\over2}\)
\(I=\int{{1\over2}(2x-3)+{5\over2}\over x^2-3x+1}dx\)
\(I={1\over2}\int {2x-3\over x^2-3x+1}dx+{5\over2}\int {1\over x^2-3x+1}dx\)
\(={1\over2}log |x^2-3x+1|+{5\over2}\int{1\over (x-{3\over2})^2-(\sqrt{5\over 2})^2}dx\)
\(={1\over2}log|x^2-3x+1|+{5\over2}{1\over 2({\sqrt{5}\over 2})}log\left| {x-{3\over 2}-{\sqrt{5}\over 2}\over x-{3\over 2}+{\sqrt{5}\over 2}} \right| +c\)
\(={1\over2}log|x^2-3x+1|+{\sqrt{5}\over 2}log |{2x-3-\sqrt{5}\over 2x-3+\sqrt{5}}|+c\)

\(\int {1\over x^2-2x+5}dx=\int {1\over x^2-2(1)x+(1)^2+4}dx\)
\(=\int {1\over (x-1)^2+2^2}dx\)
\(\int {1\over x^2-2x+5}dx={1\over 2}tan^{-1}({x-1\over 2})+c\)

Let \(\theta=\tan ^{-1} x\)
Then \(\tan \theta=x \text { and }\)
\(\frac{d \theta}{d x}=\frac{1}{1+x^2} \Rightarrow d \theta=\frac{1}{1+x^2} d x\)
\(\therefore \int e^{\tan ^{-1}} z \left(\frac{1+x-x^2}{1+x^2}\right) d x =\int e^{\operatorname{tan} ^{-1} x}\left(1+x+x^2\right) \times \frac{1}{1+x^2} d x \)
\(=\int e^{\theta}\left(1+\tan \theta+\tan ^2 \theta\right) d \theta \)
\(=\int e^{\theta}\left(\sec ^2 \theta+\tan \theta\right) d \theta\)
\(=e^\theta\tan \theta+c\)
\(=e^{tan^{-1} x} \times x+c\)
\(=x e^{\tan ^{-1} x}+c\)

Given that f'(x) = \({d\over dx}(f(x))=3x^2-4x+5\)
Integrating on both sides with respect to x, we get
\(\int f'(x)dx=\int (3x^2-4x+5)dx\)
\(f(x)=x^3-2x^2+5x+c\)
To determine the constant of integration c, we have to apply the given information f(1) = 3
\(f(1)=3\Rightarrow 3=(1)^3-2(1)^2+5(1)+c \Rightarrow c=-1\)
Thus \(f(x)=x^3-2x^2+5x-1\).

\( x =\frac{1-t^2}{1+t^2}\)
\( \frac{d x}{d t} =\frac{\left(1+t^2\right)(-2 t)-\left(1-t^2\right)(2 t)}{\left(1+t^2\right)^2} \)
\(=\frac{-2 t-2 t^3-2 t+2 t^3}{\left(1+t^2\right)^2}=\frac{-4 t}{\left(1+t^2\right)^2} \)
\( y =\frac{2 t}{1+t^2} \)
\( \frac{d y}{d t} =\frac{\left(1+t^2\right)(2)-2 t(2 t)}{\left(1+t^2\right)^2}=\frac{2+2 t^2-4 t^2}{\left(1+t^2\right)^2} \)
\(=\frac{2-2 t^2}{\left(1+t^2\right)^2}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \)
\( \frac{d y}{d x} =\frac{d y}{\frac{d t}{d t}}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \frac{\left(1+t^2\right)^2}{-t t} \)

\(\frac{d y}{d x}=\frac{\frac{d x}{d t}}{\frac{d x}{d t}}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \frac{\left(1+t^2\right)^2}{-4 t}\)
\(=\frac{\left(1-t^2\right)}{-2 t}=\frac{1}{2 t}\left(t^2-1\right)\).

We have \(x^2+y^2=4\)
As before, \({dy\over dx}=-{x\over y}\)
Hence, by the quotient rule
\({d^2y\over dx^2}=-{d\over dx}({x\over y})\)
\(=-{y.1-x.{dy\over dx}\over y^2}\)
\(=-{y-x(-{x\over y})\over y^2}\)
\(=-{x^2+y^2\over y^3}=-{4\over y^3}\).

\(y=4 \sec 5 x\)
\(Take\ u=5 x\)
\(\frac{d u}{d x}=5\)
\(y =4 \sec u\)
\(\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}=4 \sec u \tan u .(5)\)
\(=20 \sec 5 x \tan 5 x\)

Let x denote the number of kilograms bought per day and C denote the cost. Then,
\(C(x)= \begin{cases}0.16 x, & \text { if } 0 \leq x<100 \\ 0.14 x, & \text { if } x \geq 100\end{cases}\)
The sketch of this function
It is discontinuous at x = 100 since \(lim_{x\rightarrow100^-}c(x)=16\) and \(lim_{x\rightarrow100^+}c(x)=14\)
Note that C(100) = 14. Thus,\(lim_{x\rightarrow100^-}c(x)=16\neq 14=lim_{x\rightarrow100^+}C(x)=C(100).\)
Note also that the function jumps from one finite value 14 to another finite value 16.

\(lim_{x\rightarrow{0}}[{x^2+x\over x}+4x^3+3]=lim_{x\rightarrow 0}({x^2+x\over x})+lim_{x\rightarrow{0}}(4x^3+3)\)
\(lim_{x\rightarrow 0}(x+1)+lim_{x\rightarrow 0}(4x^3+3)\)
= (0 + 1) + (0 + 3)
= 4.

Given: f(-2) = 0, f(2) = 0
 \(\lim _{x \rightarrow-2} f(x)=0, \lim _{x \rightarrow 2} f(x)\) does not exist.

To sketch the graph of y = f(x)
(i) consider the points (-2, 0) and (2, 0).
(ii) Both points lie on x-axis. Therefore choose x-axis where \(x \leq 2\)
(iii) For x > 2 choose some other curve which does not pass through (2, 0).
It is better to choose y = 1 where x > 2

​\(f(x)=\begin{cases} { x^ 2 } ,\quad x \le 2 \\ 8-{ 2x } ,\quad 2 < x < 4 \\ { 4 } ,\quad x\ge 4 \end{cases}\)​
To sketch the graph of y = f(x)
(i) y = x² is an upward parabola with vertex at origin 
when x = 2, y = (2)² = 4
Draw the graph y = x² where \(x \leq 2\)
(ii) y = 8 - 2x is a straight line.
when x = 2, y = 8 - 4 = 4
when x = 4, y = 8 - 8 = 0
Join (2, 4) and (4, 0) by a line segment and delete the points (2, 4) and (4, 0)
(iii) y = 4 is a horizontal line with y intercept 4.
when x = 4, y = 4
Draw y = 4 where \(k \geq 4\)

\( \lim _{x \rightarrow 4^{-}} f(x)=0 \ and\ \lim _{x \rightarrow \mathbb{4}^{+}} f(x)=4 \)
\( \therefore \lim _{x \rightarrow 4} f(x) \) does not exist.
\( \therefore \lim _{x \rightarrow x_0} f(x) exists\ for\ x_0 \in R-\{4\} \)

Let the vertices of the triangle be A, B, C.
Then, given \(\overrightarrow { OA } =\hat { i } +2\hat { j } +3\hat { k } ,\overrightarrow { OB } =3\hat { i } -4\hat { j } +5\hat { k } \) and \(\overrightarrow { OC } =-2\hat { i } +3\hat { j } -7\hat { k } \)
\(\overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =(3\hat { i } -4\hat { j } +5\hat { k } )-(\hat { i } +2\hat { j } +3\hat { k } )=2\hat { i } -6\hat { j } +2\hat { k } \)
\(|\overrightarrow { AB } |=\sqrt { { 2 }^{ 2 }+{ (-6 })^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 4+36+4 } =\sqrt { 44 } \)
\(\overrightarrow { BC } =\overrightarrow { OC } -\overrightarrow { OB } =(-2\hat { i } +3\hat { j } -7\hat { k } )-(3\hat { i } -4\hat { j } +5\hat { k } )=-5\hat { i } +7\hat { j } -12\hat { k } \)
\(|\overrightarrow { BC } |=\overrightarrow { OC } -\overrightarrow { OB } =(-2\hat { i } +3\hat { j } -7\hat { k } )-(3\hat { i } -4\hat { j } +5\hat { k } )=-5\hat { i } +7\hat { j } -12\hat { k } \)
\(|\overrightarrow { BC } |=\sqrt { { (-5) }^{ 2 }+{ 7 }^{ 2 }+{ (-12) }^{ 2 } } =\sqrt { 25+49+144 } =\sqrt { 218 } \)
\(\overrightarrow { CA } =\overrightarrow { OA } -\overrightarrow { OC } =(\hat { i } +2\hat { j } +3\hat { k } )-(-2\hat { i } +3\hat { j } -7\hat { k } )=3\hat { i } -\hat { j } +10\hat { k } \)
\(|\overrightarrow { CA } |=\sqrt { { 3 }^{ 2 }+{ (-1) }^{ 2 }+{ 1 }0^{ 2 } } =\sqrt { 9+1+100 } =\sqrt { 110 } \)
\(\therefore\) Perimeter of \(\Delta ABC,\)
\(|\overrightarrow { AB } |+|\overrightarrow { BC } |+|\overrightarrow { CA } |=\left( \sqrt { 44 } +\sqrt { 218 } +\sqrt { 110 } \right) \) units

Let the vertices of the triangle be A( 1, 0, 0), B(0, 1, 0), C(0, 0, 1).
Let D, E, F are the mid-point of the sides BC, CA and AB respectively.

\(\therefore D \ is \ ({x_1+x_2\over 2},{y_1+y_2\over 2},{z_1+z_2\over 2})\)
\(\Rightarrow D \ is \ (0,{1\over 2},{1\over2})\) and E is \(({1\over 2},0,{1\over 2}), F \ is \ ({1\over2},{1\over 2},0)\)
Medians \(\overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}\)
\(=(0\hat{i}+{1\over2}\hat{j}+{1\over 2}\hat{k})-(\hat{i}-0\hat{j}+0\hat{k})=-\hat{i}+{1\over2}\hat{j}+{1\over2}\hat{k}\)
\(r=\sqrt{x^2+y^2+z^2}=\sqrt{1+{1\over4}+{1\over 4}}=\sqrt{4+1+1\over 4}={\sqrt{6}\over 2}\)
Hence, the direction cosines of \(\overrightarrow{AD}\) are,
\({-1\over {\sqrt{6}\over 2}},{-1\over 2\times{\sqrt{6}\over 2}},{-1\over 2\times{\sqrt{6}\over 2} } \Rightarrow {-2\over \sqrt{6}},{1\over \sqrt{6}},{1\over \sqrt{6}}\)
The median \(\overrightarrow{BE}\) are
\({-1\over 2\times{\sqrt{6}\over 2}},-{-1\over {\sqrt{6}\over 2}},{-1\over 2\times{\sqrt{6}\over 2} } \Rightarrow {1\over \sqrt{6}},{-2\over \sqrt{6}},{1\over \sqrt{6}}\)
The median \(\overrightarrow{CF}=\overrightarrow{OF}-\overrightarrow{OC}=({1\over 2}\hat{i}+{1\over2}\hat{j}+0\hat{k})\)\(-(0\hat{i}+0\hat{j}+\hat{k})={1\over2}\hat{i}+{1\over2}\hat{j}-\hat{k}\)
\(r=\sqrt{{1\over4}+{1\over4}+1}={\sqrt{6}\over 2}\)
\(\therefore\) The direction of cosines of \(\overrightarrow{CF}\) are \({1\over 2\times{\sqrt{6}\over 2}},{1\over 2\times{\sqrt{6}\over 2} },{-1\over {\sqrt{6}\over 2}} \Rightarrow {1\over \sqrt{6}},{1\over \sqrt{6}},{-2\over \sqrt{6}}\)

Let ABC be a triangle and let O be the origin.
Let D and E be the midpoints of AB and AC\(\overrightarrow{OE}={\overrightarrow{OA}+\overrightarrow{OC}\over 2}​={\overrightarrow{a}+\overrightarrow{c}\over 2}​\)

\( \text { Let } \overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}, \overrightarrow{O C}=\vec{c} \)
\(\overrightarrow{O D}= \frac{\vec{a}+\vec{b}}{2}, \quad \overrightarrow{O E}=\frac{\vec{a}+\vec{c}}{2} \)
\(\left.\therefore \overrightarrow{D E}=\overrightarrow{O E}-\overrightarrow{O D}=\left(\frac{\vec{a}+\vec{c}}{2}\right)-\left(\frac{\vec{a}+\vec{b}}{2}\right)=\frac{\vec{a}+\vec{c}-\vec{a}-\vec{b}}{2}\right) \)
\(=\frac{\vec{c}-\vec{b}}{2}=\frac{(\overrightarrow{O C}-\overrightarrow{O B})}{2}\)
\( \overrightarrow{D E}=\frac{\overrightarrow{B C}}{2}=\frac{1}{2} \overrightarrow{B C} \)
\(\therefore D E \| B C . \)
\(\text {Also } \overrightarrow{D E}=\frac{1}{2} \overrightarrow{B C} \Rightarrow|\overrightarrow{D E}|=\frac{1}{2}|\overrightarrow{B C}| \)
\(D E=\frac{1}{2} B C \)
Hence, \(D E \| B C \text { and } D E=\frac{1}{2} B C \text {. }\) 

Let \(\overrightarrow{a}\) and \(\overrightarrow{b}\) be the position vectors of the points A and B.
\(\Rightarrow \overrightarrow{OA}=\overrightarrow{a}\) and \( \overrightarrow{OB}=\overrightarrow{b}\).
Let P divides the line segment AB in the ratio 1:2 and Q divides the line segment AB in the ratio 2 : 1
\(\therefore \overrightarrow{OP}={1.(\overrightarrow{OB})+2(\overrightarrow{OA})\over 1+2}={1(\overrightarrow{b})+2(\overrightarrow{a})\over 3}={\overrightarrow{b}+2\overrightarrow{a}\over 3}\)
and \( \overrightarrow{OQ}={2(\overrightarrow{OB})+1(\overrightarrow{OA})\over 2+1}={2\overrightarrow{b}+\overrightarrow{a}\over 3}={\overrightarrow{a}+2\overrightarrow{b}\over 3}\)
Hence, the required position vectors are \({\overrightarrow{b}+2\overrightarrow{a}\over 3}\)and \({\overrightarrow{a}+2\overrightarrow{b}\over 3}\).

Let |A| = \(\begin{vmatrix} 1 &x^2 &x^3 \\ 1 & y^2 &y^3 \\1 &z^2 &z^3 \end{vmatrix}\) .
Putting x = y gives |A| = \(\begin{vmatrix} 1 &x^2 &x^3 \\ 1 & y^2 &y^3 \\1 &z^2 &z^3 \end{vmatrix}\) = 0 (since R₁ \(\equiv\) R₂).
Therefore (x - y) is a factor.
The given determinant is in cyclic symmetric form in x, y and z. Therefore (y - z) and (z - x) are also factors.
The degree of the product of the factors (x − y)( y − z)(z − x) is 3 and the degree of the product of the leading diagonal elements 1× y² × z³ is 5.
Therefore the other factor is k(x² + y² + z² ) + l(xy + yz + zx) .
Thus \(\begin{vmatrix} 1 &x^2 &x^3 \\ 1 & y^2 &y^3 \\1 &z^2 &z^3 \end{vmatrix}\)=[k(x² + y² + z² ) + l(xy + yz + zx) ] × (x − y)( y − z)(z − x) .
Putting x = 0, y = 1 and z = 2, we get
\(\begin{vmatrix} 1 &0 &0 \\ 1 & 1 & 1 \\ 1 &4 &8 \end{vmatrix}\)= [k(0+1+4 ) +l(0+2+0) ](-1)( 1 - 2)(2 - 0)
\(\Rightarrow\) (8 - 4) = [(5k + 2l)](−1)(−1)(2)
4 = 10k + 4l \(\Rightarrow\) 5k + 2l = 2.
Putting x = 0, y = −1 and z =1, We get
\(\begin{vmatrix} 1 &0 &0 \\ 1 & 1 & -1 \\ 1 &4 &8 \end{vmatrix}\) = [k(2) + l(−1)](1)(−2)(1)
\(\Rightarrow\) [(2k − l)(−2)] = 2
2k − l = - 1.
Solving (1) and (2), we get k = 0, l =1.
Thus \(\begin{vmatrix} 1 &x^2 &x^3 \\ 1 & y^2 &y^3 \\1 &z^2 &z^3 \end{vmatrix}\) = (x - y)(y - z)(z - x)(xy + yz + zx).

Given a, b, c are p^th, q^th and r^th terms of a G.P.
\(p^{\text {th }} \text { term } \Rightarrow A R^{r^{-1}}=a\)
\(q^{\text {th }} \text { term } \Rightarrow A R^{q-1}=b\)
\(r^{\text {th }} \text { term } \Rightarrow A R^{r-1}=c\)
\(A R^{P^{-1}}=a\)
Take log on both sides.
\(\log A+(p-1) \log R=\log a .\)
Similarly
\(\log A+(q-1) \log R=\log b\)
\(\log A+(r-1) \log R=\log c\)
\(L H S=\left|\begin{array}{lll} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{array}\right|\)
\(=\left|\begin{array}{lll} \log A+p \log R-\log R & p & 1 \\ \log A+q \log R-\log R & q & 1 \\ \log A+r \log R-\log R & r & 1 \end{array}\right|\)
\(=\left|\begin{array}{lll} \log A & p & 1 \\ \log A & q & 1 \\ \log A & r & 1 \end{array}\right|+\left|\begin{array}{lll} p \log R & p & 1 \\ q \log R & q & 1 \\ r \log R & r & 1 \end{array}\right|-\left|\begin{array}{ccc} \log R & p & 1 \\ \log R & q & 1 \\ \log R & r & 1 \end{array}\right|\)
Since in first and third determinants C₁ and C₃ are proportional and in second determinants C₁ and C₂ are proportional.
= 0 +0 - 0 0 = RHS.
Hence proved.

Given f(x) = \(\begin{bmatrix} cos \ x & -sin \ x & 0 \\ sin x & cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
f(x) \(\times\) f(y) = \(\begin{bmatrix} cos \ x & -sin \ x & 0 \\ sin x & cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\)\(\left[ \begin{matrix} cosy & -siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
= \(\left[ \begin{matrix} cosxcosy-sinxcosy & -cosxsiny-sinxcosy & 0 \\ sin \ x cos \ y+cos \ xsin \ y & -sin \ x \ sin \ y+cos \ x \ cos \ y & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
= \(\left[ \begin{matrix} cos(x+y) & -sin(x+y) & 0 \\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
[\(\therefore\)cos(x + y) = cos x cos y - sin x sin y]
sin(x + y) = sin x cos y + cos x sin y]
= f(x + y)

Cost matrix A = [30 15 45], Fruit matrix

Cost of packages are obtained by computing AB. That is, by multiplying cost of each item in A (cost matrix A) with number of items in B (Fruit matrix B).
\(A B=\left[\begin{array}{lll} 30 & 15 & 45 \end{array}\right]\left[\begin{array}{lll} 6 & 5 & 6 \\ 3 & 4 & 6 \\ 3 & 4 & 6 \end{array}\right]=\left[\begin{array}{lll} 360 & 390 & 540 \end{array}\right]\)
Pack-I cost Rs. 360, Pack-II cost Rs. 390, Pack-III costs Rs. 540.

Let p(n) : 2n < (n + 2)! for all k\(\in\)N.
Step1: P(1) 2.1 < (1+2)!
\(\Rightarrow\) 2 < 3!\(\Rightarrow\) 2 < 6 which is true for P(1)  [\(\because\) 3! = 3 x 2 x 1 = 6]
Step 2: P(k) : 2k < (k + 2)!. Let it be true for P(k)
Step3:P(k+1) 2(k+1) < (k+1+2)!
Since 2k < (k + 2)!  (from Step 2)
\(\Rightarrow\)2k+2 < (k+2)! +2
\(\Rightarrow\) 2(k+1) < (k+2)!+2
Also, (k+ 2)! + 2 < (k+ 3)!
\(\therefore\) 2(k+1) < (k+3)!
\(\Rightarrow\) 2(k + 1) < (k + 2 + 1)! which is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

Given h = 61.4 + 2.3 F
Given h = 160 ⇒160 = 61.4 + 2.3 F
⇒ 2.3 F = 160-61.4 = 98.6
F =\(\frac { 98.6 }{ 2.3 } \) = 42.87
Given h = 170,
170 = 61.4 + 2.3 F
⇒ 170 - 61.4 = 2.3 F
2.3 F = 108.6
⇒ F =\(\frac { 108.6 }{ 2.3 } \)= 47.23
So the ranges of values are
42.8 < x < 47.23

cosec (90° + A) = sec A, cot (90° + A) = - tan A
LHS sec A + x cos A(-tan A)
\(=\frac{1}{cosA}-xcosA\times\frac{sinA}{cosA}=\frac{1}{cosA}-xsinA\)
RHS = sin(90°+A) = cos A
\(\therefore\frac{1}{cosA}-xsinA=cosA\)
\(\Rightarrow \frac{1}{cosA}-sinA=xsinA\Rightarrow\frac{1-cos^2A}{cosA}=xsinA\)
\(\Rightarrow \frac{sin^2A}{sinAcosA}=x\)
\(x=\frac{sinA}{cosA}=tanA\)

Given that:
2x+y=5.....(i)
x+3y+8=0...(ii)
3x+4y=7 ....(iii)
Equation of any line passing through the point of intersection of equation (i) and (ii) is
(2x+y-5)+λ(x+3y+8)=0 ...(iv)     (λ=constant)
⇒ 2x+y-5+λx+3λy+8λ=0
⇒ (2+λ)x+(1+3λ)y-5+8λ=0
Slope of line m₁ (say) = \(\frac { -(2+\lambda ) }{ 1+3\lambda } \)   \(\left[ \because m=\frac { -a }{ b } \right] \)
Now slope of line 3x + 4y = 7 is
m₂(say) = -\(\frac { 3 }{ 4 } \)
If equation (iii) is parallel to equation (iv) then m₁ = m₂
⇒ \(\frac { -(2+\lambda ) }{ 1+3\lambda } =-\frac { 3 }{ 4 } \)
⇒ \(\frac { 2+\lambda }{ 1+3\lambda } =\frac { 3 }{ 4 } \) ⇒ 8+4λ=3+9λ
⇒ 9λ-4λ=5  ⇒ 5λ=5  ⇒  λ=1
On putting the value of A.in equation (iv) we get
(2x+y-5)+1(x+3y+8)=0
⇒ 2x+y-5+x+3y+8=0  ⇒ 3x+4y+3=0
Hence, the required equation is 3x+4y+3=0

Let t_k denote the k^th term of the given series.
Then \(t_k{1\over k(k+1)}.\) 
By using partial fraction we get
\({1 \over k(k+1)}={1\over k}-{1\over{k+1}}\)
Thus \(t_1+t_2+....+t_n=\left( 1-{1\over 2} \right)+\left( {1\over 2}+{1\over 3} \right)+\left( {1\over 3}-{1\over 4} \right)+...+\left( {1\over n} -{1\over n+1} \right)=1-{1\over n+1}.\)

\(\frac{2^n+3^n}{n}\)

(i) general equation is ax² + 2hxy + by² + 2gx + 2fy + C = 0
given equation is  λx²-10xy + 12y² + 5x -16y - 3 = 0
Comparing the given equation with the general equation of the second degree we have
a = λ, b = 12,  c = 3, h = -5, g = 5/2, f = -8
Now applying the condition for pair of straight lines
abc + 2fgh-af²- bg²- ch² = 0
i.e., λ(12)(-3)+2(-8)\(\left( \frac { 5 }{ 2 } \right) \)(-5)-λ(-8)²-12\(\left( \frac { 5 }{ 2 } \right) \)²-(-3)(-5)² = 0
⇒ -36λ + 200 - 64λ - 75 + 75 = 0 ⇒ λ = 2
∴ The equation is 2x²-10xy + 12y² + 5x - 16y - 3 = 0
Let us first factorize the terms of second degree terms from the above equation,
we get 2x² - 10xy + 12y² = (x - 2y) (2x - 6y)
⇒ 2x²- 10xy + 12y²+ 5x - 16y - 3 ≡ (x - 2y)(2x - 6y)
⇒ 2x²-10xy + 12y²+5x -16y - 3 ≡ (x - 2y + c₁)(2x - 6y + c₂)
Equating like terms, we get
2c₁ + c₂ = 5, 3c₁ + c₂ = 8, c₁c₂ = -3
Solving first two equation, we get c₁= 3, c₂ = -1
∴ The separate equations of the lines are x - 2y + 3 = 0 and 2x - 6y - 1 = 0
(ii) Point of intersection of the lines is given by solving the two equation of the lines, we get
(x,y) = \(\left( -10,-\frac { 7 }{ 2 } \right) \) (or use the formula \(\left( \frac { hf-bg }{ ab-h^{ 2 } } ,\frac { hg-af }{ ab-{ h }^{ 2 } } \right) \))
(iii) Angle between the lines is given by
tanፀ = \(\left| \frac { 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \right| \)
= \(\frac { 2\sqrt { 25-24 } }{ 2+12 } =\frac { 1 }{ 7 } \)
∴ \(\theta ={ tan }^{ -1 }\left( \frac { 1 }{ 7 } \right) \).

The Sine formula is, \({a\over sin A}={b\over s in B}={c\over s in C}=2R\)
LHS = \({a^2+b^2\over a ^2+c^2}={({2RsinA})^2+({2RsinB})^2\over ({2RsinA})^2+({2RsinC})^2 }\)
\(={sin^2A+sin^2B\over sin^2A+sin^2C }={1-cos^2A+sin^2B\over 1-cos^2A+sin^2C }\)
\(={1-(cos^2A-sin^2B)\over 1-(cos^2A-sin^2C) }={1-cos(A+B)cos(A-B)\over 1-cos(A+C)cos(A-C)}\)
\(={1+cos(A-B)cosC\over 1+cos(A-C)cosB}\)

Let, P(n) = 1² + 2² + 3² +...+k² = \(\frac { n(n+1)(2n+1) }{ 6 } \)
Substituting n = 1 in the statement we get, P(1) = \(\frac { 1(1+1)(2(1)+1) }{ 6 } \)  = 1.
Hence, P(1) is true. Let us assume that the statement is true for n = k. Then
P(k) = 1² + 2² + 3²+...+k² = \(\frac { k(k+1)(2K+1) }{ 6 } \)
We need to show that P(k + 1) is true. Consider
\(P(k+1) =\underbrace{1^2+2^2+3^3+...k^2}+(k+1)^2\)
= P(k) + (k+1)²
= \(\frac { k(k+1)(2K+1) }{ 6 } +(k+1)^2\)
= \(\frac { k(k+1)(2k+1)+6(k+1)^{ 2 } }{ 6 } =\frac { (k+1)(k(2k+1)+6(k+1) }{ 6 } \)
= \(\frac { (k+1)(2k^{ 2 }+7k+6) }{ 6 } =\frac { (k+1)[(k+2)(2k+3)] }{ 6 } \)
= \(\frac { (k+1)[(k+1)+1)(2(k+1)+1)] }{ 6 } \)
That is, P(k+1) = \(\frac { (k+1)((k+1)+1)(2(k+1)+1) }{ 6 } \)
This implies P(k + 1) is true. The validity of P(k + 1) follows from that of P(k). Therefore by the principle of mathematical induction,
1² +2² + 3²+...n² = \(\frac { n(n+1)(2n+1) }{ 6 } \), for all n ≥ 1.

sin 72° = sin (90° - 18°)
= cos 18°\(=\frac{1}{4}\sqrt{10+2\sqrt 5}\)

reciprocal of the other
The roots are reciprocal of the other
Let α and \({1\over α}\) be the roots
\(∴\ α+{1\over α}={-b\over a}\)
and \(α.{1\overα}={c\over a}\)
\(⇒\ 1={c\over a}\)
⇒ c = a
which is the required condition

Thrice the other
The roots are thrice the other
Let α and 3α be the roots
\(α+3α=-{b\over a}\)
\(⇒\ 4α=-{b\over a}\)
\(⇒\ α=-{b\over 4a}\)
\((α)(3α)={c\over a}\)
\(⇒\ 3α^2={c\over a}\)
\(⇒\ α^2={c\over 3a}\)
Substituting (1) in (2) we get
\(\left(-b\over 4a\right)^2={c\over 3a}\)
\(⇒\ {b^2\over 16a^2}={c\over 3a}\)
\(⇒\ {b^2\over 16a^2}={c\over 3a}\)
\(⇒\ {b^2\over 16a^2}={c\over 3}\)
⇒ 3b² = 16ac which is the required condition.

By the given data m and 3m are the slopes of the pair of lines ax² + 2hxy + by² = 0
We know that m₁ + m₂ \(=\frac { -2h }{ b } and\quad m_{ 1 }{ m }_{ 2 }=\frac { a }{ b } \)
\(\therefore \quad m+2m=\frac { -2h }{ b } and\quad m(3m)=\frac { a }{ b } \)
\(\Rightarrow \quad 4m=\frac { -2h }{ b } and\quad { 3m }^{ 2 }=\frac { a }{ b } \)
\(\Rightarrow \quad m=-\frac { 2h }{ 4b } and3{ m }^{ 2 }=\frac { a }{ b } \)
\(\Rightarrow \quad m=\frac { -h }{ 2b } ...91)\quad and\quad 3{ m }^{ 2 }=\frac { a }{ b } ...(2)\)
Substituting (1) in (2) we get,
\(3{ \left( \frac { -h }{ 2b } \right) }^{ 2 }=\frac { a }{ b } \)
\(\Rightarrow 3\left( \frac { { h }^{ 2 } }{ { 4b }^{ 2 } } \right) =\frac { a }{ b } \)
\(\Rightarrow \frac { { 3h }^{ 2 } }{ 4b } =a\)
\(\Rightarrow\) 3h² = 4ab

(i) Let the variable x represents the quantity of gas, and y represents the number of days.
By the given data,

Using two - point form, the linear relationship between quantity of gas in the cylinder to the number of days is
\(\frac { y-0 }{ 24-0 } =\frac { x-0 }{ 14.2-0 } \ \left[ \because \ \frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right] \)
\(\frac { y }{ 24 } =\frac { x }{ 14.2 } \)
\(\Rightarrow\)   14.2 y = 24x
\(\Rightarrow\)  24x - 14.2y = 0
\(\Rightarrow\)  12x - 7.1y = 0  ....(1)
Which is the required linear relation...(2)
Find the equation relating the quantity of gas in the cylinder to the days.
To find the Time taken to cross the pole, put y = 0
\(\Rightarrow \ \therefore \ 12.5x\ =\ 150\ \ \Rightarrow \ x=\frac { 150 }{ 12.5 } =12\ sec\)
(ii) Draw the graph for first 96 days

y = f(x) is a periodicfunctionwith period 24. 
Therefore, f(x) = f(x + 24)

There are 7 Indian's and 5 Americans. A committee of 5 members with majority Indian's can be formed by the following ways.

∴ Total number of ways of forming the committee
= 7C₄ \(\times \)5C₁+7C₃ \(\times \)5C₂+7C₅ \(\times \)5C₀
= 7C₃ \(\times \)5 + 7C₃\(\times \)5C₂ + 7C₂\(\times \)1
\(=\frac { 7\times 6\times 5 }{ 3\times 2\times 1 } \times 5+\frac { 7\times 6\times 5 }{ 3\times 2\times 1 } \times \frac { 5\times 4 }{ 2\times 1 } \times \frac { 7\times 6 }{ 2\times 1 } \)
= 175 + 350 + 21
= 546.

Let p = q+h
h is numerically very small and so h₁² h³ ... may be neglected
RHS \(={(n+1)p+(n-1)q\over (n-1)p+(n+1)q}={(n+1)(q+h)+(n-1)q\over (n+1)(q+h)+(n+1)q}\)
\(={nq+q-nh+h+nq-q\over nq-q+nh-h+nq+q}={2nq+(n+1)h\over2nq+(n-1)h}\)
\(={1+{n+1\over2n}.{q\over h}\over1+{n-1\over2n}.{h\over q}}=\left(1+{n+1\over2n}.{h\over q}\right)\left(1+{n-1\over2n}.{h\over q}\right)^{-1}\)
\(=\left({1+{n+1\over2n}.{h\over q}}\right)\left(1-{n-1\over2n}.{h\over q}\right)=1+\left({n+1\over 2n}-{n-1\over2n}\right){h\over q}\)
\(={1+{1\over n}}.{h\over q}\)
LHS \(={p\over q}^{1\over n}=\left(q+h\over q\right)^{1\over n}=\left(1+{h\over q}\right)^{1\over n}=1+{1\over n}.{h\over q}\)
From (1) and (2), LHS = RHS
Now \(\sqrt[8]{15\over16}={(8+1)(15)+(8-1)(16)\over(8-1)(15)+(8+1)(16)}\)  [n = 8, p = 15 and q = 16]
 \(={(9)(15)+7(16)\over(7)(15)+9(16)}={135+112\over105+144}={247\over 249}\)
\(\left( \frac { 15 }{ 16 } \right) =0.9919\)

LHS = ²ⁿC_n
= \(\frac { 2n! }{ n!(2n-n)! } =\frac { 2n! }{ n!n! } \)
= \(\frac { (2n)(2n-1)(2n-2)...4.3.2.1 }{ n!n! } =\frac { (2n)(2n-2)...4.2[(2n-1)(2n-3)...3.1] }{ n!n! } \)    [Separate the even and odd terms]
= \(\frac { { 2 }^{ n }.n(n-1)...2.1[(2n-1)(2n-3)...3.1] }{ n!n! } \)
= \(\frac { { 2 }^{ n }.(n)![1.3.5...(2n-3)(2n-1)] }{ n!n! } =\frac { { 2 }^{ n }[1.3.5...(2n-3)(2n-1)] }{ n! } \) = RHS

Let P(h, k) be the point on RQ such that ㄥORQ = θ where θ is a variable.
From triangle PLR, we get sin θ = \(\frac{k}{b}\) ...(1)
From triangle PMQ, we get cos θ = \(\frac{h}{b}\) ...(2)

Squaring and adding (1) and (2), we get,
sin²θ + cos²θ = (\(\frac{k}{b}\))² + (\(\frac{h}{a}\))²
1 = \(\frac{k^2}{b^2}+\frac{k^2}{b^2}\)
ஃ Locus of (h, k) is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

LHS = \({ \left( { x }^{ 3 }+6 \right) }^{ \frac { 1 }{ 3 } }-{ \left( { x }^{ 3 }+3 \right) }^{ \frac { 1 }{ 3 } }\)
\(={ x }^{ 3\times \frac { 1 }{ 3 } }{ \left( 1+\frac { 6 }{ { x }^{ 3 } } \right) }^{ \frac { 1 }{ 3 } }-{ x }^{ 3\times \frac { 1 }{ 3 } }{ \left( 1+\frac { 3 }{ { x }^{ 3 } } \right) }^{ \frac { 1 }{ 3 } }\)
\(=x\left[ 1+\frac { 1 }{ 3 } \left( \frac { 6 }{ { x }^{ 3 } } \right) \right] -x\left[ 1+\frac { 1 }{ 3 } \left( \frac { 3 }{ { x }^{ 3 } } \right) \right] \)
\(=x+\frac { 2 }{ { x }^{ 2 } } -x-\frac { 1 }{ { x }^{ 2 } } \)
\(=\frac { 2 }{ { x }^{ 2 } } -\frac { 1 }{ { x }^{ 2 } } =\frac { 1 }{ { x }^{ 2 } } =RHS\)
Hence proved.

(100 - 1)⁴  (a -b)ⁿ = nC₀ aⁿ b⁰ - nC₁ a^n-1 b¹ +...nC_n a⁰ bⁿ, n \(\epsilon \) N
= 100⁴ - 4C₁ (100)³ (1)¹ + 4C₂ (100)² (1)² - 4C₃ (100)(1³)+1⁴
= 100000000 + 4(1000000) + \(\frac { 4\times (3) }{ 2\times 1 } \)(10000) - 400+1
= 100000000 -4000000 +60000 - 400 +1
= 96,059,601

Let = 6 + 66 + 666 + ... upto n terms
= 6 (I + 11 + 111+ ....) upto n terms
\(={6\over9}(9+99+999+ ...)\) upto n terms
\(={63\over 6}[(10 -1) + (10^2-1) + (10^3 -1) + ...]\) upto n terms
\(={6\over 9}[(10+ 10^2 + 10^3+ ...) - (1+ 1+1...)]\) upto n terms​​​​​​​
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]\)[In a G.P with a = 10 r = 10, \(S_n={(r^n-1)\over r-1}\)]
\(={6\over9}\left[{ 10(10^n-1)\over 10-1}n\right]={6\over9}\left[ 10(10^n-1)-9n\over9\right]\)
\({ S }_{ n }=\frac { 6 }{ 81 } \left[ 10\left( { 10 }^{ n }-1 \right) -9n \right] \)

Let pen) be the statement
\({1\over 1.2.3}+{1\over 2.3.4}+...+{1\over {n(n+1)(n+2)}}={n(n+3)\over 4(n+1)(n+2)}\)
Step 1: Putting n = 1, we get
\({1\over 1.2.3}={1(1+3)\over 4(1+1)(1+2)}⇒{1\over 6}={4\over 4(2)(3)}={1\over 6}\)
∵  p(1) is true
Step 2: Let us assume that p(K) is true.
\(∵\ {1\over 1.2.3}+{1\over 2.3.4}+..+{1\over K(K+1)(K+2)}={K(K+3)\over 4(K+1)(K+2)}\)
Step 3: To prove that p(K + 1) is true
ie to P.T. \({1\over 1.2.3}+{1\over 2.3.4}+...+{1\over K(K+1)(K+2)(K+3)}={(K+1)(K+4)\over 4(K+2)(K+3)}\)
LHS = \({1\over 1.2.3}+{1\over 2.3.4}+...+{1\over K(K+1)(K+2)}+{1\over (K+1)(K+2)(K+3)}\)
\(={K(K+3)\over 4(K+1)(K+2)}+{1\over (K+1)(K+2)(K+3)}\)
\(={1\over (K+2)(K+1)}\left[{K(K+3)\over 4}+{1\over (K+3)}\right]={1\over (K+2)(K+1)}\left[K(K+3)^2+4\over 4(K+3)\right]\)
\(={1\over 4(K+1)(K+2)(K+2)}[K(K^2+6K+9)+4]\)
\(={1\over 4(K+1)(K+2)(K+2)}[K^3+6K^2+9K+4]\)
\(={1\over 4(K+1)(K+2)(K+2)}[(K+1)(K+1)(K+4)]\)
= (K + 1)(K + 1)(K + 4)
[Using Synthetic division K³ + 6K²+ 9K +4]

∵ p(K + 1) is true.
Hence, by the principle of mathematical induction
p(K + 1) is true for all values of n.

Let y = 3x -5.
\(\Rightarrow y+5=3x\Rightarrow \frac { y+5 }{ 3 } =x\)
Let g(y) = \(\frac { y+5 }{ 3 } \)
\(gof(x)=g(f(x))=g(3x-5)=\frac { 3x-5+5 }{ 3 } =\frac { 3x }{ 3 } =y\)
Also f o g(y) = f(g(y)) = \(f\left( \frac { y+5 }{ 3 } \right) =3\left( \frac { y+5 }{ 3 } \right) -5=y+5-5=y\)
Thus g o f = Ix and fog = Iy.
This implies that f and g are bijections and inverses to each other.
Hence f is a bijection and f^-1(y) = \(\frac { y+5 }{ 3 } \)
Replacing y by x we get, f^-1(x) = \(\frac { x+5 }{ 3 } \)

\({ tan }^{ -1 }\left( \frac { cos\quad x }{ 1-sin\quad x } \right) ={ tan }^{ -1 }\left( \frac { sin\left( \frac { \pi }{ 2 } -x \right) }{ 1-cos\left( \frac { \pi }{ 2 } -x \right) } \right) \)  \([Since\quad sin(90-\theta )=cos\theta \quad and\quad cos\quad (90-\theta )=sin\quad \theta ]\)
\(={ tan }^{ -1 }\left[ \frac { sin\left( \frac { \pi -2x }{ 2 } \right) }{ 1-cos\left( \frac { \pi -2x }{ 2 } \right) } \right] \)
\(={ tan }^{ -1 }\left[ \frac { 2\quad sin\left( \frac { \pi -2x }{ 4 } \right) .cos\left( \frac { \pi -2x }{ 4 } \right) }{ 2{ sin }^{ 2 }\left( \frac { \pi -2x }{ 4 } \right) } \right] \left[ \because sin\theta =2sin\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } and1-cos\quad \theta =2{ sin }^{ 2 }\frac { \theta }{ 2 } \right] \)
\(={ tan }^{ -1 }\left( \frac { cos\left( \frac { \pi -2x }{ 4 } \right) }{ sin\left( \frac { \pi -2x }{ 4 } \right) } \right) \)
\(={ tan }^{ -1 }\left( cot\left( \frac { \pi -2x }{ 4 } \right) \right) \)
\(={ tan }^{ -1 }\left( tan\left( \frac { \pi }{ 2 } -\frac { \pi -2x }{ 4 } \right) \right) \quad \quad [\because cot\quad \theta =tan(90-\theta )]\)
\(={ tan }^{ -1 }\left( tan\left( \frac { 2\pi -\pi }{ 4 } +2x \right) \right) \)
\(={ tan }^{ -1 }\left( tan\left( \frac { \pi +2x }{ 4 } \right) \right) =\frac { \pi +2x }{ 4 } \quad \quad [\because { tan }^{ -1 }(tan\theta )=\theta ]\)
\(=\frac { \pi }{ 4 } +\frac { 2\pi }{ 4 } =\frac { \pi }{ 4 } +\frac { x }{ 2 } \) which is the simplest form.

f(-3) = (-3)² - 3    \(\left[ \therefore \ f(x)={ x }^{ 2 }-3\quad when\ x=-3 \right] \)
= 9 - 3 = 6
f(5) = 5² + 3(5)-2   \(\left[ \therefore f(x)={ x }^{ 2 }+3x-2\quad when\quad x=5 \right] \)
= 25 + 15 - 2
= 38
f(2) = 2² - 3
= 4 - 3 = 1   \(\left[ \therefore \ f(x)={ x }^{ 2 }-3\ when\ x=2 \right] \)
f(-1) = (-1)² + (-1) -5   \(\left[ \therefore \ f(x)={ x }^{ 2 }+x-5\ when\ x=-1 \right] \)
= 1-1-5 = -5
f(0) = 0²-3 = -3  \(\left[ \therefore \ f(x)={ x }^{ 2 }-3\ when\ x=0 \right] \)
\(\therefore\) f(-3) = 6, f(5) = 38, f(2) = 1, f(-1) = -5, f(0) = -3

Let P be the intruder, A and B are the soldiers.
Let x be the distance between the intruder and soldier B.
In ΔABP

Given ㄥPAB
and ㄥPBC
In ΔABP, ㄥAPB
In ΔABP, using sine formula,
\(⇒\ {5\over sin\ 15^0}={x\over sin30^0}\)
\(⇒\ x{5\over sin15^0}sin30^0=5\times {1\over 2 sin15^0}\)
Now, sin 15 = sin (45 - 30)
= sin 45 cos 30 - cos 45 sin 30
\(={1\over \sqrt2}\times{\sqrt3\over 2}-{1\over \sqrt2}={\sqrt3-1\over 2\sqrt2}\)
Substituting this value in (1) we get,
\(x={5\over {2(\sqrt3-1)\over 2\sqrt2}}={5\sqrt2\over \sqrt3-1}\)

\({{x}\over{(x^2+1)(x-1)(x+2)}}={Ax+B\over (x^2+1)}+{C\over (x-1)}+{D\over (x+2)}\)
\(⇒\ \ {x\over (x^2+1)(x-1)(x+2)}x={(Ax+B)(x-1)(x+2)+C(x^2+1)(x+2)+D(x^2+x)(x-1)\over (x^2+1)(x-1)(x+2)}\)
x = (Ax + B) (x - 1) (x + 2) + C(x² + 1) (x + 2) + D(x² + 1) (x - 1) ...(1)
Putting x = 1 in (1) we get
1 = C(2)(3) ⇒ \(C={1\over 6}\)
Putting x = -2 in (1) we get
-2 = D(5)(-3)
⇒ \(D={2\over 15}\)
Putting x = 0 in (1) we get
0 = (B) (-1) (2) + C(1) (2) + D(1) (-1)
⇒ 0 = -2B + 2C - D
⇒ 0 = \(-2B+2\left(1\over 6\right)-{2\over 15}\)[substituting the values of C and D]
⇒ \(2B={2\over 6}-{2\over 15}={1\over 3}-{2\over 15}\)
⇒ \(2B={5-2\over 15}={3\over 15}={1\over 5}\)
⇒ \(B={1\over 10}\)
Equating the Co-efficient of x³ in (1) we get
0 = A + C + D
⇒ A = - C - D
⇒ \(A=-{1\over 6}-{2\over 15}\)
⇒ \(A={-5-4\over 30}={-9\over 30}={-3\over10}\)
⇒ \(A={-3\over 10}\)
\(∴\ {x\over (x^2+1)(x-1)(x+2)}={{-3\over 10}x+{1\over 10}\over x^2+1}+{{1\over 6}\over x-1}-{{2\over 15}\over x+2}\)
= \({{-3x+1}\over{10(x^2+1)}}+{{1}\over{6(x-1)}}+{{2}\over{15(x+2)}}\)

The relation is defined by aRb if a + b \(\le\) 6 for all a, b  \(\in \)N.
a+b \(\le\)6 \(\Rightarrow\) a \(\le\) 6 - b

\(\therefore\) The list of ordered pairs are (5,1) (4, 2) (3, 3) (2, 4) and (1, 5).
Symmetric : (5, 1) \(\in \) R \(\Rightarrow\) (1, 5) \(\in \) R
(4, 2) \(\in \) R \(\Rightarrow\) (2, 4) \(\in \) R
\(\therefore\) R is symmetric

Given inequality is \({x^3(x-1)\over x-2}>0\)
The critical numbers are 0, 1, 2
The possible intervals are (-∞, 0) (0, 1) (1, 2) (2, ∞)

The given inequality  \({x^2(x-1)\over x-2}>0\) is satisfied by the intervals (0, 1) and (2, ∞)
∴ Solution set is (0, 1)∪(2, ∞)
∴ Solution set is \((0,1)\bigcup(2, \infty)\)

We have A + B + C = 180°
B + C = 180° - A = 180° - 60° = 120°
\(\Rightarrow \frac { B+C }{ 2 } =60°\)...(1)
Using sine formula, \(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } =k\)
a = k sin A, b = k sin B, c = k sin C
RHS = 2a cos\(\left( \frac { B-C }{ 2 } \right) \)...(2)
= 2.K sin A cos\(\left( \frac { B-C }{ 2 } \right) \)
=2.K sin 60° cos \(\left( \frac { B-C }{ 2 } \right) \)
= 2K.sin\(\left( \frac { B+C }{ 2 } \right) \)cos\(\left( \frac { B-C }{ 2 } \right) \)
= K\(\left[ sin\left( \frac { B+C+B-C }{ 2 } \right) +sin\left( \frac { B+C-B-C }{ 2 } \right) \right] \)
= K[sin B + sin C] 
= K sin B + K sin C
= b + c [From(2)]
LHS Hence proved.

Given relation is 2a + 3b = 30 for all a, b \(\in \) N.
2a + 3b = 30 \(\Rightarrow\) 2a =30 - 3b
\(\Rightarrow a=\frac { 30-3b }{ 2 } \)

\(\therefore\) The list of ordered pairs are (12, 2) (9, 4) (6, 6) (3, 8)
Transitivity: Clearly R is not transitive.

2a + 3b = 30
R = {(3,8), (6,6), (9,4). (12,2)}
Not reflexive, Not Symmetric, transitive, hence not an equivalence relation.

Given sin θ = \(\frac{1}{25}\)
\(cos\theta =\sqrt { 1-{ sin }^{ 2 }\theta } =\sqrt { 1-\frac { 1 }{ 625 } } =\frac { \sqrt { 624 } }{ 25 } =\frac { 4\sqrt { 39 } }{ 25 } \)
Now, \(sin\frac { \theta }{ 2 } =\sqrt { \frac { 1-cos\theta }{ 2 } } =\sqrt { \frac { 1-\frac { 4\sqrt { 39 } }{ 25 } }{ 2 } } =\sqrt { \frac { 25-4\sqrt { 39 } }{ 50 } } \)
\(cos\frac { \theta }{ 2 } =\sqrt { \frac { 1+cos\theta }{ 2 } } =\sqrt { \frac { 1-\frac { 4\sqrt { 39 } }{ 25 } }{ 2 } =\sqrt { \frac { 25-4\sqrt { 39 } }{ 50 } } } \)
Consider \(sin\left( \frac { \pi }{ 4 } -\frac { \theta }{ 2 } \right) =sin\frac { \pi }{ 4 } cos\frac { \theta }{ 2 } -cos\frac { \pi }{ 4 } sin\frac { \theta }{ 2 } \)
\(sin\left( \frac { \pi }{ 4 } -\frac { \theta }{ 2 } \right) =\frac { 1 }{ \sqrt { 2 } } cos\frac { \theta }{ 2 } -\frac { 1 }{ \sqrt { 2 } } sin\frac { \theta }{ 2 } \)
\(\frac { 1 }{ \sqrt { 2 } } \left( cos\frac { \theta }{ 2 } -sin\frac { \theta }{ 2 } \right) =\frac { 1 }{ \sqrt { 2 } } \left[ \sqrt { \frac { 25+4\sqrt { 39 } }{ 50 } } -\sqrt { \frac { 25-4\sqrt { 39 } }{ 50 } } \right] \)
\(=\frac { 1 }{ 5\sqrt { 2 } } \left[ \sqrt { \frac { 25+4\sqrt { 39 } }{ 50 } } -\sqrt { \frac { 25-4\sqrt { 39 } }{ 50 } } \right] \)

Since \(\alpha \in \left( \pi ,\frac { 3\pi }{ 2 } \right) \), \(\alpha \) lies in III quadrant  
So, only tan \(\alpha \) and cot \(\alpha \) are positive
Also \(\beta \in \left( \frac { \pi }{ 2 } ,\pi \right) \) , β lies in II quadrant
So only sin β and cosec β are positive.
Given cot \(\alpha \) =\(\frac{1}{2}\)
\(\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 5 } \)
sec β = \(\frac{-5}{3}\)
\(\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } =4\)
 
ஃ tan \(\alpha \) = 2
tan β =\(\frac{-4}{3}\)
Now, \(tan\left( \alpha +\beta \right) =\frac { tan\alpha +tan\beta }{ 1-tan\alpha tan\beta } =\frac { 2-\frac { 4 }{ 3 } }{ 1-\left( 2 \right) \left( \frac { 4 }{ 3 } \right) } \)
\(=\frac { \frac { 6-4 }{ 3 } }{ 1+\frac { 8 }{ 3 } } =\frac { \frac { 2 }{ 3 } }{ \frac { 3-8 }{ 3 } } =\frac { 2 }{ 3 } \times \frac { 3 }{ 11 } =\frac { 2 }{ 11 } \)

f +g : R ⟶ R and f - g : R ⟶ R
Now (f + g) (x) = f(x) + g(x)
⇒ ( f+ g)(x) = x + |x|
\(⇒\ \left( f+g \right) x =\begin{cases}x +x\quad if\quad x\ge 0 \\ x-x \ if\quad x<0\quad \end{cases}\)
\(⇒\ \left( f+g \right) x =\begin{cases}2x\quad if\quad x\ge 0 \\ 0 \ if\quad x<0\quad \end{cases}\)
Similarly (f - g) (x) = f(x) - g(x) = x - |x|
\(⇒\ \left( f-g \right) x =\begin{cases}x -x\quad if\quad x\ge 0 \\ x-(-x )\ if\quad x<0\quad \end{cases}\)
\(⇒\ ​​\left( f-g \right) \left( x \right) =\begin{cases} 0 \quad if\quad x\ge 0 \\ 2x \ if\quad x<0\quad \end{cases}\)

\(LHS=sin(B+C-A)+sin(C+A-B)+sin(A+B-C)=4sinAsinBsinC\)
\(=sin(180-A-A)+sin(180-B-B)+sin(180-C-C)\)
\(=sin(180-2A)+sin(180-2B)+sin(180-2C)=sin\quad 2A+sin2B+sin2C\)
\(=2sin\left( \frac { 2A+2B }{ 2 } \right) cos\left( \frac { 2A-2B }{ 2 } \right) +2sinC\quad cosC\)
\(=2sin(A+B)cos(A-B)+2sinC\quad cosC=2sinC[cos(A-B)+cosC]\)
\(=2sinC[cos(A-B)+cos(180-(A+B))]\)
\(=2sinC[cos(A-B)-cos(A+B)]=2sinC2sinAsinB\)
\(=4\quad SinASinBsinC=RHS\)

Given log₂ x-3log_1/6 = 6 [using quotient rule]
\(⇒\ {1\over log_x^2}-{3\over log_x^{1\over 2}}=6\)
\(⇒\ {1\over log_x^2}-{3\over log_x^1-log_x^2}=6\) [using quotient rule]
\(⇒\ {1\over log_x^2}-{3\over 0-log_x^2}=6\)
\(⇒\ {1\over log_x^2}+{3\over log_x^2}=6\)
\(⇒\ {1\over log_x^2}(1+3)=6\)
\(⇒\ {1\over log_x^2}(4)=6\)
\(⇒\ {1\over log_x^2}={6\over 4}={3\over 2}\)
\(⇒\ log_2^x={3\over2}\)
\(⇒\ 2^{3\over2}=x\)
⇒ (2³)^1/2 = x ⇒ (8)^1/2 = x
\(⇒\ x=\sqrt8=\sqrt{4\times2}\)
⇒  \(x=x\sqrt { 2 } \)

If 3x + 5y = 45

All points bounded above x = 0, y = 0 and 3x + 5y = 45 is required region. Darkly shaded area will represents the solution set of the given linear inequalities.

LHS = sin A + sin B + sin C
= \(2sin\left( \frac { A+B }{ 2 } \right) .cos\left( \frac { A-B }{ 2 } \right) +2sin\frac { C }{ 2 } cos\frac { C }{ 2 } \)
= \(2sin\left( 90-\frac { C }{ 2 } \right) cos\left( \frac { A-B }{ 2 } \right) +2sin\frac { C }{ 2 } cos\frac { C }{ 2 } \)
= \(2cos\frac { C }{ 2 } cos\left( \frac { A-B }{ 2 } \right) +2sin\frac { C }{ 2 } cos\frac { C }{ 2 } \)
= \(2cos\frac { C }{ 2 } \left[ cos\left( \frac { A-B }{ 2 } \right) +sin\frac { C }{ 2 } \right] \)
= \(2cos\frac { C }{ 2 } \left[ cos\left( \frac { A-B }{ 2 } \right) +sin\left( 90-\left( \frac { A+B }{ 2 } \right) \right) \right] \)
= \(2cos\frac { C }{ 2 } \left[ cos\left( \frac { A-B }{ 2 } \right) +cos\left( \frac { A+B }{ 2 } \right) \right] \)
= \(2cos\frac { C }{ 2 } .cos\frac { A }{ 2 } cos\frac { B }{ 2 } \)
= \(4cos\frac { A }{ 2 } cos\frac { B }{ 2 } cos\frac { C }{ 2 } \) = RHS
Hence proved.

f(-4) = +4 + 4    [\(\therefore\) f(x) = -x + 4 when x = -4]
=8
f(1) = 1-1² [\(\therefore\) f(x) = x-x² when x = 1]
f(1) = 0
f(-2) = (-2)²-(-2) [\(\therefore\) f(x) = x²-x when x = -2]
= 4+2 = 6
f(7) = 0   [\(\therefore\) f(x) = 0 when x = 7]
f(0 = 0²-0  [\(\therefore\) f(x) = x² - x when x = 0]
=0
\(\therefore\) f(-4) = 8, f(1) = 0, f(-2) = 6, f(7) = 0 and f(0) = 1

\(LHS=sin\frac { \theta }{ 2 } sin\frac { 7\theta }{ 2 } +sin\frac { 3\theta }{ 2 } sin\frac { 11\theta }{ 2 } \)
\(=\frac { 1 }{ 2 } \left[ cos\left( \frac { \theta }{ 2 } -\frac { 7\theta }{ 2 } \right) -cos\left( \frac { \theta }{ 2 } -\frac { 7\theta }{ 2 } \right) \right] +\frac { 1 }{ 2 } \left[ cos\left( \frac { 3\theta }{ 2 } -\frac { 11\theta }{ 2 } \right) -cos\left( \frac { 3\theta }{ 2 } -\frac { 11\theta }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ cos(-3\theta )-cos(4\theta )+cos(-4\theta )-cos7\theta \right] \)
\(=\frac { 1 }{ 2 } \left[ cos3\theta -cos4\theta +cos4\theta -cos7\theta \right] \quad \quad \left[ \because cos(-\theta )=cos\theta \right] \)
\(=\frac { 1 }{ 2 } \left[ cos3\theta -cos7\theta \right] =\frac { 1 }{ 2 } \left[ 2sin\left( \frac { 3\theta +7\theta }{ 2 } \right) .sin\left( \frac { 7\theta -3\theta }{ 2 } \right) \right] \)
\(=sin5\theta .sin2\theta =RHS.\)

(i) \(y=-x^{ \left( \frac { 1 }{ 3 } \right) }\)

\(Let \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\)
\(Then \ y=-x^{ ^{ \frac { 1 }{ 3 } } }\) is the reflection of the graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)  about the x-axis.
(ii) \(y=x^{1\over 3}+1\)

Let \(y=x^{ ^{ \frac { 1 }{ 3 } } }\)
Then \(y=x^{ ^{ \frac { 1 }{ 3 } } }+1\) is the x graph of \(y=x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the upward for one unit
(iii) \(y=x^{1\over 3}-1\)

Let \(y=x^{1\over 3}\)
Then \(y=x^{1\over 3}\)-1 is the graph of \(x^{ ^{ \frac { 1 }{ 3 } } }\) shifts to the downward for one unit.
(iv)  \(y=(x+1)^{1\over 3}\)

\(y=(x+1)^{1\over 3}\) causes the graph of \({x}^{\frac{1}{3}}\), shifts to the left for one unit.

Let A, B be the land marks and C be the position of the plane,
Given ㄥACB 45°

Using cosine formula,
c² = a² + b² - 2ab cos C
c² = 2² + 1² - 2(2)(1) cos 45⁰
\(=4+1-4\left(1\over \sqrt3\right)\)

c² = 5 - 2√2
\(c^2=\sqrt{5-2\sqrt2}km\)

Let ABC be the triangles with position vectors \(\vec { OA } ,\vec { OB } \) and \(\vec { OC } \) 
(i.e.,) \(\vec { \quad OA } =\left( 4\hat { i } +5\hat { j } +6\hat { k } \right) \) ,\(\vec { OB } =5\hat { i } +6\hat { j } +4\hat { k } \) and \(\vec { OC } =6\hat { i } +4\hat { j } +5\hat { k } \) 
\(\vec { AB } =\vec { OB } -\vec { OA } =\left( 5\hat { i } +6\hat { j } +4\hat { k } \right) -\left( 4\hat { i } +5\hat { j } +6\hat { k } \right) \) 
= \(\hat { i } +\hat { j } -2\hat { k } \) 
\(\left| \vec { AB } \right| =\sqrt { 1+1+4 } =\sqrt { 6 } \) 
\(\vec { BC } =\vec { OC } -\vec { OB } =\left( 6\hat { i } +4\hat { j } +5\hat { k } \right) -\left( 5\hat { i } +6\hat { j } +4\hat { k } \right) \) 
= \(\hat { i } -2\hat { j } +\hat { k } \) 
\(\left| \vec { BC } \right| =\sqrt { 1+4+1 } =\sqrt { 6 } units\) 
\(\vec { AC } =\vec { OC } -\vec { OA } =\left( 6\hat { i } +4\hat { j } +5\hat { k } \right) -\left( 4\hat { i } +5\hat { j } +6\hat { k } \right) \) 
\(\\ 2\hat { i } -\hat { j } -\hat { k } \) 
\(\left| \vec { AC } \right| =\sqrt { 4+1+1 } =\sqrt { 6 } units\) 
Now,\(\left| \vec { AB } \right| =\left| \vec { BC } \right| =\left| \vec { AC } \right| =\sqrt { 6 } \)
⇒ ΔABC is an equilateral triangle 

\(LHS=\left| \begin{matrix} -{ a }^{ 2 } & ab & ac \\ ab & -{ b }^{ 2 } & bc \\ ac & bc & -{ c }^{ 2 } \end{matrix} \right| ={ 4a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }\) 
Taking a from R₁ b from R₂ and c from R₃ as common factors we get,
= \(\left( abc \right) \left| \begin{matrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{matrix} \right| \) 
Again taking a from C₁' b from C₂ and e from C₃ we get,
= \(\left( abc \right) \left( abc \right) \left| \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right| \) 
=  \(\left( abc \right) ^{ 2 }\left| \begin{matrix} 0 & 0 & 2 \\ 2 & 0 & 0 \\ 1 & 1 & -1 \end{matrix} \right| \begin{matrix} { R }_{ 1 }\rightarrow { { R }_{ 1 }+{ R }_{ 2 } } \\ { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 } \end{matrix}\) 
= \(\left( abc \right) { c }^{ 2 }\left[ 0+0+2\left| \begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix} \right| \right] \) 
= (abc)² (4) = 4a²b²c² = RHS

\(tan\quad \frac { A }{ 2 } =\sqrt { \frac { (s-b)(s-c) }{ s(s-a) } } ...(1)\)
\(tan\quad \frac { C }{ 2 } =\sqrt { \frac { (s-a)(s-b) }{ s(s-c) } } ...(2)\)

\(\\ \therefore \frac { s-b }{ s } =\frac { 1 }{ 3 } \)
3s - 3b = s
2s = 3b

a + b + c = 3b
a + c = -2b
∴ a, b, c are in A.P

= \(\int { \frac { { sin }^{ 6 }x+cos^{ 6 }x }{ sin^{ 2 }xcos^{ 2 }x } } \)
= \(\int { \frac { \left( { sin }^{ 2 }x \right) ^{ 3 }+\left( { cos }^{ 2 }x \right) ^{ 3 } }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } \) dx\(\left[ { a }^{ 3 }+{ b }^{ 3 }=\left( a+b \right) ^{ 3 }-3ab\left( a+b \right) \right] \)
= \(\int { \frac { \left( { sin }^{ 2 }x \right) ^{ 3 }+\left( { cos }^{ 2 }x \right) ^{ 3 }-{ 3sin }^{ 2 }x{ cos }^{ 2 }x\left( { sin }^{ 2 }x+{ cos }^{ 2 }x \right) }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } \)
= \(\int { \frac { { 1 }^{ 3 }-3{ sin }^{ 2 }x{ cos }^{ 2 }x(1) }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } \) dx = \(\int { \frac { 1-3{ sin }^{ 2 }x{ cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } \)dx
= \(\int { \left( \frac { 1 }{ { sin }^{ 2 }x{ cos }^{ 2 }x } -3 \right) } \)dx = \(\int { \frac { { sin }^{ 2 }x{ +cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } -3\)dx
= \(\int { \frac { { sin }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx+\int { \frac { { cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx-\int { 3 } \)dx
= \(\int { sec^{ 2 } } \) x dx + \(\int { cosec } ^{ 2 }xdx\) - 3 \(\int { dx } \) = tan x - cot x - 3x + c

Let A be the event that the construction job will be complete on time,
E₁: There will be a strike and
E₂: There will be no strike.
Given P(E₁) =0.65, P(E₂) = 1-P(E₁) = 1-0.65 = 0.35, P(A/E₁) = 0.32, P(AE₂) = 0.80
By the total probability theorem, we get
P(A) = P(E₁). P(A/E₁) + P(E₂).P(A/E₂₎
= 0.65 \(\times\) + 0.35 \(\times\) 0.80 = 0.208 + 0.28
P(A) = 0.488

Given f(2)=2
\(\lim _{ x\rightarrow { 2 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 2 }^{ + } }{ { x }^{ 2 } } =2(2)=4\)
\(\lim _{ x\rightarrow { 2 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 2 }^{ - } }{ 2x } =2^{ 2 }=4\)
\(\therefore \lim _{ x\rightarrow { 2 } }{ f\left( x \right) } =4\neq f\left( 2 \right)\)
\(\therefore \quad f\quad has\quad a\quad removable\quad discontinuity\quad at\quad x=2.\)

\(Put\quad x-\pi =\theta \Rightarrow x\quad =\pi +\theta \)
\(Also\quad \theta =x-\pi \rightarrow 0\quad as\quad x\rightarrow \pi \)
\(\lim _{ x\rightarrow \pi }{ \frac { \sin { x } }{ x-\pi } } =\quad \lim _{ \theta \rightarrow 0 }{ \frac { \sin { (\pi +\theta ) } }{ \theta } } = \lim _{ \theta \rightarrow 0 }{ \frac { -\sin { \theta } }{ \theta } } \)
\(=(-1).\lim _{ \theta \rightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } =(-1)(1)=-1\)

\(x=a\sec ^{ 3 }{ \theta } \)
\(\Rightarrow \frac { dx }{ d\theta } =3a\sec ^{ 2 }{ \theta } .\frac { d }{ d\theta } (\sec { \theta } )=3a\sec ^{ 2 }{ \theta } .\sec { \theta } \tan { \theta } =3a\sec ^{ 3 }{ \theta } \tan { \theta } \)
\(y=a\tan ^{ 3 }{ \theta } \)
\(\Rightarrow \frac { dy }{ d\theta } =a.3\tan ^{ 2 }{ \theta } .\frac { d }{ d\theta } (\tan { \theta } )=3a\tan ^{ 2 }{ \theta } .\sec ^{ 2 }{ \theta } \)
\(\frac { dy }{ dx } =\frac { dy }{ d\theta } /\frac { dx }{ d\theta } =\frac { 3a\tan ^{ 2 }{ \theta } \sec ^{ 2 }{ \theta } }{ 3a\sec ^{ 3 }{ \theta } \tan { \theta } } =\frac { \tan { \theta } }{ \sec { \theta } } =\frac { \sin { \theta } }{ \cos { \theta \times \frac { 1 }{ \cos { \theta } } } } =\sin { \theta } \)
\(\therefore \frac { dy }{ dx } \quad at\quad \theta =\frac { \pi }{ 3 } =\sin { \frac { \pi }{ 3 } } =\frac { \sqrt { 3 } }{ 2 } \)

Let S be the sample space and A be the event of taking 2 hundred rupee note.
THerefore, n(S) = 12c₂ = 66, n(A) = 4c₂ = 6 and \(\left( \overline { A } \right) \) = 66 - 6 = 60
Therefore, odds in favour of A is 6 : 60
That is, odds in favour of A is 1 : 10, and P(A) = \(\frac { 1 }{ 11 } \)

Ten coins are tossed simultaneously one time = one coin is tossed 10 times
Let S the sample space,
that is,

Let A be the event of getting exactly two heads,
B be the event of getting at most two heads, and
C be the event of getting at least two heads.
When ten coins are tossed, the number of elements in sample space is 2ⁿ = 2¹⁰ = 1024
n(S) = 1024
n(A) = ¹⁰C₂ = 45
n(B) = ¹⁰C₁₀ + ¹⁰C₁ + ¹⁰C₂ = 1 + 10 + 45 = 56
n(C) = ¹⁰C₂ + ¹⁰C₃ + ¹⁰C₄ + .... + ¹⁰C₁₀
= n(S) - (¹⁰C₀ + ¹⁰C₁) = 1024 - 11 = 1013
The required probabilities are
(i) P(A) = \(\frac { n(A) }{ n(S) } =\frac { 45 }{ 1024 } \)
(ii) P(B) = \(\frac { n(B) }{ n(S) } =\frac { 56 }{ 1024 } =\frac { 7 }{ 8 } \)
(iii) P(C) = \(\frac { n(C) }{ n(S) } =\frac { 1013 }{ 1024 } \)

\(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x=\int \sin ^{-1} x \times \frac{x}{\sqrt{1-x^2}} d x \)
\(Let\ u=\sin ^{-1} x\)
Differentiating on both sides
\(\frac{d u}{d x}=\frac{1}{\sqrt{1-x^2}}\)
\(d u=\frac{1}{\sqrt{1-x^2}} d x \)
\(d v=\frac{x}{\sqrt{1-x^2}} \)
\(v=\int \frac{x}{\sqrt{1-x^2}} d x\)
Take \(t=\sqrt{1-x^2}\)
\( \mathrm{t}^2 =1-x^2\)
\(2 \mathrm{td} t =-2 x \mathrm{~d} x \)
\(\mathrm{tdt} =-x \mathrm{~d} x \)

\(x \mathrm{~d} x =-\mathrm{t} \mathrm{dt}\)
\( v =\int \frac{1-1 d }{1} \)
\(=-\int d=-1 \)
\(v =-\sqrt{1-x^2} \)
\(\therefore \int \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} d x =\sin ^{-1} x\left(-\sqrt{1-x^2}\right)- \int\left(-\sqrt{1-x^2}\right) \times \frac{1}{\sqrt{1-x^2}} d x \ {\left[\because \int u d v=u v-\int v d u\right] } \)
\(=-\sqrt{1-x^2} \sin ^{-1} x+\int d x\)
\( =-\sqrt{1-x^2} \sin ^{-1} x+x+c\)

\(\int{1\over (5-4x)}dx=-{1\over4}log|(5-4x)|+c\)

We differentiate both sides of the equation,
\({d\over dx}(x^2)+{d\over dx}(y^2)={d\over dx}(1)\)
\(2x+2y{dy\over dx}=0\)
Solving for the derivative yields
\({dy\over dx}=-{x\over y}\)

\(y=5^{\frac{-1}{x}}\)
\(u =\frac{-1}{x}\)
\(\frac{d u}{d x} =1 / x^2\)
\(y =5^u\)
\(\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}\)
\(=5^u \log 5\left(\frac{1}{x^2}\right)=\frac{5^{-1 / x} \log 5}{x^2} \) \({\left[\because d\left(a^x\right)=a^x \log a\right] }\)

y = x sin x cos x
\(\frac{d y}{d x}=\frac{d}{d x}(x) \sin x \cos x+x \frac{a}{d x}(\sin x) \cos x +x \sin x \frac{d}{d x}(\cos x)\) \([\because d(u v w)=u v(d w)+u(d v) w+(d u) v w]\)
\(=(1) \sin x \cos x+x \cos x \cos x+x \sin x(-\sin x)\)
\(=\sin x \cos x+x \cos ^2 x-x \sin ^2 x\)
\(=\sin x \cos x+x\left(\cos ^2 x-\sin ^2 x\right)\)
\(=\sin x \cos x+x \cos 2 x \ \left[\because \cos ^2 A-\sin ^2 A=\cos 2 A\right] \)

\(y={tan \ x \over x}\)
\(\frac{d y}{d x}=\frac{x \frac{d}{d x}(\tan x)-\tan x \frac{d}{d x}(x)}{x^2}\)
\(=\frac{x\left(\sec ^2 x\right)-\tan x(1)}{x^2}\left[\because d\left(\frac{u}{v}\right)=\frac{v d u-u d v}{v^2}\right]\)
\(=\frac{x \sec ^2 x-\tan x}{x^2}\)

\(y={cos \ x \over x^3}\)
\({dy\over dx}={x^3(-sin \ x)-cos \ x(3x^2)\over x^6}={-x^2(x \ sin \ x+3cos \ x)\over x^6}=-{(x \ sin \ x+3cos \ x)\over x^4}\)

\(f^{\prime}\left(2^{-}\right)=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{(-x+2)-0}{x-2}\)
\(=\lim _{x \rightarrow 2^{-}} \frac{-(x-2)}{(x-2)}=-1\)
\(f^{\prime}\left(2^{+}\right)=\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2 x-4-0}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2(x-2)}{(x-2)}=2\)
\(\text {by (1) & (2), } f^{\prime}\left(2^{-}\right) \neq f^{\prime}\left(2^{+}\right)\)
\(\therefore\) It is not differentiable.

\(f(x)=\sqrt{1-x^2}\)
\(f^{\prime}\left(1^{-}\right)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x^2}-0}{x-1} \quad \begin{aligned} f(x) &=\sqrt{1-x^2} \\ f(1) &=\sqrt{1-1} \\ &=0 \end{aligned}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x^2}}{x-1}=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{(1-x)(1+x)}}{x-1}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x} \sqrt{1+x}}{x-1}\)
\(\therefore f^{\prime}(x)=\lim _{x \rightarrow 1^{-}} \frac{+\sqrt{1-x} \sqrt{1+x}}{-(1-x)}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1-x} \sqrt{1+x}}{-\sqrt{1-x} \sqrt{1-x}}\)
\(=\lim _{x \rightarrow 1^{-}} \frac{\sqrt{1+x}}{-\sqrt{1-x}} \rightarrow-\infty\)
\(\therefore f^{\prime}(x) \rightarrow-\infty^{-} \text {as } x \rightarrow 1^{-}\)
It is not diferentiable.

Given \(g(x)= \begin{cases}x^{2}-b^{2} & \text { if } x<4 \\ b x+20 & \text { if } x \geq 4\end{cases}\)
\(lim_{x\rightarrow 4^-}g(x)=lim_{x\rightarrow 4^-}x^2-b^2=(4)^2-b^2=16-b^2\)..(1)
\(lim_{x\rightarrow 4^+}g(x)=lim_{x\rightarrow 4^+}bx+20=b(4)+20=4b+20\)...(2)
Also, g(4) = bc + 20 = 4b + 20..(3)
Since g(x)is continuous on \((-\infty,\infty)\)
\(\Rightarrow lim_{x\rightarrow 4^-}g(x)=lim_{x\rightarrow 4^+}g(x)=g(4)\)
From (1), (2) and (3) we get
16-b² = 4b + 20
\(\Rightarrow b^2+4b+20-16=0\)
\(\Rightarrow b^2+4b+4=0\)
\(\Rightarrow (b+2)^2=0\)
\(\Rightarrow (b+2)=0\)
\(\Rightarrow b=-2\)