\(f^{ ' }\left( 1^{ - } \right) =\lim _{ x\rightarrow 1^{ - } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ \frac { { x }^{ 2 }-1 }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ \frac { (x+1)(x-1) }{ x-1 } } =\lim _{ x\rightarrow 1^{ - } }{ (x+1) } =1+1=2\quad ...(1)\)
\(f^{ ' }\left( 1^{ + } \right) =\lim _{ x\rightarrow 1^{ + } }{ \frac { f\left( x \right) -f\left( 1 \right) }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ \frac { { x }^{ 2 }-1 }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ \frac { (x+1)(x-1) }{ x-1 } } =\lim _{ x\rightarrow 1^{ + } }{ x+1 } =1+1=2 ...(2)\)
From (1)and (2), \(f^{ ' }\left( 1^{ - } \right) =f^{ ' }\left( 1^{ + } \right) \)
\(\therefore f\left( x \right) \)is differentiable at x = 1 and \(f^{ ' }\left( 1 \right) =2\)

\(Given A=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\)
\(A^2=A \times A=\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\)
\(A^2=\left[\begin{array}{lll} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]\)
\(A^3=A^2 \times A=\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\)
\(=\left[\begin{array}{lll} 5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39 \end{array}\right]\)
\(=\left[\begin{array}{lll} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]\)
\(A^3-6 A^2+7 A+k I=0\)
\(\left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-6\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]+7\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]+k\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=0\)
\(\left[\begin{array}{ccc} 21-30+7+k & 0+0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+k & 23-30+7+0 \\ 34-48+14+0 & 0+0+0+0 & 55-78+21+k \end{array}\right]=0\)
\(\left[\begin{array}{ccc} -2+k & 0 & 0 \\ 0 & -2+k & 0 \\ 0 & 0 & -2+k \end{array}\right]=0\)
\(-2+k=0\)
\(\therefore k=2\)
Alternative Method:
W.K.T, Product of roots \(=|A|\) ...........(1)
\(\text { Given } A^3-6 A^2+7 A+k I=0\)
\(\text {Product }=\frac{-c}{a}=\frac{-k}{1}\)
\((1) \Rightarrow \frac{-k}{1}=\left|\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right|\)
\(-k=1(6-0)-0+2(0-4)\)
\(-k=6-8\)
-k = -2
\(\therefore\) k = 2

\(cot(\frac{-15\pi}{4})=-cot(\frac{15\pi}{4})=-cot(4\pi-\frac{\pi}{4})=cot\frac{\pi}{4}=1\)

Equation of a line parallel to 3x-4y-5=0 is (1) 3x-4y+k=0
putting x = -1in 3x-4y-5=0,we get -3-4y-5=0, we get -3-4y-5=0\(\Rightarrow \)-4y=8\(\Rightarrow \)y=-2
\(\therefore \)(-1-2) is a point on 3x-4y-5=0
Given distance between equation (1)and(2)is 1.
\(\Rightarrow \)length of perpendicular from (-1-2)to 3x-4y+k is one unit
\(\therefore \)\(\left| \frac { 3\left( -1 \right) -4\left( -2 \right) +k }{ \sqrt { { 3 }^{ 2 }+{ \left( -4 \right) }^{ 2 } } } \right|=1\)
\(\Rightarrow \left| \frac { 3+8+K }{ 5 } \right| =1\Rightarrow \left| \frac { 5+K }{ 5 } \right| =1\)
\(\Rightarrow \left| 5+K \right| =5\Rightarrow 5+K=\pm 5\)
If 5 + K = 5 then K = 0
If 5 + K = -5 then K = -10
3x - 4y = 0 and 3x - 4y - 10 as the equations of the required lines.

Here T₂-T₁ = \(log\frac { { a }^{ 2 } }{ b } -log\quad a=log\frac { { a }^{ 2 }/b }{ a } \)
= \(log\frac { a }{ b } \)
T₃-T₂ = \(log\frac { a^{ 3 } }{ b^{ 2 } } -log\frac { a^{ 2 } }{ b^{ 2 } } =log\frac { a^{ 3 } }{ b^{ 2 } } +log\frac { a^{ 2 } }{ b^{ 2 } } \)
= \(log\frac { a^{ 3 } }{ b^{ 2 } } \times \frac { b }{ { a }^{ 2 } } =log\frac { a }{ b } \)
∴ T₂-T₁ = T₃-T₂ = \(log\left( \frac { a }{ b } \right) \)
∴ The given sequences is an A.P

(i) With repetition

The given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
The unit place can be filled in only one way using 3.
Since repetition is allowed, the tens place can be filled in 10 ways using any one of the digits from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
The hundreds place can be filled in 9 ways using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (excluding 0)
∴ By fundamental principle of multiplication, total number of 3 digit numbers = 9 \(\times\) 10 \(\times\) 1 = 90.
(ii) Repetition of digits is not allowed

The unit place can be filled in 1 way
Since repetition of digits is not allowed, the tens place can be filled in 8 ways.
Hundreds place can be filled in 8 ways.
∴ Total number of 3-digit numbers without repetition = 1 \(\times\) 8 \(\times\) 8 = 64

\({x^2+1\over x(x+1)^2}={A\over x}+{B\over x+1}+{c\over x(x+1)^2}\)  [Since the denominator is linear factor of repeated factors]
\(⇒\ {x^2+1\over x(x+1)^2}={A(x+)^2+Bx(x+1)+Cx\over x(x+1)^2}\)
⇒ x²+ 1 = A(x + 1)² + Bx(x + 1) + Cx
Putting x = 0, we get
1 = A ⇒ = 1
Putting x = -1, we get
1 + 1 = C(-1) ⇒ 2 = -C ⇒ C = -2
Putting x = 1 in (1) we get
2 = A(4) + B(2) + C
⇒ 24A + 2B + C   [∴ A = 1, C = -2]
⇒ 2 = 4 + 2B - 2
⇒ 2 + 2 - 4 = 2B ⇒ B =0
\(∴\ {x^2+1\over x(x+1)^2}={1\over x}+{0\over x+1}-{2\over (x+1)^2}\)
\(\frac { { x }^{ 2 }+1 }{ x\left( x+1 \right) ^{ 2 } } =\frac { 1 }{ x } \frac { 2 }{ \left( x+1 \right) ^{ 2 } } \)

Given cost of one meal = Rs. 100
Number of customers is given by the function D(x) = 200 - x
∴ Day cost function = (200 - x)(100)
= 20000 - 100x

Given \(\overrightarrow { a } \) is perpendicular to both \(\overrightarrow { b } \)  and \(\overrightarrow { c } \)
\(\overrightarrow { a } =\lambda (\overrightarrow { b } \times \overrightarrow { c } )\) for some scalar \(\lambda \)
\(\Rightarrow \quad 1={ |\overrightarrow { a } | }^{ 2 }={ \lambda }^{ 2 }|\overrightarrow { b } \times \overrightarrow { c } |^{ 2 }\quad \Rightarrow 1={ \lambda }^{ 2 }[|{ \overrightarrow { b } | }^{ 2 }{ |\overrightarrow { c } | }^{ 2 }-(\overrightarrow { b } .\overrightarrow { c } )^{ 2 }]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[1(1)-{ |\overrightarrow { b } | }^{ 2 }{ |\overrightarrow { c } | }^{ 2 }-{ cos }^{ 2 }\left( \frac { \pi }{ 6 } \right) ]\quad \Rightarrow 1={ \lambda }^{ 2 }[1-{ cos }^{ 2 }\left( \frac { \pi }{ 6 } \right) ]\)
\(\Rightarrow 1={ \lambda }^{ 2 }[1-{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }]\ \Rightarrow 1={ \lambda }^{ 2 }[1-\frac { 3 }{ 4 } ]\)
\(\Rightarrow 1={ \lambda }^{ 2 }\left( \frac { 1 }{ 4 } \right) \)
\({ \lambda }^{ 2 }=4\)
\(\lambda =\pm 2\)
\(\therefore \overrightarrow { a } =\pm 2(\overrightarrow { b } \times \overrightarrow { c } )\)

Let I = \(\int { \frac { { x }^{ 2 }{ tan }^{ -1 }\left( { x }^{ 3 } \right) }{ 1+{ x }^{ 6 } } } dx=\int { \frac { { x }^{ 2 }{ tan }^{ -1 }\left( { x }^{ 3 } \right) }{ 1+\left( { x }^{ 3 } \right) ^{ 2 } } } \)dx
put x³ = t \(\Rightarrow\) 3x² dx = dt \(\Rightarrow\) x² dx = \(\frac { dt }{ 3 } \)
I = \(\int { \frac { tan^{ -1 }(t) }{ 1+{ t }^{ 2 } } } .\frac { dt }{ 3 } =\frac { 1 }{ 3 } \int { \frac { tan^{ -1 }(t) }{ 1+{ t }^{ 2 } } } dt\)
Let tan^-1 (t) = u \(\Rightarrow\) \(\frac { dt }{ 1+{ t }^{ 2 } } \) = du
I = \(\frac { 1 }{ 3 } \int { udu } =\frac { 1 }{ 3 } \left( \frac { { u }^{ 2 } }{ 2 } \right) +c=\frac { { u }^{ 2 } }{ 6 } +c=\frac { 1 }{ 6 } \left[ { tan }^{ -1 }\left( t \right) \right] ^{ 2 }+c\) [\(\therefore\)u = tan^-1 (t)]
= \(\frac { 1 }{ 6 } \left[ { tan }^{ -1 }\left( x^{ 3 } \right) \right] ^{ 2 }+c\)  [\(\because\) t = x³]

Out of integers from 1 to 11, there are 5 even integers and 6 odd integers.
Let A; Both the numbers chosen are odd
B: Sum of he numbers chosen is even 
[Number of ways of selecting odd nos = ⁶C₂ No. of ways of getting sum as an even number = ⁵C₂ + ⁶C₂]
\(\therefore P(A/B)=\frac { p(A\cap B) }{ P(B) } =\frac { ^{ 6 }{ { C }_{ 2 } } }{ ^{ 5 }{ { C }_{ 2 }+ }^{ 6 }{ { C }_{ 2 } } } =\frac { \frac { 6\times 5 }{ 2\times 1 } }{ \frac { 5\times 4 }{ 2\times 1 } +\frac { 6\times 5 }{ 2\times 1 } } =\frac { 15 }{ 10+15 } =\frac { 15 }{ 25 } =\frac { 3 }{ 5 } \)
\(\therefore P(A/B)=\frac { 3 }{ 5 } \)

\(Let\quad u={ x }^{ x }\quad and\quad v=x\log { x }\)
\( u={ e }^{ \log _{ x }{ x } }={ e }^{ x\log _{ x }{ x } }\quad and\quad v=x\log { x } \)
\(\Rightarrow \frac { dy }{ dx } ={ e }^{ x\log { x } }\times \frac { d }{ dx } (x\log { x } )\quad and\)
\( \Rightarrow \frac { du }{ dx } ={ e }^{ x\log { x } }\left( x\frac { 1 }{ x } +\log { x } \right) \)
\(\Rightarrow \frac { dy }{ dx } ={ x }^{ x }(1+\log { x } )\quad ...(1)\\ v=x\log { x } \)
\(\frac { dv }{ dx } ={ x }.\frac { 1 }{ x } +\log { (x) } =1+\log { x } ..(2)\)
\(From\quad (1)\quad and\quad (2),\quad \frac { du }{ dv } =\frac { du }{ dx } /\frac { dv }{ dx } \)
\(\Rightarrow \frac { du }{ dv } =\frac { { x }^{ x }(1+\log { x } ) }{ 1+\log { x } } ={ x }^{ x }\)

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 = \(\frac { n(A) }{ n(S) } =\frac { 5 }{ 27 } \)

\(\int { \frac { 1 }{ 3 } cos } \left( \frac { x }{ 3 } -4 \right) +\frac { 7 }{ 7x+9 } +{ e }^{ \frac { x }{ 5 } +3 }dx\)
= \(\frac { 1 }{ 3 } \int { cos } \left( \frac { x }{ 3 } -4 \right) dx+7\quad \int { \frac { 1 }{ 7x+9 } dx } +\int { { e }^{ \frac { x }{ 5 } +3 }dx } \)
= \(\frac { 1 }{ 3 } \frac { sin\left( \frac { x }{ 3 } -4 \right) }{ \frac { 1 }{ 3 } } +7.\frac { log\left| 7x+9 \right| }{ 7 } +\frac { { e }^{ \frac { x }{ 5 } +3 } }{ \frac { 1 }{ 5 } } +c\)
= \(sin({x\over 3}-4)+log|7x+9|+5e^{{x\over 5}+3}+c\)

\( = \int\left[\frac{8}{\sqrt{1-(4 x)^2}}+\frac{27}{\sqrt{1-9 x^2}}-\frac{15}{1+25 x^2}\right] d x \)
\(= 8 \int \frac{1}{\sqrt{1-(4 x)^2}} d x+27 \int \frac{1}{\sqrt{1-9 x^2}} d x-15 \int \frac{1}{1+25 x^2} d x \)
\(= 8 \int \frac{1}{\sqrt{1-(4 x)^2}} d x+27 \int \frac{1}{\sqrt{1-(3 x)^2}} d x-15 \int \frac{1}{1+(5 x)^2} d x \)
\(= 8\left(\frac{\sin ^{-1}(4 x)}{4}\right)+27\left(\frac{\sin ^{-1}(3 x)}{3}\right)-15\left(\frac{\tan ^{-1}(5 x)}{5}\right)+c\)
\(= 2 \sin ^{-1}(4 x)+9 \sin -1(3 x)-3 \tan ^{-1}(5 x)+c\)

Given \(f(x)={x^2-1\over x^3-1}\)
The function f(x) is not defined at x = 1.
\(\therefore lim_{x\rightarrow 1}f(x)=lim_{x\rightarrow 1}{x^2-1\over x^3-1}\)

\( lim_{x\rightarrow 1}{(x+1)\over(x^2+x+1)}={1+1\over1+1+1}={2\over3}\)
To make f(x) continuous at x = 1,
\( lim_{x\rightarrow 1}f(x)=f(1)={2\over3}\)
\(\Rightarrow f(x)={2\over3}\)

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 \({\overrightarrow{a}+2\overrightarrow{b}\over 3}\).

Given \(\left|\begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2 x & 5 x^2 \end{array}\right|=0\)
\(\Rightarrow\left|\begin{array}{ccc} 1 & 4 & 20 \\ 0 & 6 & 15 \\ 0 & -2-2 x & 15-5 x^2 \end{array}\right|=0 \quad \begin{aligned} &R_1 \rightarrow R_1 \\ &R_2 \rightarrow R_1-R_2 \\ &R_3 \rightarrow R_2-R_3 \end{aligned}\)
expand by C₁
\(1\left[6\left(5-5 x^2\right)-15(-2-2 x)\right]-0+0=0\)
\(-30 x^2+30 x+60=0\)
\((\div 30)-x^2+x+2=0\)
\(x^2-x-2=0\)
\((x-2)(x+1)=0\)
\(\therefore x=2 \text { (or) } x=-1\)

Suppose that is rational
then \(\sqrt { 5 } \) = (where p and q are integer which are co-prime)
⇒ p =\(\sqrt { 5 } \)q = p² = 5q²   ....(1)
\(\frac { { p }^{ 2 } }{ 5 } \)= q² ⇒ 5 is a factor of p
So let P = 5c
substituting p = 5c in (1) we get
(5c)² = 5q² ⇒ 25² = 5q²
⇒ c²  = \(\frac { { 5q }^{ 2 } }{ 25 } =\frac { { q }^{ 2 } }{ 5 } \)
⇒ 5 is a factor of q also
⇒ So 5 is a factor of p and q which is a contradiction.
⇒ \(\sqrt { 5 } \) is not a rational number.
⇒\(\sqrt { 5 } \)  is an irrational number.

The given equation is x cos \(\alpha\) + y sin \(\alpha\) = p or \(\frac{x}{p/\cos\alpha}+\frac{y}{p/sin\alpha}=1\) ........(i)
This cuts the coordinate axes at \(A(p/\cos\alpha,0)\) and \(B(0,p/\sin\alpha)\). Let P (h, k)be the mid point of the intercept AB. Then,
\(h=\frac{p/\cos\alpha+0}{2},k=\frac{0+p/\sin\alpha}{2}\)
\(\Rightarrow h=\frac{p}{2\cos\alpha},k=\frac{p}{2\sin\alpha}\)
\(\Rightarrow \cos\alpha=\frac{p}{2h},\sin\alpha=\frac{p}{2k}\)
Here, u is a variable. to find the locus of P (h, k), we have to eliminate c.
From (i), we obtain
\(\cos^2\alpha+\sin^2\alpha=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\Rightarrow 1=\frac{p^2}{4h^2}+\frac{p^2}{4k^2}\Rightarrow \frac{4}{p^2}=\frac{1}{h^2}+\frac{1}{k^2}\)
Hence, the locus of (h, k) is \(\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}\)

sin x - 3 sin 2x + sin 3x = cos x -3 cos 2x + cos 3x
sin 3x + sin x - 3 sin 2x = cos 3x + cos x - 3 cos 2x
2sin 2x cos x - 3 sin 2x = 2 cos 2x cos x - 3 cos 2x
(sin 2x -cos2x) (2 cos x - 3) = 0
Then, sin 2x - cos 2x = 0 since 2 cosx- 3 \(\neq\) 0
sin 2x = cos 2x \(\Rightarrow\)tan 2x = 1 \(\Rightarrow x={n\pi \over2}+{\pi \over8},n\in Z\)

Given a = 5
Since the particle gets doubled, the G.P will be 5, 10,20,40, ... 1,50,000
⇒ a.r^n-1 > 1,50,000
⇒ a.(2^n-1) > 1,50,000
⇒ 5(2^n-1) > 1,50,000
⇒ \(2^{n-1}>{1,50,000\over5}\)
⇒ 2^n-1 > 30,000
⇒ 2^n-1 > 2⁴ x 1875
⇒ \({2^{n-1}\over24}>1875\)
⇒ 2^n-5 > 1875
⇒ (n - 5) log 2 > log 1875
⇒ \(n-5>{log1875\over log2}\)
⇒ \(n-5>{3.2730\over 0.3010}\)
⇒ n-5 > 10.873
⇒ n > 10.873 + 5
⇒ n > 15.873
⇒ n = 15
Hence the 15^th day, the infectious Virus particles just grow over 1,50,000 units.

Given A = {a, b,c}
(i) Let R = {(a, a) (b, b) (c,c)}
R is reflexive
R is symmetric and R is transitive \(\Rightarrow\) R is an equivalence relation.
This is the equivalence relation of smallest cardinality on A.
\(\therefore\) n(R) = 3
(ii) Let R = {(a, a) (a, b) (a,c) (b,a) (b, b) (b, c) (c, a) (c, b) (c,c)}
R is reflexive since (a, a) (b, b) and (c, c) \(\in \) R
R is symmetric since (a, b) \(\in \) R \(\Rightarrow\)(b, a) \(\in \) R
(b, c) \(\in \) R \(\Rightarrow\) (c, a) \(\in \) R
(c, a) \(\in \) R \(\Rightarrow\) (a, c) \(\in \)R
R is also transitive since (a, b) (b, c) \(\in \) R \(\Rightarrow\) (a, c) \(\in \)R
Hence R is are equivalence relation of largest cardinality on A.
\(\therefore\) n(R) = 9

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).
(4, 2) \(\in \) R and (2, 4) \(\in \) R \(\Rightarrow\) (4, 4) \(\notin \) R
\(\therefore\) R is not transitive.

Given \(\frac { cos^{ 4 }\alpha }{ { cos }^{ 2 }\beta } +\frac { { sin }^{ 4 }\alpha }{ { sin }^{ 2 }\beta } \)
⇒ cos⁴\(\alpha\)sin²β + sin⁴\(\alpha\)cos²β = cos²β sin²β
⇒ cos⁴\(\alpha\)(1-cos²β)+cos²β(1-cos²\(\alpha\))2 = cos²β(1-cos²β)

⇒ cos⁴\(\alpha\)-2cos²\(\alpha\)cos²β + cos⁴β = 0
⇒ (cos²\(\alpha\)-cos²β)² = 0
⇒ cos²\(\alpha\)-cos²β = 0
⇒ cos²\(\alpha\) = cos²β...(1)
⇒ 1-sin²\(\alpha\) = 1-sin²β
⇒ sin²\(\alpha\) = sin^-2β
sin⁴\(\alpha\)+sin⁴β = 2sin²\(\alpha\)sin²β
LHS = sin⁴\(\alpha\) + sin⁴β = (sin²-sin²β)² + 2sin²\(\alpha\)sin²β
= 2sin²\(\alpha\)sin²β = [∵ sin θ sin²\(\alpha\) = sin²β, from (2)]
= RHS
Hence proved.

\(P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)}[\because A \subset B \Rightarrow A \cap B=A]\)

WKT
\(\int e^{a x} \cos b x d x=\frac{e^{a x}}{a^{2}+b^{2}}[a \cos b x+b \sin b x]+c\)
\(\text { Putting } a=-4 \text { and } b=1\)
\(\int e^{-4 x} \cos x d x=\frac{e^{-4 x}}{17}[-4 \cos x+\sin x]+c\)

\(\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}-1}}{2 x+1}=\lim _{x \rightarrow \infty} \frac{\sqrt{1-\frac{1}{x^{2}}}}{2+\frac{1}{x}}=\frac{\sqrt{1-0}}{2+0}=\frac{1}{2}\)

\(\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32} \ldots . . =\sqrt{2}+2 \sqrt{2}+3 \sqrt{2}+4 \sqrt{2} . \)
\(=\sqrt{2}[1+2+3+\ldots .] \)
\(S_{n} =\frac{\sqrt{2}[n(n+1)]}{2} \)
\(=\frac{n(n+1)}{\sqrt{2}} \)

\({ }^{(n+5)} P_{(n+1)} \quad=\frac{11(n-1)}{2}{ }^{n+3} P_{n}\)
\(\frac{(n+5) !}{(n+5-n-1) !} =\frac{11(n-1)}{2} \times \frac{(n+3) !}{(n+3-n) !} \)
\(\frac{(n+5)(n+4)(n+3) !}{4 !} =\frac{11(n-1) \times(n+3) !}{2 \times 3 !} \)
\(\frac{(n+5)(n+4)}{4 \times 3 !} =\frac{11(n-1)}{2 \times 3 !} \)
\(n^{2}+9 n+20 =22 n-22 \)
\(n^{2}-13 n+42 =0 \)
\((n-6)(n-7) =0 \)
\(n=6 \text { or } 7 \)

The value of
\(\log _{3} 11 \log _{11} 13 \log _{13} 15 \log _{15} 27 \log _{27} 81=\log _{3} 81 \)
\(=\log _{3} 3^{4}=4 \log _{3} 3=4 \)

\(4 \sin ^{2} x+3 \cos ^{2} x+\sin \frac{x}{2}+\cos \frac{x}{2} \)
\(=3\left(\sin ^{2} x+\cos ^{2} x\right)+\sin ^{2} x+\sin \frac{x}{2}+\cos \frac{x}{2} \)
\(=3+\sin ^{2} x+\sin \frac{x}{2}+\cos \frac{x}{2} \)
\(\text { Maximum of } \sin x=1\)
\(\text { Maximum value of } \sin ^{2} x=1\)
\(\text { Maximum value of } \sin \frac{x}{2} \text { occurs where } x=45^{\circ}\)
\(\text { Maximum value is } 3+1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=4+\frac{2}{\sqrt{2}}=4+\sqrt{2}\)

\((A \times B) \cap(A \times C) =A \times(B \cap C) \)
\(n[(A \times B) \cap(A \times C)] =n(A) \times n(B \cap C) \)
\(\Rightarrow 8 =n(A) \times 2 \Rightarrow n(A)=4 \)

Probability of getting odd number = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
Probability of getting even numbers = 1-\(\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
Now, probability of getting no odd number when the die is tossed twice = Probability of getting even number when the die is tossed thrice.
 \(\Rightarrow \quad \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =\frac { 1 }{ 8 } \)
\(\therefore\)  P(Odd number atleast once) = \(1-\frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)

Given f(0) = \(\frac { 1 }{ 2 } \)
\(\lim _{ x\rightarrow 0 }{ f\left( x \right) } =\lim _{ x\rightarrow 0 }{ \frac { 1-\cos { x } }{ { x }^{ 2 } } } =\lim _{ x\rightarrow 0 }{ \frac { 2\sin { ^{ 2 }\frac { x }{ 2 } } }{ { x }^{ 2 } } } =\lim _{ x\rightarrow 0 }{ \frac { 2 }{ 4 } } { \left[ \frac { \sin { \frac { x }{ 2 } } }{ \frac { x }{ 2 } } \right] }^{ 2 }=\frac { 2 }{ 4 } { (1) }^{ 2 }=\frac { 1 }{ 2 } .\)

\(\int \sqrt{1+\cos 2 x} d x=\int \sqrt{2 \cos ^2 x} d x=\sqrt{2} \int \cos x d x=\sqrt{2} \sin x+c\)

\(y=\sin \left(e^x\right)\)
Take \(u=e^x \Rightarrow \frac{d u}{d x}=e^x\)
\(y =\sin u\)
\(\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}=\cos u\left(e^x\right)\)
\(=\cos \left(e^x\right) e^x=e^x \cos \left(e^x\right)\)

\(lim_{x\rightarrow 0}{1-cosx\over x^2}=lim_{x\rightarrow 0}{2sin^2{x \over 2}\over x^2}[\because cos 2x=1-2sin^2x]\)
\(=2.lim_{x\rightarrow0}{sin^2({x\over 2})\over {x^2\over 4}\times 4}={2\over4}[lim{{x\over 2}\rightarrow 0}{sin({x\over2})\over ({x\over2})}]^2\)
\(={2\over4}\times 1^2={2\over4}={1\over2}\)  \([\because lim_{\theta \rightarrow o}{sin \theta \over \theta}=1]\)
\(\therefore lim_{x\rightarrow 0}{1-cosx\over x^2}={1\over2}\)

k = 2, 2x - y + 2 = 0, 6x - 2y + 1 = 0, \(\theta={\tan}^{-1}\left({1\over 7} \right)\)

60, 108

Given series is  1² + 3² + 5² +...
Let Tn be the nth term
T_n = (nth term of 1, 3, 5,...)²
= [1+(n-1)2]² = (1 + 2n - 2)² = (2n-1)²
= 4n² + 1 - 4n
∴ Sum of n terms =  \(\sum { 4{ n }^{ 2 } } -4n+1=4\sum { n^{ 2 } } -4\sum { n } +n\)
= \(4\frac { (n)(n+1)(2n+1) }{ 6 } -\frac { 4n(n+1)+n }{ 2 } \)
= \(\frac { n }{ 2 } \) [2(n + 1)(n + 1) - 6(n + 1) + 3]
= \(\frac { n(4{ n }^{ 2 }-1) }{ 3 } \)

12x + 5y = 7 and 12x + 5y + 7 = 0
Given parallel lines are 12x + 5y = 7 and 12x + 5y + 7 = 0
Here a = 12, b = 5, c₁ = -7  and c₂ = 7
Distance between parallel lines = \(\left| \frac { { c }_{ 1 }-{ c }_{ 2 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \right| \)
= \(\left| \frac { -7-7 }{ \sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } } \right| =\left| \frac { -14 }{ \sqrt { 169 } } \right| \)
Distance between parallel lines = \(\frac { 14 }{ 13 } \)