Let \(\overrightarrow{BC}=\overrightarrow{a},\overrightarrow{AC}=\overrightarrow{b},\overrightarrow{BA}=\overrightarrow{c}\)

Then \(|\overrightarrow{a}|=a,|\overrightarrow{b}|=b,\) and \(|\overrightarrow{c}|=c\)
Since \(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\)
We have \(\overrightarrow{a}=\overrightarrow{c}+\overrightarrow{b}\) and angle between \(\overrightarrow{c}\) and \(\overrightarrow{b}\) is (180-A)
\(=|\overrightarrow{a}|^2=|\overrightarrow{c}+\overrightarrow{b}|^2=|\overrightarrow{c}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{c}||\overrightarrow{b}|cos (180-A)\)
\(=c^2+b^2+2cb \ cos (180-A)\)
\(=b^2+c^2-2cb \ cos A\)    \([\because cos(180-A)=-cos \ A]\)

Let I = \(\int { \frac { { e }^{ x }dx }{ { e }^{ 2x }+6{ e }^{ x }+5 } } \) = \(\int { \frac { { e }^{ x }dx }{ \left( { e }^{ x } \right) ^{ 2 }+6{ e }^{ x }+5 } } \)
Put t= e^x \(\Rightarrow\) dt = e^x dx
I = \(\int { \frac { dt }{ { t }^{ 2 }+6t+5 } =\int { \frac { dt }{ { t }^{ 2 }+6t+9-9+5 } } =\int { \frac { dt }{ \left( t+3 \right) ^{ 2 }-4 } } \int { \frac { dt }{ \left( t+3 \right) ^{ 2 }-{ 2 }^{ 2 } } } } \)
= \(\frac { 1 }{ 2\left( 2 \right) } log\left| \frac { t+3-2 }{ t+3+2 } \right| +c\)
= \(\frac { 1 }{ 4 } log\left| \frac { { e }^{ x }+1 }{ { e }^{ x }+5 } \right| +c\)

Given P(Simultaneous occurrence of atleast one of A and B) =p
\(\Rightarrow P(AUB)\) = p and P (occurrence of exactly one of A and B) = q
\(\Rightarrow P(AUB)-P(A\cap B)=q\)
\(\therefore p-P(A\cap B)=q\)
\(\Rightarrow P(A\cap B)=p-q\)
\(\Rightarrow 1-P(\overline { A\cap B) } =p-q\)   \(\left[ \because P(A\cap B)+P(\overline { A\cap B } )=1 \right] \)
\(\Rightarrow 1-P(\bar { A } \cup \bar { B } )=p-q\)
\(\Rightarrow P(\bar { A } \cup \bar { B } )=p-q\)
\(\Rightarrow P(\bar { A } )+P(\bar { B } )-P(\bar { A } \cap \bar { B } )=1-p+q\)
\(\Rightarrow P(\bar { A } )+(P\bar { B } )=1-p+q+p(\bar { A } \cap \bar { B } )=1-p+q+P(\overline { A\cup B } )\)
= 1 - p + q +1 - p     \(\left[ \because P\overline { A\cup B } =1-P(AUB)=1-p \right] \)
= 2 - p+q
\(\therefore P(\bar { A } )+P(\bar { B } )\) = 2 - 2 p+q   Hence proved.

Using definition of derivative
We have \(f^{ ' }\left( 2 \right) =\lim _{ x\rightarrow 2 }{ \frac { xf\left( x \right) -f(2) }{ x-2 } } \)
\(\Rightarrow \lim _{ x\rightarrow 2 }{ \frac { xf\left( x \right) -f(2) }{ x-2 } } =1 \quad \left[ \because \quad f^{ ' }\left( 2 \right) =1 \right] ...(1)\)
\(Now,\quad \lim _{ x\rightarrow 2 }{ \frac { xf\left( 2 \right) -2f(x) }{ x } } =\lim _{ x\rightarrow 2 }{ \frac { xf\left( 2 \right) -2f(2)+2f\left( 2 \right) -2f(x) }{ x-2 } } =\lim _{ x\rightarrow 2 }{ \frac { (x-2)(f.(2))-2(f\left( x \right) -f\left( 2 \right) ) }{ x-2 } } \)
\(=\lim _{ x\rightarrow 2 }{ \frac { (x-2)f\left( 2 \right) }{ x-2 } } -2.\lim _{ x\rightarrow 2 }{ \frac { f\left( x \right) -f\left( 2 \right) }{ x-2 } } =\lim _{ x\rightarrow 2 }{ f(2)-2.f^{ ' }\left( 2 \right) } \)
\(=f(2)-2.f^{ ' }\left( 2 \right) =4-2(1)=4-2=2.\)

Let A be the oil change, P(A) = 0.3
B be the filter change, P(B) = 0.4
And \(P(A \cap B)=0.15\)
\(\text { (i) } P(B / A) =\frac{P(A \cap B)}{P(A)}\)
\(=\frac{0.15}{0.3}=\frac{15}{30}=0.5\)
\(\text { (ii) } P(A / B) =\frac{P(A \cap B)}{P(B)}\)
\(=\frac{0.15}{0.4}=\frac{15}{40}=\frac{3}{8}=0.375\)

\(\lim _{v \rightarrow c^{-}}(m)=\lim _{v \rightarrow c^{-}} \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{m_0}{\sqrt{\lim _{v \rightarrow c^{-}}\left(1-\frac{v^2}{c^2}\right)}}\)
For h > 0, c- h < v < c . This implies,(c-h)² \(<v^2<c^2\)
That is, \(\frac{(c-h)^2}{c^2}<\frac{v^2}{c^2}<1\) 
That is, \(\lim _{h \rightarrow 0} \frac{(c-h)^2}{c^2}<\lim _{h \rightarrow 0} \frac{v^2}{c^2}<\lim _{h \rightarrow 0} 1\)
That is, \(1<\lim _{h \rightarrow 0} \frac{v^2}{c^2}<1\)
That is  \(1<\lim _{v \rightarrow c^{-}} \frac{v^2}{c^2}<1\). By Sandwich theorem, \(lim_{v\rightarrow c^-}=1.\)
Therefore, \(lim_{v-c^-}(m)\rightarrow \infty.\)
That is, the mass becomes very very large (infinite) as \(v\rightarrow c^-.\)
The left hand limit is necessary.
Otherwise as\(v \rightarrow c^+\) makes \(1-{v^2\over c^2}<0\) and consequently we cannot find the mass.

The given inequality is x+y ≥ 4
Draw the graph of the line x+y =4

Table of values satisfying the equation
x+y =4

Putting (0, 0) in the given inequation, we have 0+0≥4⇒0≥4, which is false.
∴ Half plane of x+y≥4 is away from origin.
Also, the given inequality is 2x-y>0
Draw the graph of the line 2x - y =0
Table of values satisfying the equation
2x-y =0

Putting (3, 0) in the given inequation, we have 2\(\times\)3-0>0⇒6> 0, which is true.
∴ Half plane of 2x-y≥0 containing (3,0)

The given equation of line is 5x - 12y = 3 and the given point is (3, -2).
Let (a, b) be any moving point.
\(\therefore\) Distance between (a, b) and the point (3, -2) = \(\sqrt{(a-3)^2+(b+2)^2}\) and the distance of (a, b) from the line 5x - 12y = 3 = \(|\frac{5a-12b-3}{\sqrt{25+144}}|=|\frac{5a-12b-3}{13}|\)
According to the question, we have \([\sqrt{(a-3)^2+(b+2)^2}]=|\frac{5a-12b-3}{13}|\)
\(\Rightarrow (a-3)^2(b+2)^2=\frac{5a-12b-3}{13}\)
Taking numerical values only, we have \((a-3)^2(b+2)^2=\frac{5a-12b-3}{13}\)
\(\Rightarrow a^2-6a+9+b^2+4b+4=\frac{5a-12b-3}{13}\)
\(\Rightarrow a^2+b^2-6a+4b+13=\frac{5a-12b-3}{13}\)
\(\Rightarrow\) 13a² + 13b² - 78a + 52b + 169 = 5a - 12b - 3
\(\Rightarrow\) 13a² + 13b² - 83a + 64b + 172 = 0
So, only locus of the point is 13x² + 13y² - 83x + 64y + 172 = 0
Hence, the value of the filter is 13x² + 13y² - 83x + 64y + 172 = 0.

Let the x-axis be the time in seconds the y-axis be the distance in metres.
Let the engine be at the origin O.
Therefore the length of the train 560m is the negative y-intercept.
The uniform speed 12.5 m/s is the slope of the motion of the train.
\((speed=\frac{distance}{time})\)

Since we are given slope and y-intercept, the equation of the line is y = mx + b
(i) The equation of the motion of the train,
when m = 12.5 and b = 560,
is y = 12.5x - 560
(ii) When the engine touches the other side of the bridge (island)
y = 2065 and b = 0
x = 165.2 seconds.
(iii) When y = 0, the last coach cross the entry point of the bridge,
0 = 12.5x - 560
x = 44.8 seconds.
(iv) When y = 2065, the time taken for the train to cross the other end of the bridge is given by
2065 = 12.5x - 560
x = 210 seconds

\(\sqrt{3}tan^2\theta+(\sqrt{3}-1)tan\theta-1=0\)
\(\sqrt{3}tan^2\theta+\sqrt{3}tan\theta-tan\theta-1=0\)
\((\sqrt{3}tan\theta-1)(tan\theta+1)=0\)
Thus, either \(\sqrt{3}tan\theta-1=0(or)tan\theta+1=0\)

From (i) and (ii) we have the general solution.

(i) Let the x - axis represent the weight and the y-axis represent the length

(ii) Consider the point \(\begin{pmatrix} { x }_{ 1 } & { y }_{ 1 } \\ 2 & 3 \end{pmatrix}\begin{pmatrix} { x }_{ 2 } & { y }_{ 2 } \\ 4 & 4 \end{pmatrix}\)
Equation of the straight line using two point form is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\Rightarrow \quad \frac { y-3 }{ 4-3 } =\frac { x-2 }{ 4-2 } \)
\(\Rightarrow \quad \frac { y-3 }{ 1 } =\frac { x-2 }{ 2 } \)
\(\Rightarrow\) 2y - 6 = x-2
 \(\Rightarrow\) x - 2y + 4 = 0
(iii) To find the actual length of the spring,
Put x = 0
\(\Rightarrow\) 0 - 2y + 4 = 0
\(\Rightarrow\) -2y = -4
\(\Rightarrow\) y = 2 cm
(iv) Put y = 9 in (1) we get
\(\Rightarrow\) x - 18 + 4 = 0
\(\Rightarrow\) x - 14 = 0
\(\Rightarrow\) x = 14 kg
\(\therefore\) 14 Kg must be added
(v) Put x = 6 in (1) we get,
\(\Rightarrow\)  6 - 2y + 4 = 0
\(\Rightarrow\) 10 - 2y = 0
\(\Rightarrow\) 10 = 2y 
\(\Rightarrow\) y = 5
\(\therefore\) Length of the spring = 5 cm

Let T_n be the nth term of the given series
Then T_n = 1 + 4 + 4² + 4³ + ...
\(=1\left(4^n-1\over 4-1\right)\)
\(={4^n-1\over 3}\)
Let S_n be the sum to n terms of the given series
Then \(S_n={\sum_{k=1}^n}T_k=\sum_{k=1}^n{4^n-3\over3}\)
\(⇒\ S_n={1\over3}\left[ \sum_{k=1}^n4^n-\sum_{k=1}^n3\right]\)
\(⇒\ S_n= {1\over3}[4^1+4^]+...+4^n-3^n\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]\)
\(⇒\ S_n={1\over3}\left[ 4{(4^n-1)\over4-1}-3n\right]={1\over 3}\left[4{(4^n-1)-9n\over3}\right]\)

\({ S }_{ n }=\frac { 4 }{ 9 } \left[ \left( { 4 }^{ n }-1 \right) -n/3 \right] \)

Let p(n) denote the statement
\(1^2+3^2+5^2+...+(2n-1)^2={n(2n-1)(2n+1)\over 3}\)

Step1: Putting n = 1
\(1^2={1(2-1)(2+1)\over3}={3\over 3}=1\)
∵ p(1) is true
Step 2: Let us assume that p(K) is true
i.e. \(1^2+3^2+5^2+..+(2K-1)^2={K(2K-1)\over3}\)
To prove that p(K + 1) is true
\(∵\ 1^2+3^2 +5^2 + ..+ (2K-1)^2 + [2(K + 1)-1]^2={(K+1)(2(K+1)-1)(2(K+1)+1)\over 3}\)
i.e. to P.T. \(1^2 + 3^2 + 5^2 + ... + (2K -1)^2(2K + 1)^2 ={(K+1)(2K+1)(2K+3)\over 3}\)
LHS = 1²+ 3² + 5² + ... + (2K - 1)² + (2K + 1)²
\(={K(2K-1)(2K+1)\over3}+(2K+1)^2\)
\(=(2K+1)\left[{K(2K-1)\over 3}+2K+1\right]\)
\(=(2K+)\left[2K^2-K+6K+3\over3\right]={(2K+1)(2K^2+5K+3)\over3}\)
\(={(2K+1)(2K^2+5K+3)\over3}\)=RHS
∵ p(K+1) is true
∵ By mathematical induction, P(n) is true for all values of n.

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.

Let C be the position of the target and A and B be the positions of the fighter jet

Given ㄥBAC = 30, ㄥABC = 45
ஃ ㄥC = 180 - (30 - 45) = 180 - 75 = 105
Given AB = 100 km
Using sine formula,
\(\frac { a }{ sinA } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { a }{ sin30° } =\frac { 100 }{ sin105° } \)
\(\Rightarrow \frac { a }{ \frac { 1 }{ 2 } } =\frac { 100 }{ sin105° } \Rightarrow 2a=\frac { 100 }{ sin105° } \Rightarrow a=\frac { 50 }{ sin105° } \)
Now, sin 105° = sin (60 + 45) = sin 60 cos 45 + cos 60 sin 45
= \(\frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \)
Substituting (2) in (1) we get,
a = \(\frac { 50 }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } \Rightarrow a=\frac { 50\left( 2\sqrt { 2 } \right) }{ \sqrt { 3 } +1 } =\frac { 100\sqrt { 2 } }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
a = \(\frac { 100\left( \sqrt { 6 } -\sqrt { 2 } \right) }{ 3-1 } =50\left( \sqrt { 6 } -\sqrt { 2 } \right) km\)

Let y = \(\log _{ 7 }{ u }\quad and\quad u=\log _{ 7 }{ x } \)
\(\therefore \frac { dy }{ du } =\frac { 1 }{ u\quad \log _{ 7 }{ e } } \ and\ \frac { du }{ dx } =\frac { 1 }{ x\quad \log _{ e }{ 7 } } \)
\( Now\quad \frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)
\( \Rightarrow \frac { dy }{ dx } =\frac { 1 }{ u\quad \log _{ 7 }{ e } } \times \frac { 1 }{ x\quad \log _{ e }{ 7 } } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { 1 }{ \log _{ 7 }{ x. } \log _{ e }{ 7 } \times x\quad \log _{ e }{ 7 } } =\frac { 1 }{ x\log _{ 7 }{ x. } { \left( \log _{ e }{ 7 } \right) }^{ 2 } } \)
 

\(\int e^{x \log 2} e^x d x=\int e^{\log 2^x} e^x d x=\int 2^x e^x d x\)
\(=\int(2 e)^x d x=\frac{(2 e)^x}{\log (2 e)}+c\)

(ii) AB and BA is a symmetric matrices
(AB - BA)^T = (AB)^T + (BA)^T
= B^TA^T + A^TB^T
= BA - AB
(AB - BA)^T = -(AB - BA)
\(\therefore\)(AB - BA) is a skew-symmetric matrix.

\(\sqrt [ 3 ]{ 1003 } ={ (1003) }^{ \frac { 1 }{ 3 } }={ (1000+3) }^{ 1/3 }={ (1000) }^{ 1/3 }\left[ 1+\frac { 3 }{ 1000 } \right] ^{ 1/3 }=10[1+0.003]^{ 1/3 }\)
\(=10\left[ 1+\frac { 1 }{ 3 } (0.003)+\frac { \left( \frac { 1 }{ 3 } \right) \left( -\frac { 2 }{ 3 } \right) }{ 1.2 } { (0.003) }^{ 2 }+... \right] \)
= 10 [1 + 0.001 - 0.0000001 + ...] = 10.00999 = 10.0100

As m - m = 0,
m - m is divisible by 7 \(\Rightarrow\)  mRm
\(\therefore\) R is reflexive.
Let mRn. Then m - n = 7k for some integer k 
Thus n-m = 7 (-k) and hence nRm
\(\therefore\) R is symmetric.
Let mRn and nRp
\(\Rightarrow\) m-n = 7k and n - p = 7l for some
\(\Rightarrow\) m = 7k + n and - p = 7l- n integers k and l
so m-p = 7k+n+7l-n
\(\Rightarrow\) m - p = 7(k+l) \(\Rightarrow\) mRp
\(\therefore\) R is transitive.
Thus, R is an equivalence relation.

Given  \({{1}\over{2}}\) log₁₀ \((11+4\sqrt{7})\) = log₁₀ (2 + x).
\(\Rightarrow\) \({{1}\over{2}} \) log₁₀ \((7+4+4\sqrt{7})\) = log₁₀(2 + x)
\(\Rightarrow \) \({{1}\over{2}}\) log₁₀ \(((\sqrt{7})^2+2^2+2.2\sqrt{7})\) = log₁₀(2 + x)
\(\Rightarrow\)  \({{1}\over{2}}\) log₁₀\({(\sqrt{7}+2)}^{2}\)= log₁₀(2 + x)
\(\Rightarrow\)  log₁₀\((\sqrt{7}+2)\) = log₁₀(2 + x)
\(\Rightarrow\)  \((\sqrt{7}+2)\)= (2 + x)
\(\Rightarrow\)   \(x=\sqrt{7}\)

\(P(X / Y)=\frac{1}{2} \Rightarrow \frac{P(X \cap Y)}{P(Y)}=\frac{1}{2} \)
\(P(Y / X)=\frac{1}{3} \Rightarrow \frac{P(X \cap Y)}{P(X)}=\frac{1}{3} \)
\(\text {But } P(X \cap Y)=\frac{1}{6} \Rightarrow \frac{\frac{1}{6}}{P(Y)}=\frac{1}{2}\)
\(P(Y)=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{3}\)
\(\text {and } P(X)=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{2}\)
\(P(X \cup Y)=P(X)+P(Y)-P(X \cap Y)\)
\(=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{3+2-1}{6} =\frac{2}{3} \)

\(\int \sqrt{\frac{1-x}{1+x}} d x =\int \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x \)
\(=\int \sqrt{\frac{(1-x)^{2}}{1^{2}-x^{2}}} d x \)
\(=\int \frac{1-x}{\sqrt{1-x^{2}}} d x \)
\(=\int\left(\frac{1}{\sqrt{1-x^{2}}}-\frac{x}{\sqrt{1-x^{2}}}\right) d x \)
\(=\sin ^{-1} x-\left(-\sqrt{1-x^{2}}\right)+c \)
\(=\sin ^{-1} x+\sqrt{1-x^{2}}+c \)

\(\text { Given } f(0)=5\)
\(\lim _{x \rightarrow 0} \frac{f(x)-5}{x} =4 \)
\(\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} =4 \)
\(f^{\prime}(0) =4 \)
\(g(x) =\left(x^{2}+2 x+1\right) f(x) \)
\(g^{\prime}(x) =\left(x^{2}+2 x+1\right) f^{\prime}(x)+(2 x+2) f(x) \)
\(g^{\prime}(0) =(1) f^{\prime}(0)+2 f(0) \)
\(=1(4)+2(5) \)
\(=4+10=14 \)

\(\lim _{x \rightarrow 0} \frac{\sin p x}{\tan 3 x}=4 \Rightarrow \frac{p}{3}=4 \Rightarrow p=12\)

Perpendicular distance from origin to the given line is
\(\frac{1}{\sqrt{\frac{1}{3^{2}}+\frac{1}{4^{2}}}}=\frac{1}{\sqrt{\frac{1}{9}+\frac{1}{16}}}=\frac{1}{\sqrt{\frac{16+9}{144}}}=\frac{12}{5}\)

No. of ways of answering = 5³ = 125
Correct answer - 1
.'. Number of incorrect answer = 125 - 1 = 124

\(x^{2}-\mathrm{k} x+\mathrm{c}=0\)
a and b are the roots
\(\therefore a+b=k, a b=c\)
To find
\(\sqrt{(a-b)^{2}+0^{2}}=a-b=\sqrt{(a+b)^{2}-4 a b}=\sqrt{k^{2}-4 c}\)
 

\(\left(2 \cos ^{2} \frac{\pi}{16}\right)\left(2 \cos ^{2} \frac{3 \pi}{16}\right)\left(2 \cos ^{2} \frac{5 \pi}{16}\right)\left(2 \cos ^{2} \frac{7 \pi}{16}\right) \)
\(=2^{4}\left[\cos \frac{\pi}{16} \cos \frac{3 \pi}{16} \cos \frac{5 \pi}{16} \cos \frac{7 \pi}{16}\right]^{2} \)
\(=\left(\cos \frac{8 \pi}{16}+\cos \frac{6 \pi}{16}\right)^{2}\left(\cos \frac{8 \pi}{16}+\cos \frac{2 \pi}{16}\right)^{2} \)
\(=\left(\cos \frac{6 \pi}{16}\right)^{2}\left(\cos \frac{2 \pi}{16}\right)^{2} \)
\(=\left\{\frac{1}{2}\left[\cos \frac{8 \pi}{16}+\cos \frac{4 \pi}{16}\right]\right\}^{2} \)
\(=\frac{1}{4} \cdot \frac{1}{2}=\frac{1}{8} \)

Let A = {1, 2,3,4,8} = {5, 2,3,6}
A and B have two elements in common.
Number of elements common to A \(\times\) B and B x A = 2 \(\times\)2 = 2²
Similarly here we have 17² element in common

Let R be the point which divides the line joining the points P and Q externally in the ratio 2:1
=\(\frac { 2(-\hat { i } +\hat { j } +\hat { k } )-1(\hat { i } +2\hat { j } -\hat { k } ) }{ 2-1 } \)
=
=\(\frac { -3\hat { i } +3\hat { k } }{ 1 } =-3\hat { i } +3\hat { k } \)

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) = \(1-\frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)

\(Given,\quad \lim _{ x\rightarrow 2 }{ \frac { { x }^{ n }-{ 2 }^{ n } }{ x-2 } } =80\)
\( \Rightarrow n.{ 2 }^{ n-1 }=80\)
By trial method, put n 5
\(\Rightarrow 5({ 2 }^{ 5-1 })=80\)
\(\Rightarrow 5({ 2 }^{ 4 })=80\)
\( \Rightarrow 5(16)=80\)
\( \Rightarrow 80=80\)
\(\therefore n=5\)

\( \int \sin x d x =-\cos x+c \)
\(\therefore \int \sin 3 x d x =\frac{-\cos 3 x}{3}+c\)

\( \int\left(1-x^2\right)^{-\frac{1}{2}} d x =\int \frac{1}{\left(1-x^2\right)^{1 / 2}} d x \)
\(=\int \frac{1}{\sqrt{1-x^2}} d x=\sin ^{-1} x+c\)

\(f(x)=-x^2+2\)
\(f(x+h)=-(x+h)^2+2=-x^2-h^2-2 x h+2\)
\(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{-x^2-h^2-2 x h+2+x^2-2}{h}\)
\(=\lim _{h \rightarrow 0} \frac{+h(-h-2 x)}{h}\)
= -0 - 2x
\(f^{\prime}(x)=-2 x\)

\(|x|= \begin{cases}-x & \text { if } x<0 \\ 0 & \text { if } x=0 \\ x & \text { if } x>0\end{cases}\)
If x > 0,then |x| =  x, which tends to 0 as
\(x \rightarrow 0\) from the right of 0. That is, \(\lim _{ x\rightarrow0^+}{|x| } =0\)
If x < 0, then |x| = - x which again tends to 0 as x\(\rightarrow\)0. from the left of 0. That is, \(\lim _{ x\rightarrow0^-}{|x| } =0\).
Thus, \(\lim _{ x\rightarrow0^-}{|x| } =0=\lim _{ x\rightarrow0^+}{|x| }.\)
Hence \(\lim _{ x\rightarrow0}{|x| } =0\).

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

x +y + 2 = 0

720, 4320

The general solution of sin \(\theta\) = sin \(\alpha,\alpha \in [-{\pi\over2},{\pi \over 2}],is \ \theta =n\pi +(-1)^n \alpha,n \in Z\)
\(sin \theta =-{\sqrt{3}\over2}=sin(-{\pi\over 3}),\)
Thus the general solution is 
\(\theta =n\pi +(-1)^n(-{\pi\over3})=n\pi+(-1)^{n+1}{\pi\over 3};n \in Z ....(i)\)

Let T_n be the n^th term of the sequence
Then, given T₁ = 3, T₂ = 3, T₃ = 6 and 
T_n = T_n-1 + T_n-2, n > 2.
T₃ = T₂ + T₁ = 3 + 3 = 6
T₄ = T₃ + T₂ = 6 + 3 = 9
T₅ = T₄ + T₃ = 9 + 6 = 15.

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.