We shall prove the result by using principle of mathematical induction.
Let \(P\left( n \right) { :A }^{ n }=\left[ \begin{matrix} { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \end{matrix} \right] \)
Now, \(P\left( 1 \right) { :A }^{ 1 }=\left[ \begin{matrix} { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \)
The result is true for n = 1.
Let the result be true for n = k.
So, \({ A }^{ k }=\left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
Now, we prove that P(k + 1) is true.
Now, A^k+1 = A. A^k
\(=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \end{matrix} \right] \)
= A^k+1
Hence, it is true n = k + 1.
Hence, by principle of mathematical induction P(n) is true for all \(n\epsilon N\)

GIven, \(P(A)=\frac{2}{5}, P(B)=\frac{1}{3}, P(C)=\frac{1}{2}, P(A \cap C)=\frac{1}{5}\) 
and \(P(B \cap C)=\frac{1}{4}\) 
\(\therefore P\left(\frac{C}{B}\right)=\frac{P(B \cap C)}{P(B)}=\frac{1 / 4}{1 / 3}=\frac{3}{4}\) 
and \(P\left(A^{\prime} \cap C^{\prime}\right)=1-P(A \cup C)\) 
\(=1-[P(A)+P(C)-P(A \cap C)]\) 
\(=1-\left(\frac{2}{5}+\frac{1}{2}-\frac{1}{5}\right)=1-\left(\frac{4+5-2}{10}\right)\) 
\(=1-\frac{7}{10}=\frac{3}{10}\)

(i) Line is normal to the plane and its DR's are
(1+ 2, -3+1, 3+3) i.e. (3, - 2, 6)
(ii) Find the equation of plane passing through the point (1, - 3, 3) and perpendicular to a line having DR's 3, - 2, 6.
\(=3 x-2 y+6 z-27=0\)

\((5,-2,0) ;(5 \hat{i},-2 \hat{j}, 0 \hat{k})\)

(i) \(\int e^{2 y} d y=\int 1 d x\) 
(ii) Put y = 3, to get the required value of 
\(=\frac{e^{6}+9}{2}\)

Let \(I=\int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}=\int_{0}^{1} \frac{e^{x}}{e^{2 x}+1} d x\) 
Now, put \(e^{x}=t \Rightarrow e^{x} d x=d t\) 
Upper limit When x = 1, then t = e¹ = e
Lower limit When x = 0, then t = e^o = 1
\(\therefore \ I=\int_{1}^{e} \frac{d t}{t^{2}+1}=\left[\tan ^{-1} t\right]_{1}^{e}\) 
\(=\tan ^{-1} e-\tan ^{-1} 1=\tan ^{-1} e-\frac{\pi}{4}\) 

On putting sin x = t, given integral reduces to
\(\int_{0}^{1} e^{t} d t .[\text { Ans. } e-1]\)

On putting 1+x² = t, given integral reduces to \(\frac{1}{2} \int_{1}^{2} \frac{d t}{\sqrt{t}}\) 
=\(\sqrt{2}-1\)

Let the radius of spherical ball of the salt be r.
\(\therefore \text { Volume of the ball, } V=\frac{4}{3} \pi r^{3}\) 
and surface area, \(S=4 \pi r^{2}\) 
\(\therefore \frac{d V}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=-\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}=-4 \pi r^{2} \frac{d r}{d t}\) 
[here, we take negative sign because salt is dissolving]
According to the given condition,
\(\frac{d V}{d t} \propto S \Rightarrow \frac{d V}{d t}=k S\) wehre - k )ISproportio.na II' t)' constant.
\(\Rightarrow -4 \pi r^{2} \frac{d r}{d t}=k \cdot 4 \pi r^{2} \Rightarrow \frac{d r}{d t}=-k\) 
Hence, the radius of ball is decreasing at constant rate.

Let r be the radius, A be the area and P be the perimeter.
Then, we have \(\frac{d A}{d t}=\text { constant }=k(\text { say })\)

\(\begin{array}{l} f(x)=-\cos x, \text { if } \frac{\pi}{2}

\( \text {Hint } \lim _{x \rightarrow 0} f(x)=f(0) \)
\(\Rightarrow \lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \cos x=k \)
\(\Rightarrow 1+1=k \)

A=0

(A³)^-1=(A-1)³

Given matrix equation is
\( \left[\begin{array}{ll} 2 x & 3 \end{array}\right]\left[\begin{array}{rr} 1 & 2 \\ -3 & 0 \end{array}\right]\left[\begin{array}{l} x \\ 8 \end{array}\right]=O \)
\(\Rightarrow \left[\begin{array}{ll} 2 x & 3 \end{array}\right]\left[\left[\begin{array}{rr} 1 & 2 \\ -3 & 0 \end{array}\right]\left[\begin{array}{l} x \\ 8 \end{array}\right]\right)=O \) 
[by associative law of multiplication]
\(\Rightarrow \left[\begin{array}{ll}2 x & 3\end{array}\right]\left[\begin{array}{c}x+16 \\ -3 x\end{array}\right]=0\)
\(\Rightarrow [2 x(x+16)-9 x]=[O]\)
\(\Rightarrow\) 2 x^{2} + 32 x - 9x = 0
\(\Rightarrow\)\(2 x^{2}+23 x=0\)
\(\Rightarrow x(2 x+23)=0\)
\(\Rightarrow\) x = 0 and x = -23 / 2

Given, AB = BA
Now, (A + B)² = (A + B)·(A + B)
= A·(A + B) + B·(A + B)
= A² + AB + BA + B²
= A² + AB + AB + B² [ஃ BA = AB, given]
= A ² + 2AB + B²

(i) 3 x 3
(ii) 9
(iii)a₂₃ = x² - y, a31 = 0, a₁₂ = 1]

\(\text {Here, } \tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}, x y<1\)
\(\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{4}\)
\(\frac{x+y}{1-x y}=1\)
\( x+y=1-x y\)
\(x+y+x y=1\)
Therefore, the value of \( x+y+x y \text { is } 1 \text { . } \)

\( \tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5} \)
\(\text {Now, } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2} \)
\(\tan ^{-1} x=\frac{\pi}{2}-\cot ^{-1} x \)
\(\therefore \frac{\pi}{2}-\cot ^{-1} x+\frac{\pi}{2}-\cot ^{-1} y=\frac{4 \pi}{5} \)
\(-\cot ^{-1} x-\cot ^{-1} y+\pi=\frac{4 \pi}{5} \)
\(\cot ^{-1} x+\cot ^{-1} y=\pi-\frac{4 \pi}{5} \)
\(=\frac{\pi}{5} \)

(i) R is reflexive, as (a, a) ∈ R,\(\vee a \in A\)
(ii) R is symmetric, as (0, 1) e R => (1, 0) e R and (0, 3) e R
⇒ (3, 0)∈ R.
(iii) R is not transitive, as (3, 0), (0, 1) ∈ R ⇏ (3, 1) ∈ R.
[Ans. Reflexive, symmetric and not transitive]

Let \(\theta\) denote the angle at instant t
\(\frac{d\theta}{dt}=2\frac{d}{dt}(\sin \theta)\)
\(\frac{d\theta}{dt}=2\cos\theta.(\frac{d\theta}{dt})\)
\(1=2\cos\ \theta\)
\(2\cos\theta=1 cos\theta=\frac{1}{2}\)
\(\Rightarrow \theta=\cos^{-1}(\frac{1}{2})\)
Hence required angles is \(\frac{\pi}{3}\)

\(f^{\prime}(x)=\frac{1+2 x^{2}}{1+x^{2}}-\frac{1}{\sqrt{1+x^{2}}}=\frac{1+2 x^{2}-\sqrt{1+x^{2}}}{1+x^{2}}\) 
Now,\(f^{\prime}(x) \geq 0\) 
\( \Rightarrow \frac{1+2 x^{2}-\sqrt{1+x^{2}}}{1+x^{2}} \geq 0\)
\(\Rightarrow 1+2 x^{2} \geq \sqrt{1+x^{2}}\)
\( \Rightarrow 1+4 x^{4}+4 x^{2} \geq 1+x^{2} \) 
\(\Rightarrow 4 x^{4}+3 x^{2} \geq 0\) which is true for all \(x \in R\) .
So,\(f^{\prime}(x) \geq 0, \forall x \in R\)