\(\frac { dy }{ dx } ={ e }^{ -2\theta }\left( \frac { { \theta }^{ 2 }+1-{ \theta }^{ 3 }+\theta }{ { \theta }^{ 2 }-1+{ \theta }^{ 3 }+\theta } \right) \)

\(\frac { d }{ dx } \left( { 2 }^{ cos^{ 2 }x } \right) ={ 2 }^{ cos^{ 2 }x }.log_{ e }2.\left( -2sinx\quad cosx \right) =-{ 2 }^{ cos^{ 2 }x }log_{ e }2. sin2x.\)

(i) sin x, sin 2x are continuous, so f (x) is continuous in [0,\(\pi\)]
(ii) c= cos^-1 \(\left( \frac { 1\pm \sqrt { 33 } }{ 8 } \right) \)

(i) Taking log on both sides, we get
p log x + q log y = (p + q) log (x + y)
Differentiating both sides w.r.t, x, we get
\(p \cdot \frac{1}{x}+q \cdot \frac{1}{y} \cdot \frac{d y}{d x}=(p+q) \cdot \frac{1}{x+y}\left(1+\frac{d y}{d x}\right)\)
\(\Rightarrow \frac{p}{x}-\frac{p+q}{x+y}=\frac{d y}{d x}\left[\frac{p+q}{x+y}-\frac{q}{y}\right]\)
\(\Rightarrow \frac{p x+p y-p x-q x}{x(x+y)}=\frac{d y}{d x}\left[\frac{p y+q y-q x-q y}{y(x+y)}\right]\)
\(\Rightarrow \frac{p y-q x}{x(x+y)}=\frac{p y-q x}{y(x+y)} \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{y}{x}\)
\(\text { (ii) } x \frac{d y}{d x}=y \Rightarrow x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=\frac{d y}{d x} \Rightarrow \frac{d^{2} y}{d x^{2}}=0\)

\(\therefore \ f\left( \frac { \pi }{ 4 } \right) =\frac { 1 }{ 2 } \)

Performing R₂ \(\rightarrow\) R₂ + 4R₁ and R₃ \(\rightarrow\) R₃ - 7R₁ we get
\(\left|\begin{array}{ll} 1 & 1 & \sin 3 \theta \\ 0 & 7 & \cos 2 \theta+4 \sin 3 \theta \\ 0 & -14 & -2-7 \sin 3 \theta \end{array}\right|=0 \)
\(\Rightarrow-14-49 \sin 3 \theta+14 \cos 2 \theta+56 \sin 3 \theta=0 \)
\(\Rightarrow \sin 3 \theta+2 \cos 2 \theta-2=0 \)
\(\Rightarrow 3 \sin \theta-4 \sin ^{3} \theta-4 \sin ^{2} \theta=0 \)
\(\Rightarrow \sin \theta\left[3-4 \sin ^{2} \theta-4 \sin \theta\right]=0 \)
\(\Rightarrow \sin \theta=0 \text { or } 4 \sin ^{2} \theta+4 \sin \theta-3=0 \)
\(\Rightarrow \theta=n \pi ; n \in Z \text { or } 4 \sin ^{2} \theta+6 \sin \theta-2 \sin \theta-3=0 \)
\(\Rightarrow \theta=n \pi ; n \in Z \text { or }(2 \sin \theta-1)(2 \sin \theta+3)=0 \)
\(\Rightarrow \theta=n \pi ; n \in Z \text { or } \sin \theta=\frac{1}{2} \)
\(\text { or } \sin \theta=-\frac{3}{2}\)
\(\text { Hence, } \theta=n \pi ; n \in Z \text { or } \theta=n \pi+(-1)^{n} \frac{\pi}{6}, n \in Z\)

\(A^{2}-4 A \left.+7 I=\left[\begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array}\right]\left[\begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array}\right]-4\left[\begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array}\right]+7 \mid \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\(=\left[\begin{array}{rr} 4-3 & 6+6 \\ -2-2 & -3+4 \end{array}\right]-\left[\begin{array}{rr} 8 & 12 \\ -4 & 8 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \)
\(=\left[\begin{array}{rr} 1 & 12 \\ -4 & 1 \end{array}\right]-\left[\begin{array}{rr} 8 & 12 \\ -4 & 8 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \)
\(=\left[\begin{array}{rr} 1-8+7 & 12-12+0 \\ -4+4+0 & 1-8+7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\)
\(\Rightarrow A^{2} =4 A-7 I \Rightarrow A^{5}=4 A^{4}-7 I A^{3} \)
\(A^{5} =4(4 A-7 I)(4 A-7 I)-7(4 A-7 I) A \)
\(=4\left(16 A^{2}-28 \mathrm{~A}-28 \mathrm{~A}+49 I^{2}\right)-28 A^{2}+49 A \)
\(=64 A^{2}-224 A+1961-28 A^{2}+49 A \)
\(=36 A^{2}-175 A+196 I \)
\(=36(4 A-7 I)-175 A+196 I \)
\(=144 A-252 I-175 A+196 I=-31 A-56 I \)
\(=-31\left[\begin{array}{rr} 2 & 3 \\ -1 & 2 \end{array}\right]-56\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\(=\left[\begin{array}{rr} -62 & -93 \\ 31 & -62 \end{array}\right]-\left[\begin{array}{rr} 56 & 0 \\ 0 & 56 \end{array}\right] \)
\(=\left[\begin{array}{cr} -62-56 & -93-0 \\ 31-0 & -62-56 \end{array}\right]=\left[\begin{array}{rr} -118 & -93 \\ 31 & -118 \end{array}\right]\)

\(\text { Let } E=\sin \left(2 \tan ^{-1} \frac{1}{3}\right)+\cos \left(\tan ^{-1} 2 \sqrt{2}\right.\)
\(\text { Now, } 2 \tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right) \) 
\(\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right] \)
\(\Rightarrow \quad 2 \tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{2 / 3}{8 / 9}\right)=\tan ^{-1}\left(\frac{3}{4}\right) \)
\(\text { Put } \tan ^{-1}\left(\frac{3}{4}\right)=\theta \Rightarrow \tan \theta=\frac{3}{4} \) 
\(\text { Then, } \sin \theta=\frac{3}{5} \Rightarrow \theta=\sin ^{-1}\left(\frac{3}{5}\right) \)
\(\Rightarrow \tan ^{-1}\left(\frac{3}{4}\right)=\sin ^{-1}\left(\frac{3}{4}\right)\) 
From Eqs. (i) and (ii), we get
2\( \tan ^{-1}\left(\frac{1}{3}\right)=\sin ^{-1}\left(\frac{3}{5}\right)\)
Now, put \(\tan ^{-1} 2 \sqrt{2}=\phi \Rightarrow \tan \phi=2 \sqrt{2}\)
Then, \(\cos \phi=\frac{1}{3} \Rightarrow \phi=\cos ^{-1}\left(\frac{1}{3}\right)\)
Now, \(\tan ^{-1} 2 \sqrt{2}=\cos ^{-1}\left(\frac{1}{3}\right)\)
\(\tan ^{-1}\left(\frac{3}{4}\right)=\sin ^{-1}\left(\frac{3}{5}\right)\) 
\( \therefore E =\sin \left(2 \tan ^{-1} \frac{1}{3}\right)+\cos \left(\tan ^{-1} 2 \sqrt{2}\right) \)
\(=\sin \left(\sin ^{-1} \frac{3}{5}\right)+\cos \left(\cos ^{-1} \frac{1}{3}\right) \)
\(=\frac{3}{5}+\frac{1}{3}=\frac{9+5}{15}=\frac{14}{15} \)

We know tan x is increasing for \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)\(\left[y=\tan x \Rightarrow y^{\prime}=\sec ^{2} x>0\right]\)
\(\text {Also } 1>\frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1}>\tan \frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1} >1>\frac{\pi}{4} \)
\(\Rightarrow \tan ^{-1}>1>\tan ^{-1}(1) . \)

For \(x_{ 1 },x_{ 2 }\in R,f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \) \(\frac { x_{ 1 } }{ x_{ 1 }^{ 2 }+1 } =\frac { x_{ 2 } }{ x_{ 2 }^{ 2 }+1 } \)
\(\Rightarrow \) \(x_{ 1 }x_{ 2 }^{ 2 }+x_{ 1 }=x_{ 2 }x_{ 1 }^{ 2 }+x_{ 2 }\)
\(\Rightarrow \) \(x_{ 1 }x_{ 2 }(x_{ 1 }-x_{ 2 })=x_{ 2 }-x_{ 1 }\)
\(\Rightarrow \) \(x_{ 1 }=x_{ 2 }\quad or\quad x_{ 1 }x_{ 2 }=1.\)
Here there are points \(x_{ 1 }\) and \(x_{ 2 }\) with \(x_{ 1 }\neq x_{ 2 }\) and \(f(x_{ 1 })=f(x_{ 2 })\).
\(\left[ For\ Ex.\ Take\ x_{ 1 }=2,x_{ 2 }=\frac { 1 }{ 2 } \right] \)
then \(f(x_{ 1 })=\frac { 2 }{ 4+1 } =\frac { 2 }{ 5 } \) and \(f(x_{ 2 })=\frac { 1/2 }{ 1+1/4 } =\frac { 2 }{ 5 } \)
Thus \(f(x_{ 1 })=f(x_{ 2 })=\frac { 2 }{ 5 } \) but \(x_{ 1 }\neq x_{ 2 }\quad i.e.f(x)=\frac { 1 }{ 2 } ]\)
Hence 'f' is not one-one.
Also 'f' is not onto.
For if so, then for \(1\in R\) , there exists \(x\in R\) such that f(x) = 1,
Which gives \(\frac { x_{ 2 } }{ x_{ 2 }^{ 2 }+1 } =1\)
But there is no such x in the domain R.
[ ∵ x² − x + 1 = 0 does not give any real value of x]   

Since, X = Number of four seen
On tossing two dice, X = 0, 1, 2
Also, \(P(4)=\frac{1}{10} \text { and } P(\operatorname{not} 4)=\frac{9}{10}\) 
So, \(P(X=0)=P(\operatorname{not} 4) \cdot P(\operatorname{not} 4)=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100}\) 
\(P(X=1)=P(\operatorname{not} 4) \cdot P(4)+P(4) \cdot P(\operatorname{not} 4)\) 
\(=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100}\) 
\(P(X=2)=P(4) \cdot P(4)=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100}\) 
Thus, we get the following table
\(\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & 81 / 100 & 18 / 100 & 1 / 100 \\ \hline \end{array}\) 
\(\begin{array}{c|c|c|c} \hline X P(X) & 0 & 18 / 100 & 2 / 100 \\ \hline X^{2} P(X) & 0 & 18 / 100 & 4 / 100 \\ \hline \end{array}\) 
\(\therefore \operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}=\Sigma X^{2} P(X)-\left[\sum X P(X)\right]^{2}\) 
\(=\left(0+\frac{18}{100}+\frac{4}{100}\right)-\left(0+\frac{18}{100}+\frac{2}{100}\right)^{2}\) 
\(=\frac{22}{100}-\left(\frac{20}{100}\right)^{2}=\frac{11}{50}-\frac{1}{25}\) 
\(=\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18\)

Given that L represents the number of litres of water in the pool, t seconds after the pool has been plugged off todrain, then 
\(L=200(10-t)^{2}\) 
ஃ Rate at which the water is running out
\(\frac{d L}{d t}=-200 \cdot 2(10-t) \cdot(-1)=400(10-t)\) 
[here, we take negative sign, because the water is decreasing]
Rate at which the water is running out at the end of 5s
= 400(10 - 5)= 2000 L /s = Final rate
Initial rate \((\text { at } t=0)=\left(\frac{d L}{d t}\right)_{t=0}=4000 \mathrm{~L} / \mathrm{s}\) 
\( \therefore \text { Average rate during } 5 \mathrm{~s} =\frac{\text { Initial rate }+\text { Final rate }}{2} \)
\(=\frac{4000+2000}{2}=3000 \mathrm{~L} / \mathrm{s} \) 
Hence, the water flows out during the first 5s withaverage rate of 3000 L / s

\(= tan^{-1}e-{\pi\over4}\)

Given, \(\vec{a}=\hat{i}-\hat{j}+7 \hat{k} \text { and } \vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}\)
Now, \(\vec{a}+\vec{b}=6 \hat{i}-2 \hat{j}+(7+\lambda) \hat{k}\)
and \(\vec{a}-\vec{b}=-4 \hat{i}+(7-\lambda) \hat{k}\)
\(\because(\vec{a}+\vec{b}) \text { and }(\vec{a}-\vec{b})\) are orthogonal.
\(\begin{aligned} \therefore & (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b}) =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & {[6 \hat{i}-2 \hat{j}+(7+\lambda) \hat{k}] \cdot[-4 \hat{i}+(7-\lambda) \hat{k}] } =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & -24+49-\lambda^2 =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & \lambda^2 =25 \end{aligned}\)
\(\begin{aligned} \Rightarrow & \lambda = \pm 5 \end{aligned}\)

\({2\over3}sin^{-1}\sqrt{x^3\over a^3}+c\)

\({2\over 3\sqrt2}tan^{-1}{x\over\sqrt2}+{1\over 6}log|{x-1\over x+1}|+c\)

\(log\ |{x^2+2\over\sqrt{x^2+1}}|+c\)

\({tan^{11}\ x\over 11}+{tan^9\ x\over 9}+c\)

Given curve, \(y^{2}=4 x\)   ..(i)
is a parabola having vertex (0, 0) and open right sides and the curve x² = 4y  ...(ii)
is a parabola having vertex (0, 0) and open upward.

Also, draw the square bounded by the lines x = 0,x = 4
y = 4 and y = 0.To prove, Area A₁ = Area A₂ = Area A ₃
Now, the point of intersection of curves y² = 4x and
\(x^{2}=4 y \text { is given by }\left(\frac{y^{2}}{4}\right)^{2}=4 y\) 
\(\Rightarrow y^{4}=64 y \Rightarrow y\left(y^{3}-64\right)=0\) 
\(\Rightarrow y=0, y^{3}=64 \Rightarrow y=0, y=4\) 
When \(y=0, \text { then } 0^{2}=4 x \Rightarrow x=0\) 
When \(y=4, \text { then } 4^{2}=4 x \Rightarrow x=4\) 
So, the points of intersection are 0(0, 0) and B(4, 4). 
Now, the area of the region bounded by curves y² = 4x and x² = 4y is
\(A_{2}=\int_{0}^{4}[y\{\text { parabola }(\mathrm{i})\}-y\{\text { parabola }(\mathrm{ii})\}] d x\) 
\(=\int_{0}^{4}\left(\sqrt{4 x}-\frac{x^{2}}{4}\right) d x=\int_{0}^{4}\left(2 \sqrt{x}-\frac{x^{2}}{4}\right) d x\) 
\( =\left[2 \times \frac{x^{3 / 2}}{3 / 2}-\frac{x^{3}}{12}\right]_{0}^{4}=\left[\frac{4}{3}(4)^{3 / 2}-\frac{4^{3}}{12}-(0-0)\right. \)
\(=\left[\frac{4}{3}(2)^{3}-\frac{64}{12}\right] \)
\(=\frac{32}{3}-\frac{16}{3}=\frac{16}{3} \mathrm{sq} \text { units } \) 
Now, the area of the region bounded by the curves 
\(x^{2}=4 y, x=4 \text { and } X \text { -axis is } A_{3}=\int_{0}^{4} y d x=\int_{0}^{4}\left(\frac{x^{2}}{4}\right) d x\) 
\( =\left[\frac{x^{3}}{12}\right]_{0}^{4}=\frac{(4)^{3}}{12}-0 \)
\(=\frac{16}{3} \text { sq units } \) 
Similarly, the area of the region bounded by the curve 
\(y^{2}=4 x, Y \text { -axis and } y=4 \text { is }\) 
\(A_{1}=\int_{0}^{4} x d y=\int_{0}^{4} \frac{y^{2}}{4} d y=\left[\frac{y^{3}}{12}\right]_{0}^{4}=\frac{1}{12}\left[4^{3}-0\right]\) 
\(=\frac{1}{12}(64)=\frac{16}{3} \text { sq units }\) 
Here, we see that
\(A_{1}=A_{2}=A_{3}=\frac{16}{3} \mathrm{sq} \text { units }\) 
Hence proved.