By formula of conditional probability, we know that
\( \mathrm{P}\left(\mathrm{E}_i \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{A} \cap \mathrm{E}_i\right)}{\mathrm{P}(\mathrm{A})}\)
\( =\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{A|E}_i\right)}{\mathrm{P}(\mathrm{A})} \)(by multiplication rule of probability)
\(=\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{A|E}_i\right)}{\sum_{j=1}^n \mathrm{P}\left(\mathrm{E}_j\right) \mathrm{P}\left(\mathrm{A|E}_j\right)}\) (by the result of theorem of total probability)

\(\left(x^{x}\right)^{x}(x+2 x \log x)\)

Let y = a^x_.
Taking log on both sides, we get log y = x log a
On differentiating both sides w.r.t. x, we get
\(\frac{1}{y} \frac{d y}{d x}=\log a \Rightarrow \frac{d y}{d x}=y \log a\) 
Thus,\(\frac{d}{d x}\left(a^{x}\right)=a^{x} \log a\)

If E and F are independent events, then \(P\left( E\cap F \right) =P(E)\times P(F)\)
Now \(P(E\cup \bar { F } )=P(E)+P(F)-P(E\cap F)\)
\(\Rightarrow P(E\cup F)=P(E)+P(F)-P(E)\times P(F)\)
\(\Rightarrow P(E\cup F)=\frac { 1 }{ 3 } +\frac { 1 }{ 4 } -\frac { 1 }{ 3 } \times \frac { 1 }{ 4 } \)
\(\Rightarrow P(E\cup F)=\frac { 1 }{ 3 } +\frac { 1 }{ 4 } -\frac { 1 }{ 12 } =\frac { 4+3-1 }{ 12 } \)
\(\Rightarrow P(E\cup F)=\frac { 6 }{ 12 } \)
\(\therefore P(E\cup F)=\frac { 1 }{ 2 } \)

\(L.H.S.={ sin }^{ -1 }\frac { 5 }{ 13 } +{ cos }^{ -1 }\frac { 3 }{ 5 } \)
\(\because { sin }^{ -1 }x={ tan }^{ -1 }\left( \frac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) \)
\({ sin }^{ -1 }\frac { 5 }{ 13 } ={ tan }^{ -1 }\left( \frac { 5/13 }{ \sqrt { 1-25/169 } } \right) \)
\(={ tan }^{ -1 }\frac { (5/13) }{ \sqrt { \frac { 144 }{ 169 } } } ={ tan }^{ -1 }\frac { 5 }{ 12 } \)
\(and\quad { cos }^{ -1 }x={ tan }^{ -1 }\left( \frac { \sqrt { 1-{ x }^{ 2 } } }{ x } \right) \)
\(\therefore \quad { cos }^{ -1 }\frac { 3 }{ 5 } ={ tan }^{ -1 }\left( \frac { \sqrt { 1-\frac { 9 }{ 25 } } }{ \frac { 3 }{ 5 } } \right) \)
\(={ tan }^{ -1 }\left( \frac { \sqrt { \frac { 16 }{ 25 } } }{ \frac { 3 }{ 5 } } \right) ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
\({ tan }^{ -1 }\frac { 5 }{ 12 } +{ tan }^{ -1 }\frac { 4 }{ 3 } ={ tan }^{ -1 }\left[ \frac { \frac { 5 }{ 12 } +\frac { 4 }{ 2 } }{ 1-\frac { 5\times 4 }{ 12\times 3 } } \right] \)
\(\left[ { tan }^{ -1 }x+{ tan }^{ -1 }y={ tan }^{ -1 }\frac { x+y }{ 1-xy } \right] \)
\(={ tan }^{ -1 }\left| \frac { \frac { 15+48 }{ 36 } }{ \frac { 36-20 }{ 36 } } \right| \)
\(={ tan }^{ -1 }\left( \frac { 63 }{ 16 } \right) =R.H.S.\)

Given R=[(a,a² ) : a is a prime number less than 5}
⇒ R =[(2,8),(3,27)]
Range=[8,27]

f'(x) 3x²- 6x + 3 = 3(x² - 2x + 1)
= 3 (x - 1)² > 0
Hence, function is increasing in R.

Let \(u=\sec ^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)\)
On putting \(x=\sin \theta\)
and let \(v=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)\)
On putting \(x=\sin \theta\)
\(\Rightarrow \theta=\sin ^{-1} x\)
\( \therefore u=\sec ^{-1}\left(\frac{1}{\sqrt{1-\sin ^2 \theta}}\right)=\sec ^{-1}\left(\frac{1}{\sqrt{\cos ^2 \theta}}\right) \)
\(=\sec ^{-1}(\sec \theta)=\theta=\sin ^{-1} x\)
\(\therefore \frac{d u}{d x}=\frac{1}{\sqrt{1-x^2}}\)
\(\Rightarrow \quad \theta=\sin ^{-1} x \)
\(\therefore \quad v=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right)\)
\(=\sin ^{-1}(2 \sin \theta \cdot \cos \theta) \)
\(=\sin ^{-1}(\sin 2 \theta)=2 \theta=2 \sin ^{-1} x\)
\(\therefore \frac{d v}{d x}=\frac{d}{d x}\left(2 \sin ^{-1} x\right)=\frac{2}{\sqrt{1-x^2}}\)
\(\therefore \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{1 / \sqrt{1-x^2}}{2 / \sqrt{1-x^2}} \Rightarrow \frac{d u}{d v}=\frac{1}{2}\)

Not reflexive, as (1, 1) \(\notin\) R.
Not symmetric, as (2, 3) \(\in\) R but (3, 2) \(\notin\) R.
Transitive, as relation R in a non empty set containing one element is transitive

Let A be the event of getting an odd number on the first die and B be the event of getting a multiple of 3 on the second die.
On throwing two dice, total outcomes = 36
Outcomes favourable to \(A \cap B\) i.e an odd number on first die and a multiple of 3 on the other die are 
\( \{(1,3),(1,6),(3,3),(3,6),(5,3),(5,6)\} \therefore \quad n(A \cap B)=6 \) 
Then, required probability \(P(A \cap B)=\frac{6}{36}=\frac{1}{6}\) 
Now, A = {1, 3, 5}, B = {3, 6}
So, n (A) = 3, n (B) = 2
\(\therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{3}{36}=\frac{1}{12}\) 
and \(P(B)=\frac{2}{36}=\frac{1}{18}\) 
Here, \(P(A) \cdot P(B)=\frac{1}{12} \times \frac{1}{18}=\frac{1}{216} \neq P(A \cap B)\) 
Hence, A and B are dependent events.

\(\text { Given: } P(E)=0.6, P(F)=0.3, P(E \cap F)=0.2 \)
\(P(E \mid F)=\frac{P(E \cap F)}{P(E)}=\frac{0.2}{0.3}=\frac{2}{3} \)
\(P(F \mid E)=\frac{P(E \cap F)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}\)

 We have, \(\cos \left(\tan ^{-1} \frac{3}{4}\right)\)
Let \(x=\tan ^{-1} \frac{3}{4}, then \tan x=\frac{3}{4}\)
\(\because \sec x=\sqrt{1+\tan ^{2} x}=\sqrt{1+\left(\frac{3}{4}\right)^{2}}=\sqrt{1+\frac{9}{16}} \)\(=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(\Rightarrow \cos x=\frac{4}{5} \quad\left[\because \cos x=\frac{1}{\sec x}\right]\)
\(\Rightarrow x=\cos ^{-1}\left(\frac{4}{5}\right)\)
\(\Rightarrow \quad \tan ^{-1}\left(\frac{3}{4}\right)=\cos ^{-1}\left(\frac{4}{5}\right) \quad\left[\because x=\tan ^{-1} \frac{3}{4}\right]\\ \therefore \cos \left(\tan ^{-1} \frac{3}{4}\right)=\cos \left(\cos ^{-1} \frac{4}{5}\right)=\frac{4}{5} \)
\( \left[\because \cos \left(\cos ^{-1} x\right)=x, \forall x \in[-1,1]\right]\)

X=Y

\(P(\text { head })=4 P(\text { tail }) \Rightarrow P(H)=\frac{4}{5}, P(T)=\frac{1}{5}\) 
\(\begin{array}{c|c|c|c|c} \hline \begin{array}{c} X \\ \text { (Number of tails) } \end{array} & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{64}{125} & \frac{48}{125} & \frac{12}{125} & \frac{1}{125} \\ \hline X P(X) & 0 & \frac{48}{125} & \frac{24}{125} & \frac{3}{125} \\ \hline X^{2} P(X) & 0 & \frac{48}{125} & \frac{48}{125} & \frac{9}{125} \\ \hline \end{array}\) 
\(\therefore \text { Mean }=\Sigma X P(X)=\frac{75}{125}\) 
\(\text { Variance }=\Sigma X^{2} P(X)-\left[\sum X P(X)\right]^{2}=\frac{105}{125}-\frac{9}{25}\) 
\(\text { Mean }=\frac{3}{5}, \text { variance }=\frac{12}{25}\)

Given, \(f(x)=\frac{3}{10} x^{4}-\frac{4}{5} x^{3}-3 x^{2}+\frac{36 x}{5}+11\) 
On differentiating both sides w.r.t. x, we get
\(f^{\prime}(x)=\frac{12 x^{3}}{10}-\frac{12 x^{2}}{5}-6 x+\frac{36}{5}+0\) 
Put \(f^{\prime}(x)=0\) 
\(\Rightarrow \frac{12 x^{3}}{10}-\frac{12 x^{2}}{5}-6 x+\frac{36}{5}=0\)
\(\Rightarrow \frac{6 x^{3}-12 x^{2}-30 x+36}{5}=0\)
\(\Rightarrow\)\(x^{3}-2 x^{2}-5 x+6=0\left[\right. divide by \left.\frac{6}{5}\right]\)
\(\Rightarrow (x-1)\left(x^{2}-x-6\right)=0\)
\(\Rightarrow (x-1)(x+2)(x-3)=0\)
\(\Rightarrow\) x - 1 = 0
x + 2 = 0 or x - 3 = 0
x = -2, 1, 3
Now, we find the intervals in which f(x) is strictly increasing or strictly decreasing.
\(\begin{array}{ccc} \hline \text { Interval } & \begin{array}{c} \text { Sign of } f^{\prime}(x) \\ f^{\prime}(x)=(x-1)(x+2)(x-3) \end{array} & \begin{array}{c} \text { Nature of } \\ \text { function } \end{array} \\ \hline(-\infty,-2] & (-)(-)(-)=(-)<0 & \begin{array}{l} \text { Strictly } \\ \text { decreasing } \end{array} \\ \hline[-2,1] & (-)(+)(-)=(+)>0 & \text { Strictly increasing } \\ \hline[1,3] & (+)(+)(-)=(-)<0 & \begin{array}{l} \text { Sthictly } \\ \text { decreasing } \end{array} \\ \hline[3, \infty) & (+)(+)(+)=(+)>0 & \text { Strictly increasing } \\ \hline \end{array}\) 
(a) f(x) is strictly increasing in the interval \((-2,1) \cup(3, \infty)\) 
(b) f(x) is strictly decreasing in the interval \((-\infty,-2) \cup(1,3)\)

Here g(x) = x - [x]
Let a be an integer and h is very small h > 0, then
\( [a-h]=a-1,[a+h]=a \)
\(\text {and } [a]=a \) 
\(\text {At } x=a, \mathrm{LHL}=\lim _{x \rightarrow a^{-}} g(x)=\lim _{x \rightarrow a^{-}}(x-[x])\)
\(Put x=a-h ; when x \rightarrow a^{-}, then\ h \rightarrow 0\)
\( \mathrm{LHL} =\lim _{h \rightarrow 0}(a-h-[a-h]) \)
\(=\lim _{h \rightarrow 0}(a-h-(a-1)) \)

y = a sin t
\(\frac { dy }{ dt } =a\quad cost\)
\(x=a\left( -sint+\frac { sec^{ 2 }\frac { t }{ 2 } }{ 2tan\frac { 1 }{ 2 } } \right) \)
\(\Rightarrow \frac { dx }{ dt } =a\left( -sint+\frac { 1 }{ sint } \right) \)
\(\Rightarrow \frac { dx }{ dt } =\frac { a(1-sin^{ 2 }t) }{ sint } \)
\(=\frac { acos^{ 2 }t }{ sint } \)
\(\frac { dy }{ dx } =\frac { dy }{ dt } x\frac { dt }{ dx } \)
=acost \(\times \frac { sint }{ acos^{ 2 }t } =tant\)
\(\frac { d^{ 2 }y }{ dx^{ 2 } } =sec^{ 2 }\times \frac { dt }{ dx } \)
\(\frac { sect\times sint }{ acost } =\frac { 1 }{ a } sec^{ 3 }t-tant\)
\(\left\lfloor \frac { d^{ 2 }y }{ dx^{ 2 } } \right\rfloor =\frac { 1 }{ a } sec^{ 3 }\frac { \pi }{ 3 } tan\frac { \pi }{ 3 } \)
\(=\frac { 1 }{ a } 8\sqrt { 3 } =\frac { 8\sqrt { 3 } }{ a } \)

Let (a, a) \(\in \)  R  and a are members of joint family, \(\forall a\in A\)
\(\therefore\) R is reflexive. 
Let (a, b) \(\in \) R  \(\rightarrow \) a and b are members of joint family
\(\Rightarrow \) b and a are members of joint family
\(\Rightarrow (b,a)\in R,\forall a,b\in A\)
R is symmetric
Let  (a, b) \(\in \)  R \(\Rightarrow \)  a and b are members of joint family ...(i)
And (b, c) \(\in \)  R \(\Rightarrow \)   band c are members of joint family ...(ii)
from (i) and (ii) \(\Rightarrow \)   a, band c are members of joint family
\(\Rightarrow (a,c)\in R,\forall a,b,c\in A\)
R is transitive
R is an equivalence relation
Values:
(i) Love and concern for Grandparents,
(ii) Respect for Grandparents,
(iii) Tolerance

\(\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1\)
\(\Rightarrow \sin \left(\sin ^{-1} \frac{1}{5}\right) \cos \left(\cos ^{-1} x\right)+\cos \left(\sin ^{-1} \frac{1}{5}\right) \sin \left(\cos ^{-1} x\right)=1\)
\([\sin (A+B)=\sin A \cos B+\cos A \sin B]\)
\(\Rightarrow \frac{1}{5} \times x+\cos \left(\sin ^{-1} \frac{1}{5}\right) \sin \left(\cos ^{-1} x\right)=1\)
\(\Rightarrow \frac{x}{5}+\cos \left(\sin ^{-1} \frac{1}{5}\right) \sin \left(\cos ^{-1} x\right)=1\)
\(\text {Now let } \sin ^{-1} \frac{1}{5}=y \)
\(\text {Then, } \sin y=\frac{1}{5} \Rightarrow \cos y=\sqrt{1-\left(\frac{1}{5}\right)^{2}}=\frac{2 \sqrt{6}}{5} \Rightarrow y=\cos ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)\)
\(\therefore \sin ^{-1} \frac{1}{5}=\cos ^{-1}\left(\frac{2 \sqrt{6}}{5}\right) \ldots .(2)\)
\(\text {Let } \cos ^{-1} x=z\)
\(\text {Then } \cos z=x \Rightarrow \sin z=\sqrt{1-x^{2}} \Rightarrow z=\sin ^{-1}\left(\sqrt{1-x^{2}}\right)\)
\(\therefore \cos ^{-1} x=\sin ^{-1}\left(\sqrt{1-x^{2}}\right)\)
 From 1, 2 and 3 we have
\(\frac{x}{5}+\cos \left(\frac{\cos ^{-1}(2 \sqrt{6})}{5}\right) \cdot \sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)=1 \)
\(\Rightarrow \frac{x}{5}+\frac{2 \sqrt{6}}{5} \cdot \sqrt{1-x^{2}}=1\)
\(\Rightarrow x+2 \sqrt{6} \sqrt{1-x^{2}}=5\)
\(=2 \sqrt{6} \sqrt{1-x^{2}}=5-x\)
On squaring both sides, we get:
\((4)(6)\left(1-x^{2}\right)=25+x^{2}-10 x\)
\(\Rightarrow 24-24 x^{2}=25+x^{2}-10 x\)
\(\Rightarrow 25 x^{2}-10 x+1=0\)
\(\Rightarrow(5 x-1)^{2}=0\)
\(\Rightarrow(5 x-1)=0\)
\(\Rightarrow x=\frac{1}{5}\)
Hence, the value of  x  is \( \frac{1}{5}\)

\( \tan \left(\cos ^{-1} \frac{3}{5}+\tan ^{-1} \frac{1}{4}\right) \)
\(\text { Let } \cos ^{-1} \frac{3}{5}=x \Rightarrow \cos x=\frac{3}{5} \)
\(\therefore \ \tan x=\frac{4}{3} \Rightarrow x=\tan ^{-1} \frac{4}{3} \)
\(\therefore \ \tan \left(\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{4}\right) \)
\(\therefore \ \tan \left[\tan ^{-1}\left\{\frac{\left(\frac{4}{3}+\frac{1}{4}\right)}{1-\frac{4}{3} \times \frac{1}{4}}\right\}\right]=\frac{19}{8}\)

One-one onto mapping is possible only ifn(A) = n(B)

(i) (b) : Given, perimeter of window = 10 m 
\(\therefore\) x + y + y + perimeter of semicircle = 10
\(\Rightarrow x+2 y+\pi \frac{x}{2}=10\) 
(ii) (b) : \(A=x \cdot y+\frac{1}{2} \pi\left(\frac{x}{2}\right)^{2}\) 
\(=x\left(5-\frac{x}{2}-\frac{\pi x}{4}\right)+\frac{1}{2} \frac{\pi x^{2}}{4}\left[\because \text { From }(\mathrm{i}), y=5-\frac{x}{2}-\frac{\pi x}{4}\right]\) 
\(=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{4}+\frac{\pi x^{2}}{8}=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{8}\) 
(iii) (c) : We have, \(A=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{8}\) 
\(\Rightarrow \quad \frac{d A}{d x}=5-x-\frac{\pi x}{4}\) 
Now, \(\frac{d A}{d x}=0 \Rightarrow 5=x+\frac{\pi x}{4}\) 
\(\Rightarrow x(4+\pi)=20 \Rightarrow x=\frac{20}{4+\pi}\) 
\(\left[\text { Clearly, } \frac{d^{2} A}{d x^{2}}<0 \text { at } x=\frac{20}{4+\pi}\right]\) 
(iv) (d) : At \(x=\frac{20}{4+\pi}\) 
 \(A=5\left(\frac{20}{4+\pi}\right)-\left(\frac{20}{4+\pi}\right)^{2} \frac{1}{2}-\frac{\pi}{8}\left(\frac{20}{4+\pi}\right)^{2}\) 
\(=\frac{100}{4+\pi}-\frac{200}{(4+\pi)^{2}}-\frac{50 \pi}{(4+\pi)^{2}}\) 
\(=\frac{(4+\pi)(100)-200-50 \pi}{(4+\pi)^{2}}=\frac{400+100 \pi-200-50 \pi}{(4+\pi)^{2}}\) 
\(=\frac{200+50 \pi}{(4+\pi)^{2}}=\frac{50(4+\pi)}{(4+\pi)^{2}}=\frac{50}{4+\pi}\) 
(v) (a) : We have, \(y=5-\frac{x}{2}-\frac{\pi x}{4}=5-x\left(\frac{1}{2}+\frac{\pi}{4}\right)\) 
\(=5-x\left(\frac{2+\pi}{4}\right)=5-\left(\frac{20}{4+\pi}\right)\left(\frac{2+\pi}{4}\right)\) 
\(=5-5 \frac{(2+\pi)}{4+\pi}=\frac{20+5 \pi-10-5 \pi}{4+\pi}=\frac{10}{4+\pi}\)

(i) (a) : Clearly (1, 1), (2, 2), (3, 3), \(\in\) R. So, R is reflexive on A.
Since, \((1,2) \in R \text { but }(2,1) \notin R\) So, R is not symmetric on A.
Since, \((2,3), \in R\) and \((3,1) \in R\) but \((2,1) \notin R\) .So, R is not transitive on A.
(ii) (b) : Since, (1,1), (2, 2) and (3, 3) are not in R. So, R is not reflexive on A.
Now, \((1,2) \in R \Rightarrow(2,1) \in R\) 
and \((1,3) \in R \Rightarrow(3,1) \in R\) 
So, R is symmetric
Clearly,\((1,2) \in R \text { and }(2,1) \in R \text { but }(1,1) \notin R\) 
So, R is not transitive on A.
(iii) (c) : We have, \(R=\{(x, y): y=x+5 \text { and } x<4\}\) ,where \(x, y \in N\) .
\(\therefore R=\{(1,6),(2,7),(3,8)\}\) 
Clearly, (1, 1), (2, 2) etc. are not in R. So, R is not reflexive.
Since, \((1,6) \in R\) but \((6,1) \notin R\) So, R is not symmetric. 
Since, \((1,6) \in R\) R and there is no order pair in R which has 6 as the first element. Same is the case for (2, 7) and (3, 8). So, R is transitive.
(iv) (d) : We have,R = {(x, y) : 3x - y = 0}, where \(x, y \in A=\{1,2, \ldots \ldots, 14\}\) .
\(\therefore\) R = {(I, 3), (2, 6), (3, 9), (4, 12)}
Clearly,\((1,1) \notin R\) So, R is not reflexive on A.
Since, \((1,3) \in R\) but \((3,1) \notin R\) .So, R is not symmetric on A.
Since, \((1,3) \in R\) and \((3,9) \in R\)  but  \((1,9) \notin R\) So, R is not transitive on A.
(v) (d) : Clearly, (1, 1), (2, 2), (3, 3) ∈  R. So, R is reflexive on A.
We find that the ordered pairs obtained by interchanging the components of ordered pairs in R are also in R. So, R is symmetric on A.
For \(1,2,3 \in A\)  such that (1, 2) and (2, 3) are in Rimplies that (1, 3) is also, in R. So, R is transitive on A. Thus, R is an equivalence relation.

(c) Assertion (A) Maximum value of \(\cos ^{-1} x\) is \(\pi\), which occurs at x = -1
Therefore Maximum value of \(\left(\cos ^{-1} x\right)^2\) is \(\pi^2\).
So, assertion is correct.
Reason (R) The principal value branch of \(\cos ^{-1} x\) is [0, \(\pi\)]
So, reason is incorrect.

(b) Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion