WBJEE 2016 Paper 1 (Mathematics) Question Paper with Solutions

Given,

$𝑥\frac{𝑑𝑦}{𝑑𝑥}+𝑦=𝑥{𝑒}^{𝑥}$

$\Rightarrow \frac{𝑑𝑦}{𝑑𝑥}+\frac{𝑦}{𝑥}={𝑒}^{𝑥}$ .... (i)

On comparing Eq. (i) by $\frac{𝑑𝑦}{𝑑𝑥}+𝑃𝑦=𝑄$, we get

$𝑃=\frac{1}{𝑥}$ and $𝑄={𝑒}^{𝑥}$

$\therefore$ $𝐼𝐹={𝑒}^{\int \frac{1}{𝑥}𝑑𝑥}$

$={𝑒}^{\log 𝑥}=𝑥$

Hence, solution of differential equation,

$𝑦.𝑥=\int 𝑥{𝑒}^{𝑥}𝑑𝑥+𝐶$

$\Rightarrow 𝑥𝑦=𝑥{𝑒}^{𝑥}-\int {𝑒}^{𝑥}𝑑𝑥+𝐶$

$\Rightarrow 𝑥𝑦=𝑥{𝑒}^{𝑥}-{𝑒}^{𝑥}+𝐶$

$\Rightarrow 𝑥𝑦={𝑒}^{𝑥}(𝑥-1)+𝐶$ .... (ii)

$\because$ $𝑥𝑦={𝑒}^{𝑥}𝜙\left(𝑥\right)+𝐶$ [given] ... (iii)

On comparing Eqs. (ii) and (iii), we get

$𝜙\left(𝑥\right)=(𝑥-1)$

The general equation of the parabola is x = ay² + b, where a and b are arbitrary constants.

So, the differential equation is of order 2.

The equation of the line is y = x + $𝜆$ ....... (i)

On comparing it with y = mx + c, we get,

m = 1 and c = $𝜆$.

The equation of the ellipse is

2x² + 3y² = 1

or $\frac{{𝑥}^2}{{\left(\frac{1}{\sqrt{\surd 2}}\right)}^2}+\frac{{𝑦}^2}{{\left(\frac{1}{\sqrt{\surd 3}}\right)}^2}=1$

On comparing it with $\frac{{𝑥}^2}{{𝑎}^2}+\frac{{𝑦}^2}{{𝑏}^2}=1$, we get

${𝑎}^2=\frac{1}{2}$ and ${𝑏}^2=\frac{1}{3}$

If the line touches the ellipse, then

${𝑐}^2={𝑎}^2{𝑚}^2+{𝑏}^2\Rightarrow {𝜆}^2=\frac{1}{2}.1+\frac{1}{3}=\frac{5}{6}$

$\Rightarrow {𝜆}^2=\frac{5}{6}\Rightarrow 𝜆=\sqrt{\surd \frac{5}{6}}$

The graphs of $𝑦=|𝑥-1|$ and $𝑦=\sqrt{\surd 5-{𝑥}^2}$ are shown in the figure and the shaded region is the required region bounded by the two curves.



Let A be the area bounded by the given curves. Then,

$𝐴={\int }_1^{-1}(\sqrt{\surd 5-{𝑥}^2}+𝑥-1)𝑑𝑥+{\int }_2^1(\sqrt{\surd 5-{𝑥}^2}+𝑥+1)𝑑𝑥$

$={\int }_1^{-1}\sqrt{\surd 5-{𝑥}^2}𝑑𝑥+{\int }_2^1\sqrt{\surd 5-{𝑥}^2}𝑑𝑥+{\int }_1^{-1}(𝑥-1)𝑑𝑥+{\int }_2^1(-𝑥+1)𝑑𝑥$

$={\int }_2^{-1}\sqrt{\surd 5-{𝑥}^2}𝑑𝑥+{\left[\frac{{𝑥}^2}{2}-𝑥\right]}_1^{-1}+{\left[-\frac{{𝑥}^2}{2}+𝑥\right]}_2^1$

$={\left[\frac{1}{2}\times \sqrt{\surd 5-{𝑥}^2}+\frac{5}{2}{\sin }^{-1}\frac{𝑥}{\sqrt{\surd 5}}\right]}_2^{-1}-\frac{5}{2}$

$=\frac{1}{2}+\frac{5}{2}{\sin }^{-1}\frac{2}{\sqrt{\surd 5}}+1+\frac{5}{2}{\sin }^{-1}\frac{1}{\sqrt{\surd 5}}-\frac{5}{2}$

$=-\frac{1}{2}+\frac{5}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{\surd 5}}\times \sqrt{\surd 1-\frac{1}{5}}+\frac{1}{\sqrt{\surd 5}}\sqrt{\surd 1-\frac{4}{5}}\right)$

$=-\frac{1}{2}+\frac{5}{2}{\sin }^{-1}\left(1\right)=\frac{5𝜋}{2}-\frac{1}{2}$ sq units

Sum of distances of point P from point (8, 0) and (0, 12) will be minimum, if points are collinear.

$\therefore$ Equation of line at point (8, 0) and (0, 12).

$\frac{𝑥}{8}+\frac{𝑦}{12}=1$

$\Rightarrow \frac{𝑦}{12}=1-\frac{𝑥}{8}$

$\Rightarrow 𝑦=12-\frac{𝑥}{8}\times 12$

$\Rightarrow 𝑦=12-\frac{3}{2}𝑥$

$\therefore$ Points (x, y) = (2, 9), (4, 6) and (6, 3). Hence, the number of such points P in S is 3.

Given, $𝑇=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}$

On differentiating w.r.t. I, we get

$\frac{𝑑𝑇}{𝑑𝑙}=\frac{2𝜋}{\sqrt{\surd 𝑔}}.\frac{1}{2\sqrt{\surd 1}}$

$\therefore$ $\Delta 𝑇=\frac{𝑑𝑇}{𝑑𝑙}\Delta 𝐼=\frac{𝜋}{\sqrt{\surd 𝑔𝑙}}.\left(\frac{2𝐼}{100}\right)$

$=2𝜋\sqrt{\surd \frac{𝑙}{𝑔}}.\frac{1}{100}=\frac{𝑇}{100}$

$\Rightarrow \frac{\Delta 𝑇}{𝑇}=\frac{1}{100}=1\%$

Hence, approximate change in the time period is 1%.

Let the direction ratio's of two diagonals of a cube are (1, 1, 1) and ($-$ 1, 1, 1).

$\therefore$ $\cos 𝜃=\frac{{𝑎}_1{𝑎}_2+{𝑏}_1{𝑏}_2+{𝑐}_1{𝑐}_2}{\sqrt{\surd {𝑎}_2^1+{𝑏}_2^1+{𝑐}_2^1}\sqrt{\surd {𝑎}_2^2+{𝑏}_2^2+{𝑐}_2^2}}$

$=\frac{\left(1\right)\times (-1)+\left(1\right)\times \left(1\right)+\left(1\right)\times \left(1\right)}{\sqrt{\surd {\left(1\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}\sqrt{\surd {(-1)}^2+{\left(1\right)}^2+{\left(1\right)}^2}}$

$=\frac{-1+1+1}{\sqrt{\surd 3}\sqrt{\surd 3}}=\frac{1}{3}$

Given, log_ax = log_bx and log_cx are in AP.

$\therefore$ 2log_bx = log_ax + log_cx

$\Rightarrow 2{\log }_{𝑏}𝑥=\frac{1}{{\log }_{𝑥}𝑎}+\frac{1}{{\log }_{𝑥}𝑐}$

$\Rightarrow \frac{2}{{\log }_{𝑥}𝑏}=\frac{{\log }_{𝑥}𝑎𝑐}{{\log }_{𝑥}𝑎{\log }_{𝑥}𝑐}$

$\Rightarrow 2{\log }_{𝑥}𝑐=\frac{{\log }_{𝑥}𝑏}{{\log }_{𝑥}𝑎}\left({\log }_{𝑥}𝑎𝑐\right)$

$\Rightarrow 2{\log }_{𝑥}𝑐={\log }_{𝑎}𝑏.{\log }_{𝑥}𝑎𝑐$

$\Rightarrow {\log }_{𝑥}{𝐶}^2={\log }_{𝑥}(𝑎𝑐{)}^{{\log }_{𝑎}𝑏}$

$\therefore$ ${𝑐}^2=(𝑎𝑐{)}^{{\log }_{𝑎}𝑏}$

We have,

$1+(1+𝑎)𝑥+(1+𝑎+{𝑎}^2){𝑥}^2+...\infty$

$=\sum _{𝑛=1}^{\infty }(1+𝑎+{𝑎}^2+...+{𝑎}^{𝑛-1}){𝑥}^{𝑛-1}$

$=\sum _{𝑛=1}^{\infty }\left(\frac{1-{𝑎}^{𝑛}}{1-𝑎}\right){𝑥}^{𝑛-1}$

$=\sum _{𝑛=1}^{\infty }\frac{{𝑥}^{𝑛-1}}{1-𝑎}-\sum _{𝑛=1}^{\infty }\frac{{𝑎}^{𝑛}{𝑥}^{𝑛-1}}{1-𝑎}$

$=\frac{1}{1-𝑎}\sum _{𝑛=1}^{\infty }{𝑥}^{𝑛-1}-\frac{𝑎}{1-𝑎}\sum _{𝑛=1}^{\infty }{\left(𝑎𝑥\right)}^{𝑛-1}$

$=\frac{1}{1-𝑎}\left[\right(1+𝑥+{𝑥}^2+...\infty \left)\right]-\frac{𝑎}{1-𝑎}\left[\right\{1+𝑎𝑥+\left(𝑎{𝑥}^2\right)+...\infty \left\}\right]$

$=\frac{1}{(1-𝑎)}.\frac{1}{(1-𝑥)}-\frac{𝑎}{(1-𝑎)(1-𝑎𝑥)}$

$=\frac{1}{(1-𝑎𝑥)(1-𝑥)}$

Given, ${\log }_{0.3}(𝑥-1)<{\log }_{0.09}(𝑥-1)$

$\Rightarrow 𝑙𝑜{𝑔}_{0.3}(𝑥-1)<{\log }_{{\left(0.3\right)}^2}(𝑥-1)$

$\Rightarrow {\log }_{0.3}(𝑥-1{)}^2<{\log }_{0.3}(𝑥-1)$

$\Rightarrow (𝑥-1{)}^2>𝑥-1$ [$\because$ $0.3<1$]

$\Rightarrow {𝑥}^2+1-2𝑥-𝑥+1>0$

$\Rightarrow {𝑥}^2-3𝑥+2>0$

$\Rightarrow (𝑥-1)(𝑥-2)>0$

$\Rightarrow 𝑥<1,𝑥>2\Rightarrow 𝑥>2$ [Image]

$\therefore$ $(2,\infty )$

Hence, x lies in the interval $(2,\infty )$.

We have,

$\sum _{𝑛=1}^{13}({𝑖}^{𝑛}+{𝑖}^{𝑛+1})$

$\sum _{𝑛=1}^{13}{𝑖}^{𝑛}+\sum _{𝑛=1}^{13}{𝑖}^{𝑛+1}$

$𝑖\left(\frac{1-{𝑖}^{13}}{1-𝑖}\right)+{𝑖}^2\left(\frac{1-{𝑖}^{13}}{1-𝑖}\right)$

$=𝑖\left(\frac{1-{𝑖}^{13}}{1-𝑖}\right)+{𝑖}^2\left(\frac{1-{𝑖}^{13}}{1-𝑖}\right)$

$=𝑖\left(\frac{1-𝑖}{1-𝑖}\right)-\left(\frac{1-𝑖}{1-𝑖}\right)$

$=𝑖-1$

We have,

$\left|{𝑧}_1\right|=\left|{𝑧}_2\right|=\left|{𝑧}_3\right|=1$

$\Rightarrow |{𝑧}_1{|}^2=|{𝑧}_2{|}^2=|{𝑧}_3{|}^2=1$

$\Rightarrow {𝑧}_1{\overline{𝑧}}_1={𝑧}_2{\overline{𝑧}}_2={𝑧}_3{\overline{𝑧}}_3=1$

$\Rightarrow \frac{1}{{𝑧}_1}={\overline{𝑧}}_1,\frac{1}{{𝑧}_2}={\overline{𝑧}}_2,\frac{1}{{𝑧}_3}={\overline{𝑧}}_3$

Now, $\mid \frac{1}{{𝑧}_1}+\frac{1}{{𝑧}_2}+\frac{1}{{𝑧}_3}\mid =1$

$\Rightarrow \left|{\overline{𝑧}}_1+{\overline{𝑧}}_2+{\overline{𝑧}}_3\right|=1$

$\Rightarrow \left|\overline{{𝑧}_1+{𝑧}_2+{𝑧}_3}\right|=1$

$\therefore$ $|{𝑧}_1+{𝑧}_2+{𝑧}_3|=1$

Given, p, q are roots of equation x² + px + q = 0

$\therefore$ p + q = $-$ p $\Rightarrow$ 2p + q = 0 .... (i)

and pq = q

$\Rightarrow$ p = 1

Put p = 1 in Eq. (i), we get

2 $\times$ 1 + q = 0

$\Rightarrow$ q + 2 = 0

$\Rightarrow$ q = $-$ 2

Hence, p = 1, q = $-$2

Given equation is

x² $-$ 3x + k = 0

As, $-\frac{𝑏}{2𝑎}=\frac{3}{2}\notin (0,1)$

Since, both roots cannot lie between 0 and 1.

So, no value of k is possible.

Given word is 'ARRANGE'.

If R's occur together, number of letters in word = 6

To arrange R's together number of ways = $\frac{2!}{2!}$ = 1

$\therefore$ Number of ways to permuted 'ARRANGE' = $\frac{6!}{2!}$

Given, $\frac{1}{{}^5{𝐶}_{𝑟}}+\frac{1}{{}^6{𝐶}_{𝑟}}=\frac{1}{{}^4{𝐶}_{𝑟}}$

$\Rightarrow \frac{𝑟!(5-𝑟)!}{5!}+\frac{𝑟!(6-𝑟)!}{6!}=\frac{𝑟!(4-𝑟)!}{4!}$

$\Rightarrow \frac{(5-𝑟)!}{5}+\frac{(6-𝑟)!}{30}=(4-𝑟)!$

$\Rightarrow \frac{(5-𝑟)}{5}+\frac{(6-𝑟)(5-𝑟)}{30}=1$

$\Rightarrow 30-6𝑟+30-11𝑟+{𝑟}^2=30$

$\Rightarrow {𝑟}^2-17𝑟+30=0$

$\Rightarrow {𝑟}^2-15𝑟-2𝑟+30=0$

$\Rightarrow 𝑟(𝑟-15)-2(𝑟-15)=0$

$\Rightarrow (𝑟-2)(𝑟-15)=0$

$\therefore$ $𝑟=15,2$

Hence, the value of r is 2.

For n = 1, 2, 3, ..., we find that n³ + 2n takes values 3, 12, 33, ..., which are divisible by 3.

Coefficient of x¹⁷ in (x $-$ 1) (x $-$ 2) .... (x $-$ 18)

= $-$ (1 + 2 + 3 + ... + 18)

= $-$ (171) = $-$ 171

Given,

$1+{}^{𝑛}{𝐶}_1\cos 𝜃+{}^{𝑛}{𝐶}_2\cos 2𝜃+...+{}^{𝑛}{𝐶}_{𝑛}\cos 𝑛𝜃$

which is real part of complex number.

$({}^{𝑛}{𝐶}_0+{}^{𝑛}{𝐶}_1{𝑒}^{𝑖𝜃}+...)$

i.e. $=\mathrm{Re}({}^{𝑛}{𝐶}_0+{}^{𝑛}{𝐶}_1{𝑒}^{𝑖𝜃}+...)$

$=\mathrm{Re}\left(2{\cos }^2𝑛\frac{𝜃}{2}+𝑖2\sin \frac{𝜃}{2}.\cos \frac{𝜃}{2}\right)$

$\because$ $\left[\begin{array}{c}1+\cos 𝜃=2{\cos }^2\frac{𝜃}{2}𝑎𝑛𝑑\\ \sin 𝜃=2\sin \frac{𝜃}{2}.\cos \frac{𝜃}{2}\end{array}\right]$

$={\left(2\cos \frac{𝜃}{2}\right)}^{𝑛}\mathrm{Re}{\left(\cos \frac{𝜃}{2}+𝑖\sin \frac{𝜃}{2}\right)}^{𝑛}$

$={\left(2\cos \frac{𝜃}{2}\right)}^{𝑛}\mathrm{Re}\left(\cos \frac{𝑛𝜃}{2}+𝑖\sin \frac{𝜃}{2}\right)$ [by De Moivre's theorem]

$={\left(2\cos \frac{𝜃}{2}\right)}^{𝑛}.\cos \frac{𝑛𝜃}{2}$

Let $\Delta =\mid \begin{array}{ccc}1& {\log }_{𝑥}𝑦& {\log }_{𝑥}𝑧\\ {\log }_{𝑦}𝑥& 1& {\log }_{𝑦}𝑧\\ {\log }_{𝑧}𝑥& {\log }_{𝑧}𝑦& 1\end{array}\mid$

$=\mid \begin{array}{ccc}\frac{\log 𝑥}{\log 𝑥}& \frac{\log 𝑦}{\log 𝑥}& \frac{\log 𝑧}{\log 𝑥}\\ \frac{\log 𝑥}{\log 𝑦}& \frac{\log 𝑦}{\log 𝑦}& \frac{\log 𝑧}{\log 𝑦}\\ \frac{\log 𝑥}{\log 𝑧}& \frac{\log 𝑦}{\log 𝑧}& \frac{\log 𝑧}{\log 𝑧}\end{array}\mid$

On taking $\frac{1}{\log 𝑥}.\frac{1}{\log 𝑦}.\frac{1}{\log 𝑧}$ common from R₁, R₂ and R₃, we get

$=\frac{1}{\log 𝑥}.\frac{1}{\log 𝑦}.\frac{1}{\log 𝑧}\mid \begin{array}{ccc}\log 𝑥& \log 𝑦& \log 𝑧\\ \log 𝑥& \log 𝑦& \log 𝑧\\ \log 𝑥& \log 𝑦& \log 𝑧\end{array}\mid$ [$\because$ all rows are identical]

$\therefore$ $\Delta =0$

We have, | B | = 64

$\Rightarrow$ | adj A | = 64

$\Rightarrow$ | A |² = 64

$\therefore$ | A | = $\pm$ 8

Given,

$𝑄=\left[\begin{array}{cc}\cos \frac{𝜋}{4}& -\sin \frac{𝜋}{4}\\ \sin \frac{𝜋}{4}& \cos \frac{𝜋}{4}\end{array}\right]$ and $𝑥=⎡⎣⎢\begin{array}{c}\frac{1}{\sqrt{\surd 2}}\\ \frac{1}{\sqrt{\surd 2}}\end{array}⎤⎦⎥$

Let $𝑄\left(𝜃\right)=\left[\begin{array}{cc}\cos 𝜃& -\sin 𝜃\\ \sin 𝜃& \cos 𝜃\end{array}\right]$

$\because$ ${𝑄}^3\left(𝜃\right)=𝑄\left(3𝜃\right)$

$\therefore$ ${𝑄}^3\left(\frac{𝜋}{4}\right)=\left[\begin{array}{cc}\cos \frac{3𝜋}{4}& -\sin \frac{3𝜋}{4}\\ \sin \frac{3𝜋}{4}& \cos \frac{3𝜋}{4}\end{array}\right]=\left[\begin{array}{cc}-1/\sqrt{\surd 2}& -1/\sqrt{\surd 2}\\ 1/\sqrt{\surd 2}& -1/\sqrt{\surd 2}\end{array}\right]$

Now, ${𝑄}^3\left(\frac{𝜋}{4}\right)𝑥=⎡⎣⎢\begin{array}{cc}\frac{-1}{\sqrt{\surd 2}}& -\frac{1}{\sqrt{\surd 2}}\\ \frac{1}{\sqrt{\surd 2}}& -\frac{1}{\sqrt{\surd 2}}\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{c}\frac{1}{\sqrt{\surd 2}}\\ \frac{1}{\sqrt{\surd 2}}\end{array}⎤⎦⎥=\left[\begin{array}{c}-\frac{1}{2}-\frac{1}{2}\\ \frac{1}{2}-\frac{1}{2}\end{array}\right]$

$\therefore$ ${𝑄}^3\left(\frac{𝜋}{4}\right)𝑥=\left[\begin{array}{c}-1\\ 0\end{array}\right]$

Reflexivity For reflexive (x, x) $\in$ R.

x + 2x = 3x

which is divisible by 3.

$\Rightarrow$ xRx $\in$ R

Hence, xRy is reflexive.

Symmetric Let x + 2y = 3$𝜆$

$\Rightarrow$ x = 3$𝜆$ $-$ 2y

Now, y + 2x = y + 2 (3$𝜆$ $-$ 2y)

= y + 6$𝜆$ $-$ 4y

= 6$𝜆$ $-$ 3y

$\Rightarrow$ y + 2x = 3 (2$𝜆$ $-$ y)

which is divisible by 3.

$\therefore$ (x, y) $\in$ R $\Rightarrow$ (y, x) $\in$ R

Hence, xRy is symmetric.

Transitive x + 2y = 3$𝜆$ .... (i)

and y + 2z = 3$𝜇$ .... (ii)

On adding Eqs. (i) and (ii), we get

x + 3y + 2z = 3$𝜆$ + 3$𝜇$

$\Rightarrow$ x + 2z = 3($𝜆$ + $𝜇$ $-$ y)

which is divisible by 3.

$\because$ (x, y)$\in$ R and (y, z)$\in$ R

$\Rightarrow$ (x, z)$\in$ R

Hence, xRy is transitive.

$\because$ R is reflexive, symmetric and transitive.

$\therefore$ R is an equivalence relation.

$𝐴=5^{𝑛}-4𝑛-1=(1+4{)}^{𝑛}-4𝑛-1$

$=({}^{𝑛}{𝐶}_0+{}^{𝑛}{𝐶}_1\times 4+{}^{𝑛}{𝐶}_2\times 4^2+...+{}^{𝑛}{𝐶}_{𝑛}\times 4^{𝑛})-(4𝑛+1)$

$=4^2({}^{𝑛}{𝐶}_2+{}^{𝑛}{𝐶}_3\times 4+...+{}^{𝑛}{𝐶}_{𝑛}\times 4^{𝑛-2})$

$\therefore$ A contains some multiplies of 16.

Clearly, B contains all multiplies of 16 including 0.

$\therefore$ A $\subseteq$ B.

Given f(x) = (x² + 1)³⁵, $\forall$x $\in$R.

Since, f(x) is even function.

Hence, the function is not one-one.

And f(x) > 0, $\forall$ x $\in$ R.

Hence, the function is not onto.

$\therefore$ Given, the function is neither one-one nor onto.

Let x_i = a₁, a₂, a₃, ...., a_n

and y_i = $𝜆$a₁, $𝜆$a₂, $𝜆$a₃, ......, $𝜆$a_n

$\therefore$ y_i = $𝜆$x_i

$\because$ $\overline{𝑦}=\frac{\sum {𝑦}_{𝑖}}{𝑁}$

$=\frac{\sum 𝜆{𝑥}_{𝑖}}{𝑛}$ [$\because$ ${𝑦}_{𝑖}=𝜆{𝑥}_{𝑖},𝑁=𝑛$]

$=𝜆\frac{\sum {𝑥}_{𝑖}}{𝑛}$

$=𝜆\overline{𝑥}$

$\therefore$ ${𝜎}_{𝑦}=\sqrt{\surd \frac{\sum {({𝑦}_{𝑖}-\overline{𝑦})}^2}{𝑁}}$

$=\sqrt{\surd \frac{\sum {(𝜆{𝑥}_{𝑖}-𝜆\overline{𝑥})}^2}{𝑛}}$

$=\sqrt{\surd \frac{{𝜆}^2\sum {({𝑥}_{𝑖}-\overline{𝑥})}^2}{𝑛}}$

$=\sqrt{\surd {𝜆}^2{𝜎}_2^{𝑥}}=\left|𝜆\right|{𝜎}_{𝑥}=\left|𝜆\right|𝜎$

Given, $𝑃(𝐴\cap 𝐵)=\frac{1}{6}$

$𝑃(𝐴\cup 𝐵)=\frac{31}{45}$

and $𝑃\left(\overline{𝐵}\right)=\frac{7}{10}$

$\therefore$ $𝑃\left(𝐵\right)=1-\frac{7}{10}=\frac{3}{10}$

Now, $𝑃(𝐴\cup 𝐵)=𝑃\left(𝐴\right)+𝑃\left(𝐵\right)-𝑃(𝐴\cap 𝐵)$

$\Rightarrow \frac{31}{45}=𝑃\left(𝐴\right)+\frac{3}{10}-\frac{1}{6}$

$\therefore$ $𝑃\left(𝐴\right)=\frac{31}{45}+\frac{1}{6}-\frac{3}{10}$

$\therefore$ $=\frac{5}{9}$

Then, $𝑃\left(\frac{𝐵}{𝐴}\right)=\frac{𝑃(𝐴\cap 𝐵)}{𝑃\left(𝐴\right)}=\frac{\frac{1}{6}}{\frac{5}{9}}=\frac{3}{10}>\frac{1}{6}$

and $𝑃\left(\frac{𝐴}{𝐵}\right)=\frac{𝑃(𝐴\cap 𝐵)}{𝑃\left(𝐵\right)}=\frac{\frac{1}{6}}{\frac{3}{10}}=\frac{5}{9}>\frac{1}{6}$

and, $𝑃\left(𝐴\right)\times 𝑃\left(𝐵\right)=\frac{5}{9}\times \frac{3}{10}=\frac{1}{6}=𝑃(𝐴\cap 𝐵)$

Hence, A and B are independent.

We have, $\cos {15}^{\circ }.\cos 7\frac{1^{\circ }}{2}.\sin 7\frac{1^{\circ }}{2}$

$=\frac{1}{2}\cos {15}^{\circ }\left(2\cos 7\frac{1^{\circ }}{2}.\sin 7\frac{1^{\circ }}{2}\right)$

$=\frac{1}{2}\cos {15}^{\circ }\left[\sin 2\left(7\frac{1^{\circ }}{2}\right)\right]$ [$\because$ 2sin $𝜃$ . cos$𝜃$ = sin2$𝜃$]

$=\frac{1}{2}\cos {15}^{\circ }\sin {15}^{\circ }$

$=\frac{1}{2}\times \frac{1}{2}(2\sin {15}^{\circ }\cos {15}^{\circ })$

$=\frac{1}{2}\sin {30}^{\circ }=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$

Given equation is

tan x $-$ x = 0

$\Rightarrow$ tan x = x

Solutions are abscissae of points of intersection of the curves y = tan x and y = x. From the figure, it is clearly visible that solution lies in $\left(𝜋,\frac{3𝜋}{2}\right)$

Let circumradius of $\Delta$DEF be R'. We know $\angle$FDE = 180${}^{\circ }$ $-$ 2A and FE = Rsin2A



Now, by sine rule in $\Delta$DEF

$2𝑅=\frac{𝐸𝐹}{\sin \angle 𝐹𝐷𝐸}=\frac{𝑅\sin 2𝐴}{\sin ({180}^{\circ }-2𝐴)}$

$\therefore$ ${𝑅}^{\prime }=\frac{𝑅}{2}$