JEE Main 26 June 2022 Shift 2 question paper with solutions

Given $f(x) = x - 1$ and $g(x) = \frac{x^2}{x^2 - 1}$. $f(g(x)) = g(x) - 1$ $$= \frac{x^2}{x^2 - 1} - 1 = \frac{x^2 - x^2 + 1}{x^2 - 1}$$ $f(g(x)) = \frac{1}{x^2 - 1}$; $x \neq \pm 1$, even function. Hence $f(g(x))$ is many one function. Let $y = \frac{1}{x^2 - 1}$. Then $y \cdot x^2 - y = 1$. $$x^2 = \frac{1 + y}{y}$$ $$\left( \frac{1 + y}{y} \right) \geq 0$$ Range: $y \in (-\infty, -1] \cup (0, \infty)$. Hence, Range $\neq$ Co-domain $\Rightarrow f(g(x))$ is into function.

For infinitely many solutions, $$\Delta = 0 = \Delta_x = \Delta_y = \Delta_z$$ $$\Delta = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{vmatrix} = 0$$ This implies $$\alpha(10 - 9) - 1(5 - 3) + 1(3 - 2) = 0$$ $$\Rightarrow \alpha - 2 + 1 = 0$$ $$\Rightarrow \alpha = 1$$ Now, $$\Delta_x = \begin{vmatrix} 5 & 1 & 1 \\ 4 & 2 & 3 \\ \beta & 3 & 5 \end{vmatrix} = 0$$ This implies $$5(10 - 9) - 1(20 - 3\beta) + 1(12 - 2\beta)$$ $$\Rightarrow 5 - 20 + 3\beta + 12 - 2\beta$$ $$\Rightarrow -3 + \beta = 0$$ $$\Rightarrow \beta = 3$$

Given $$A = \left( \frac{1}{2} + \frac{1}{4^2} + \frac{1}{2^3} + \frac{1}{4^4} + \ldots \infty \right)$$ We can rewrite $$A = \left( \frac{1}{2} + \frac{1}{2^3} + \ldots \infty \right) + \left( \frac{1}{4^2} + \frac{1}{4^4} + \ldots \infty \right)$$ This becomes $$A = \frac{\frac{1}{2}}{1 - \frac{1}{4}} + \frac{\frac{1}{16}}{1 - \frac{1}{16}}$$ Thus, $$\Rightarrow A = \frac{1}{2} \times \frac{4}{3} + \frac{1}{16} \times \frac{16}{15} \Rightarrow A = \frac{11}{15}$$ Now, consider $$B = \left( -\frac{1}{2} + \frac{1}{4^2} - \frac{1}{2^3} + \frac{1}{4^4} + \ldots \infty \right)$$ We can rewrite $$B = \left( -\frac{1}{2} + -\frac{1}{2^3} + \ldots \infty \right) + \left( \frac{1}{4^2} + \frac{1}{4^4} + \ldots \infty \right)$$ This becomes $$B = \frac{-\frac{1}{2}}{1 - \frac{1}{4}} + \frac{\frac{1}{16}}{1 - \frac{1}{16}}$$ Thus, $$\Rightarrow B = -\frac{1}{2} \times \frac{4}{3} + \frac{1}{16} \times \frac{16}{15}$$ Therefore, $$B = -\frac{9}{15}$$ Now, calculate $$\frac{A}{B} = \frac{\frac{11}{15}}{\frac{15}{-9}}$$ Thus, $$\frac{A}{B} = -\frac{11}{9}$$

Given the limit $\lim_{x\to0} \frac{\cos(\sin x)-\cos x}{x^4}$ which is of the form $\frac00$. We rewrite it as $\lim_{x\to0} \frac{ 2\sin\left(\frac{x+\sin x}{2}\right) \sin\left(\frac{x-\sin x}{2}\right) }{x^4}$ This can be expanded to $\lim_{x\to0} 2\left( \frac{ \sin\left(\frac{x+\sin x}{2}\right) }{ \left(\frac{x+\sin x}{2}\right) } \right) \left( \frac{ \sin\left(\frac{x-\sin x}{2}\right) }{ \left(\frac{x-\sin x}{2}\right) } \right) \frac{ \left(\frac{x+\sin x}{2}\right) }{ x^4 } \left( \frac{x-\sin x}{2} \right)$ This simplifies to $\lim_{x\to0} \frac{x^2-\sin^2x}{4x^4}$ which is again of the form $\frac00$. Apply L'Hopital Rule: $\lim_{x\to0} \frac{2x-2\sin x\cos x} {16x^3}$ Again apply L'Hopital rule: $\lim_{x\to0} \frac{2-\cos(2x)} {24x^2}$ This simplifies to $\lim_{x\to0} \frac{2(1-\cos(2x))} {24(2x^2)} \times2$ $\Rightarrow \frac2{24} \times\frac12 \times2$ $\Rightarrow \frac16$

No. of non-differentiable points = 1 (m) No. of not continuous points = 0 (n) (m, n) = (1, 0)

Given $y^2 = 8x$; $y^2 = 16(3-x)$. Finding their intersection points. $y^2 = 8x$ and $y^2 = -16(x-3)$ $$8x = -16x + 48$$ $$24x = 48$$ $$x = 2; \; y = \pm 4$$ $$A = 2 \int_{0}^{4} (x_R - x_L) \, dy$$ # Required Area $$= 2 \int_{0}^{4} \left( 3 - \frac{y^2}{16} - \frac{y^2}{8} \right) \, dy$$ $$= 2 \left[ 3y - \frac{y^3}{3 \times 16} - \frac{y^3}{3 \times 8} \right]_{0}^{4}$$ $$= 2 \left[ 3 \times 4 - \frac{4 \times 4 \times 4}{3 \times 16} - \frac{4 \times 4 \times 4 \times 2}{3 \times 8 \times 2} \right]$$ $$= 2 \left[ 12 - \frac{4}{3} - \frac{8}{3} \right] = 2 \times 12 \left[ 1 - \frac{1}{3} \right] = 2 \times 12 \times \frac{2}{3} = 16$$

Given $$\int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} \, dx = g(x) + c$$. Put $$x = \cos 2\theta$$. $$dx = -2\sin 2\theta \cdot d\theta$$ $$= \int \frac{1}{\cos 2\theta} \tan \theta (-4 \sin \theta \cdot \cos \theta) d\theta$$ $$= \int \frac{1}{\cos 2\theta} (-4 \sin^2 \theta) d\theta$$ $$= -2 \int \frac{1 - \cos 2\theta}{\cos 2\theta} d\theta$$ $$= -\frac{2}{2} \ln |\sec 2\theta + \tan 2\theta| + 2\theta + c$$ $$= \ln |\sec 2\theta - \tan 2\theta| + 2\theta + c$$ $$= \ln \left| \frac{1 - \sin 2\theta}{\cos 2\theta} \right| + \cos^{-1} x + c$$ $$= \ln \left| \frac{1 - \sqrt{1-x^2}}{x} \right| + \cos^{-1} x + c$$ Therefore, $$g(1) = 0$$ $$g(x) = \ln \left| \frac{1 - \sqrt{1-x^2}}{x} \right| + \cos^{-1} x$$ $$g\left( \frac{1}{2} \right) = \ln \left| 2 - \sqrt{3} \right| + \frac{\pi}{3}$$ $$g\left( \frac{1}{2} \right) = \ln \left| \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \right| + \frac{\pi}{3}$$

Given $x \frac{dy}{dx} + 2y = xe^x$. Rewriting, $\frac{dy}{dx} + \frac{2y}{x} = e^x$. The integrating factor (I.F.) is $x^2$. Then $y \cdot x^2 = \int x^2 e^x \, dx$. This equals $\int e^x (x^2 + 2x - 2x - 2 + 2) \, dx$. So, $yx^2 = e^x (x^2 - 2x + 2) + c$. Given $y(1) = 0$, $0 = e(1 + 0) + c$. Thus, $c = -e$. Now, $z(x) = x^2 y(x) - e^x$. This becomes $e^x (x^2 - 2x + 2) - e - e^x$. Simplifying, $= e^x (x - 1)^2 - e$. Differentiating, $\frac{dz}{dx} = e^x \cdot 2(x - 1) + e^x (x - 1)^2 = 0$. This gives $x^x (x - 1) (2 + x - 1) = 0$. Or $e^x (x - 1) (x + 1) = 0$. Thus, $x = -1, 1$. At $x = -1$, there is a local maxima. Then the maximum value is $z(-1) = \frac{4}{e} - e$.

I.F. = e^{$\int e^{x^2-2}\,dx$} = e^{$\int e^{x^2-2x+2}\,dx$} = e^{e^{x^2-2x}}. y $\cdot$ e^{e^{x^2-2x}} = $\int e^{e^{x^2-2x}} e^x (x^2-2x)e^x\,dx$. Let $e^{x^2-2x}=t$. So, y $\cdot$ e^{e^{x^2-2x}} = $\int e^t \cdot t\,dt$. $= t\cdot e^t - e^t + c$. At $x=0$, $t=0$. $0\cdot1=0-1+c$ $\Rightarrow c=1$ For $x=2$, $t=0$. $y\cdot1=0-1+1=0$ $y(2)=0$

Given $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$, equation of tangent to the ellipse is $y = mx \pm \sqrt{a^2m^2 + b^2}$ $y = mx \pm \sqrt{16m^2 + 9}$ $\hfill \ldots\text{(i)}$ $$x^2 + y^2 = 12$$ Equation of tangent to the circle is $y = mx \pm \sqrt{12}\sqrt{1 + m^2}$ $\hfill \ldots\text{(ii)}$ For common tangent, equate eq. (i) and (ii) $$\Rightarrow 16m^2 + 9 = 12(1 + m^2)$$ $$16m^2 - 12m^2 = 3$$ $$4m^2 = 3$$ $$12m^2 = 9$$

Given the equation of the ellipse: $$\frac{x^2}{4} + \frac{y^2}{2} = 1$$ with points P(4,3) and Q $$(2 \cos \theta, \sqrt{2} \sin \theta)$$. Coordinate of D is $$\left( \frac{2 \cos \theta + 4}{2}, \frac{\sqrt{2} \sin \theta + 3}{2} \right) \equiv (h, k)$$. From this, we have: $$\frac{2h - 4}{2} = \cos \theta ....(i)$$ $$\frac{2k - 3}{\sqrt{2}} = \sin \theta ....(ii)$$ Using $$(i)^2 + (ii)^2$$, then we get: $$\left( \frac{2h - 4}{2} \right)^2 + \left( \frac{2k - 3}{\sqrt{2}} \right)^2 = 1 \implies \frac{(x - 2)^2}{1} + \frac{\left( y - \frac{3}{2} \right)^2}{\left( \frac{1}{2} \right)} = 1$$ Therefore, the required eccentricity is $$e = \sqrt{\frac{1}{1 - \frac{1}{2}}} = \frac{1}{\sqrt{2}}$$

Given $\($ $\frac{x^2}{a^2}$ - $\frac{y^2}{9}$ = 1 $\)$. The point $\($ (8, 3$\sqrt{3}$) $\)$ lies on the hyperbola, then $$ \frac{64}{a^2} - \frac{27}{9} = 1 \implies a^2 = \frac{64}{4} = 16 $$ The equation of the normal at $\($ (8, 3$\sqrt{3}$) $\)$ is: $$ \frac{16x}{8} + \frac{9y}{3\sqrt{3}} = 16 + 9 $$ $$ 2x + \sqrt{3}y = 25 $$ Check options.

Given $(2x + y - 5z) + \lambda(3x - y + 4z - 7) = 0$. Rotated by $\pi/2$. $$(2 + 3\lambda)x + (1 - \lambda)y + (-5 + 4\lambda)z - 7\lambda = 0$$ $$2x + y - 5z = 0$$ $$2(2 + 3\lambda) + (1 - \lambda) - 5(-5 + 4\lambda) = 0$$ $$\Rightarrow 4 + 6\lambda + 1 - \lambda + 25 - 20\lambda = 0$$ $$30 = 15\lambda$$ $$\lambda = 2$$ Required plane: $-8x - y + 3z - 14 = 0$. Check options.

Given $\mathbf{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda (3 \hat{j} - \hat{k})$ (L1) and $\mathbf{r} = (\alpha \hat{i} - \hat{j}) + \mu (2 \hat{i} - 3 \hat{k})$ (L2). L1 and L2 are coplanar. $$\begin{vmatrix} 0 & 3 & -1 \\ 2 & 0 & -3 \\ (1 - \alpha) & 0 & 1 \end{vmatrix} = 0$$ $$-3 (2 + 3 (1 - \alpha)) = 0$$ $$2 + 3 - 3 \alpha = 0$$ $$3 \alpha = 5$$ $$\Rightarrow \alpha = \frac{5}{3}$$ Now, $$\mathbf{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & -1 \\ 2 & 0 & -3 \end{vmatrix} = \hat{i} (-9) - \hat{j} (2) + \hat{k} (-6)$$ $$(9, 2, 6)$$ Equation of plane: $$9 (x - 1) + 2(y + 1) + 6(z - 1) = 0$$ $$9x + 2y + 6z - 13 = 0$$ Perpendicular distance from $(\cdot, 0, 0)$ $$= \left| \frac{9 \cdot \frac{5}{3} + 0 + 0 - 13}{\sqrt{81 + 36 + 4}} \right| = \frac{2}{\sqrt{121}} = \frac{2}{11}$$

Given $\vec{v} = \lambda \vec{a} + \mu \vec{b}$. $\vec{v} = \lambda (1,1,2) + \mu (2,-3,1)$. $\vec{v} = (\lambda + 2\mu, \lambda - 3\mu, 2\lambda + \mu)$. $\vec{v} \cdot \vec{j} = 7$. $\lambda - 3\mu = 7$. $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{2}{\sqrt{3}}$. $\vec{v} \cdot \vec{c} = 2$. $\lambda + 2\mu - \lambda + 3\mu + 2\lambda + \mu = 2$. $2\lambda + 6\mu = 2$. $\lambda + 3\mu = 1$. $\lambda - 3\mu = 7$. $2\lambda = 8$. $\lambda = 4$. $\mu = -1$. We get $\vec{v} = (2,7,7)$.

No. of observations: 50 mean $\bar{x} = 15$ Standard deviation ($\sigma$) = 2 Let incorrect observation is $x_1$ and correct observation is $(x_1')$ Given $x_1 + x_1' = 70$ $$\bar{x} = \frac{x_1 + x_2 + \ldots + x_{50}}{50} = 15 (given)$$ $$\Rightarrow x_1 + x_2 + \ldots + x_{50} = 750 \ldots (i)$$ Now Mean of correct observation is 16 $$\frac{x_1' + x_2 + \ldots + x_{50}}{50} = 16$$ $$x_1' + x_2 + x_3 + \ldots x_{50} = 16 \times 50 \ldots (ii)$$ Subtracting equation (i) from equation (ii): $$x_1' - x_1 = 16 \times 50 - 15 \times 50$$ $$x_1' - x_1 = 50 \& x_1 + x_1' = 70$$ $$x_1' = 60$$ $$x_1 = 10$$ $$\Rightarrow 4 = \frac{x_1^2 + x_2^2 + \ldots + x_{50}^2}{50} - 15^2 \ldots (iii)$$ $$\Rightarrow \sigma^2 = \frac{x_1'^2 + x_2^2 + \ldots + x_{50}^2}{50} - 16^2 \ldots (iv)$$ From (iii) $$\Rightarrow 4 = \frac{(10)^2 + x_2^2 + x_3^2 + \ldots + x_{50}^2}{50} - 225$$ $$\Rightarrow 4 = 2 - 225 + \frac{x_2^2 + x_3^2 + \ldots + x_{50}^2}{50}$$ $$\Rightarrow 227 = \frac{x_2^2 + x_3^2 + \ldots + x_{50}^2}{50}$$ From (iv) $$\sigma^2 = \frac{(60)^2}{50} + \frac{x_2^2 + x_3^2 + \ldots + x_{50}^2}{50} - (16)^2$$ $$\sigma^2 = \frac{60 \times 60}{50} + 227 - 256$$ $$\sigma^2 = 72 + 227 - 256$$ $$\sigma^2 = 43$$

Given $16 \sin 20^\circ \sin 40^\circ \sin 80^\circ$. This is equal to $16 \sin 40^\circ \sin 20^\circ \sin 80^\circ$. We can rewrite it as $4(4 \sin (60 - 20) \sin (20) \sin (60 + 20))$. This simplifies to $4 \times \sin (3 \times 20^\circ)$. Since $\sin 3\theta = 4 \sin(60 - \theta) \times \sin \theta \times \sin (60 + \theta)$, we have $4 \times \sin 60^\circ$. This equals $4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}$.

Let $\tan^{-1}\!\left(\frac{4}{3}\right)=\theta \Rightarrow \tan\theta=\frac{4}{3}$ $E=\cos^{-1}\!\left( \frac{3}{10}\cos\theta +\frac{2}{5}\sin\theta \right)$ $=\cos^{-1}\!\left( \frac{3}{10}\times\frac{3}{5} +\frac{2}{5}\times\frac{4}{5} \right)$ $=\cos^{-1}\!\left( \frac{9}{50} +\frac{8}{25} \right)$ $=\cos^{-1}\!\left(\frac{25}{50}\right)$ $=\cos^{-1}\!\left(\frac12\right)$ $=\frac{\pi}{3}$

Put $y = 2$ $f(x + y) = 2^y \cdot f(y) + 4^y \cdot f(x)$ $f(x + 2) = 2^2 \cdot 3 + 16f(x)$ $f'(x + 2) = 16f'(x) + 3 \cdot 2^2 \ln 2$ $f'(4) = 16f'(2) + 12\ln 2$ $\hfill \ldots\text{(i)}$ $f(y + 2) = 4f(y) + 3 \cdot 4^y \ln 4$ $f'(4) = 4f'(2) + 96\ln 2$ $\hfill \ldots\text{(ii)}$ Solving eq. (i) and (ii), we get $f'(2) = 7\ln 2$ From equation (i), we get $f'(4) = 124\ln 2$ Now, $\Rightarrow 14 \cdot \dfrac{f'(4)}{f'(2)}$ $= 14 \times \dfrac{124\ln 2}{7\ln 2}$ $= 248$

Given $p + q = 3$ and $p^4 + q^4 = 369$. $$\left(\frac{1}{p} + \frac{1}{q}\right)^{-2} = (p + q)^2 = 9$$ $$p^2 + q^2 = 9 - 2pq$$ $$\frac{1}{\left(\frac{1}{p} + \frac{1}{q}\right)^2} = \frac{(pq)^2}{(q + p)^2} = \frac{(pq)^2}{9}$$ $$p^4 + q^4 = (p^2 + q^2)^2 - 2p^2q^2$$ $$369 = (9 - 2pq)^2 - 2(pq)^2$$ $$369 = 81 + 4p^2q^2 - 36pq - 2p^2q^2$$ $$288 = 2p^2q^2 - 36pq$$ $$144 = p^2q^2 - 18pq$$ $$(pq)^2 - 2 \times 9 \times pq + 9^2 = 144 + 9^2$$ $$(pq - 9)^2 = 225$$ $$pq - 9 = \pm 15$$ $$pq = \pm 15 + 9$$ $$pq = 24, -6$$ (24 is rejected because $p^2 + q^2 = 9 - 2pq$ is negative) $$\frac{(pq)^2}{9} = \frac{16 \times 16}{9} = 4$$

Given $z^2 + z + 1 = 0 \Rightarrow z = w, \ w^2$. $$\left| \sum_{n=1}^{15} \left( z^n + (-1) \frac{1}{z^n} \right) \right|^2 = \left| \sum_{n=1}^{15} \left( z^{2n} + \frac{1}{z^{2n}} + 2(-1)^n \right) \right|$$ $$= \left| \sum_{n=1}^{15} w^{2n} + \frac{1}{w^{2n}} + 2(-1)^n \right|$$ $$= \left| \frac{w^2 (1 - w^{30})}{1 - w^2} + \frac{1}{w^2} \left( \frac{1 - \frac{1}{w^{30}}}{1 - \frac{1}{w^2}} \right) + 2(-1) \right|$$ $$= \left| \frac{w^2 (1 - 1)}{1 - w^2} + \frac{1}{w^2} \frac{(1 - 1)}{1 - \frac{1}{w^2}} - 2 \right|$$ $$= |0 + 0 - 2| = 2$$

Given $$X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, X^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ $$Y = \begin{bmatrix} \alpha & \beta & \gamma \\ 0 & \alpha & \beta \\ 0 & 0 & \alpha \end{bmatrix}, Z = \begin{bmatrix} \alpha^2 & -\alpha \beta & \beta^2 - \alpha \gamma \\ 0 & \alpha^2 & -\alpha \beta \\ 0 & 0 & \alpha^2 \end{bmatrix}$$ $$Y \cdot Y^{-1} = I$$ $$\begin{bmatrix} \alpha & \beta & \gamma \\ 0 & \alpha & \beta \\ 0 & 0 & \alpha \end{bmatrix} \begin{bmatrix} \frac{1}{5} & -\frac{2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ $$\frac{\alpha}{5} = 1 \implies \alpha = 5$$ $$-\frac{2}{5} \alpha + \frac{\beta}{5} = 0 \implies \beta = 10$$ $$\frac{\alpha}{5} - \frac{2 \beta}{5} + \frac{\gamma}{5} = 0 \implies \gamma = 15$$ Therefore, $$(\alpha - \beta + \gamma)^2 = (5 - 10 + 15)^2 = 100$$

$36=2\times2\times3\times3$ Number should be odd multiple of $2$ and does not have factors $3$ and $9$ Odd multiples of $2$ are $102,106,110,114,\ldots,998$ $(225\ \text{numbers})$ No. of multiples of $3$ are $102,114,126,\ldots,990$ $(75\ \text{numbers})$ which are also included in multiples of $9$ Hence, Required $=225-75$ $=150$

Given $$^{40}C_0 + ^{41}C_1 + ^{42}C_2 + \ldots + ^{59}C_{19} + ^{60}C_{20}$$ $$\left(\frac{1}{41} + 1\right)^{41}C_1 + ^{42}C_2 + \ldots$$ $$\left[\frac{42}{41}\left(\frac{2}{42}\right) + 1\right]^{42}C_2 + ^{43}C_3 + \ldots$$ $$\left(\frac{2}{41} + 1\right)^{42}C_2 + ^{43}C_3 + \ldots$$ $$\left(\frac{43}{41} \times \frac{3}{43} + 1\right)^{43}C_3 + ^{44}C_4 + \ldots$$ $$\frac{3 + 41}{41} \cdot ^{43}C_3 + \ldots$$ Similarly: $$\frac{20 + 41}{41}$$ $$\Rightarrow m = 61; \; n = 41$$ $$m + n = 102$$

Given $a_1 > 0$, $a_2$, $a_3$, $a_4$, $a_5 \to G.P.$ $3a_2 + a_3 = 2a_4$ $3ar + ar^2 = 2ar^3$ $3 + r = 2r^2$ $2r^2 - r - 3 = 0$ $r = -1$ and $r = \frac{3}{2}$ $a_2 + a_4 = 2a_3 + 1$ $ar + ar^3 = 2ar^2 + 1$ $a \left( r + r^3 - 2r^2 \right) = 1$ $a \left( \frac{3}{2} + \frac{27}{8} - \frac{18}{4} \right) = 1$ $a = \frac{8}{3}$ When $r = -1$, $a = -\frac{1}{4}$ (rejected, $a_1 > 0$) $r = \frac{2}{3}$, $a = \frac{8}{3}$ (selected) Now $a_2 + a_4 + 2a_5$ $= \frac{8}{3} \times \frac{3}{2} + \frac{8}{3} \times \frac{27}{8} + 2 \times \frac{8}{3} \times \frac{81}{16}$ $= 4 + 9 + 27 = 40$

Given the integral $$\frac{24}{\pi} \int_0^{\sqrt{2}} \frac{(2-x^2)}{(x^2+2)\sqrt{4+x^4}} \, dx$$ we can rewrite it as $$\frac{24}{\pi} \int_0^{\sqrt{2}} \frac{x^2 \left( \frac{2}{x^2} - 1 \right)}{\left( x + \frac{2}{x} \right) \times x \sqrt{\frac{4}{x^2} + x^2}} \, dx$$ which simplifies to $$\frac{24}{\pi} \int_0^{\sqrt{2}} \frac{\left( \frac{2}{x^2} - 1 \right)}{\left( x + \frac{2}{x} \right) \sqrt{\left( x + \frac{2}{x} \right)^2 - 4}} \, dx$$ Let $$x + \frac{2}{x} = t$$ then $$dt = \left( 1 - \frac{2}{x^2} \right) \, dx$$ The integral becomes $$I = -\frac{24}{\pi} \int \frac{dt}{t \sqrt{t^2 - 4}}$$ Evaluating this gives $$= -\frac{24}{\pi} \times \frac{1}{2} \sec^{-1} \left( \frac{x + \frac{2}{x}}{2} \right) \bigg|_0^{\sqrt{2}}$$ $$= -\frac{12}{\pi} \left[ \sec^{-1} \left( \frac{2\sqrt{2}}{2} \right) - \sec^{-1}(\infty) \right]$$ $$= -\frac{12}{\pi} \left[ \frac{\pi}{4} - \frac{2\pi}{2} \right] = -\frac{12}{\pi} \left[ \frac{-\pi}{4} \right]$$

$\dfrac{x\sec\theta}{4}+\dfrac{y\tan\theta}{2}=1$ $m_1=\dfrac{\sec\theta/2}{\tan\theta} =\dfrac{\sec\theta}{2\tan\theta}$ $m_2=\dfrac{k}{h}$ $m_1m_2=-1$ $\dfrac{k}{h}\cdot\dfrac{\sec\theta}{2\tan\theta}=-1$ $\dfrac{k}{2h\sin\theta}=-1$ $\sin\theta=-\dfrac{k}{2h}$ $\cos\theta=\dfrac{\sqrt{4h^2-k^2}}{2h}$ Also, $\dfrac{h\sec\theta}{4}-\dfrac{k\tan\theta}{2}=1$ $\dfrac{h}{4}\cdot \dfrac{2h}{\sqrt{4h^2-k^2}} -\dfrac{k}{2}\cdot \dfrac{-k}{\sqrt{4h^2-k^2}} =1$ $h^2+k^2=2\sqrt{4h^2-k^2}$ $(x^2+y^2)^2=4(4x^2-y^2)$ $(x^2+y^2)^2=\dfrac{16x^2-4y^2}{4}$ $\alpha=16,\ \beta=-4$ $\alpha+\beta=12$

Given $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$. Divisible by 21 when divided by 3. Case – I: All 1 $\rightarrow (1)$. Case – II: All 8 $\rightarrow (1)$. Case – III: 3 ones and 3 eights $$\frac{6!}{3! \times 3!} = 20$$ Required probability $\therefore p = \frac{22}{64}$ $$96p = 96 \times \frac{22}{64} = 33$$

Given $e_2$: induced emf in secondary coil, $i_1$: Current in primary coil, $M$: Mutual inductance. $$e_2 = -M \frac{di_1}{dt}$$ $$M = -\frac{e_2}{\frac{di_1}{dt}}$$ $$[M] = \left[ \frac{e_2}{\frac{di_1}{dt}} \right] = \left[ \frac{\frac{W}{q}}{\frac{AT}{AT^{-1}}} \right] = \left[ ML^2T^{-2} \right]$$ $$= [ML^2T^{-2}A^{-2}]$$

Using constraint $$\sum \vec{T} \cdot \vec{a} = 0$$ $$-4T a_1 - 2T a_2 - T a_3 - T a_4 = 0$$ $$4a_1 + 2a_2 + a_3 + a_4 = 0$$

Work done = area under $F-x$ curve. Area below $x$-axis is negative and area above $x$-axis is positive. So $$W_3 > W_2 > W_1 > W_4$$

The total kinetic energy $K_{total}$ is the sum of rotational and translational kinetic energies. $$K_{total} = K_{rotational} + K_{Translational}$$ The expression for total kinetic energy is given by: $$K_{total} = \frac{1}{2} I_{cm} \omega^2 + \frac{1}{2} m V_{cm}^2$$ For pure rolling, the velocity of the center of mass $v_{cm}$ is $R \omega$. The moment of inertia about the center of mass $I_{cm}$ is: $$I_{cm} = \frac{2}{5} m R^2$$ The rotational kinetic energy $K_{Rot}$ is: $$K_{Rot} = \frac{1}{2} I_{cm} \omega^2 = \frac{1}{2} \times \frac{2}{5} m R^2 \times \frac{V_{cm}^2}{R^2} = \frac{1}{5} m V_{cm}^2$$ The total kinetic energy $K_{Total}$ is: $$K_{Total} = \frac{1}{5} m V_{cm}^2 + \frac{1}{2} m V_{cm}^2 = \frac{7}{10} m V_{cm}^2$$ The ratio of rotational kinetic energy to total kinetic energy is: $$\frac{K_{Rot}}{K_{Total}} = \frac{\frac{1}{5} m V_{cm}^2}{\frac{7}{10} m V_{cm}^2} = \frac{2}{7}$$

Effective acceleration due to gravity is the resultant of $g$ and $rw^2$ whose direction and magnitude depends upon $\theta$. Hence assertion is false. When $\theta = 0^\circ$ (at equator), effective acceleration is radially inward.

Reynold's number is given by $$\frac{\rho v d}{\eta}$$

Average K.E./molecule $= \frac{f}{2}kT$. So, $\frac{K_{\mathrm{Ar}}}{K_{\mathrm{O}_2}} = \frac{\frac{3}{2}kT}{\frac{5}{2}kT} = \frac{3}{5}$.

The capacitors are in series, so the equivalent capacitance is calculated as follows: $$\frac{1}{C_{eq}} = \frac{1}{10} + \frac{1}{15} + \frac{1}{20} = \frac{12 + 8 + 6}{120} = \frac{26}{120}$$ Thus, the equivalent capacitance is: $$C_{eq} = \frac{60}{13} \, \mu\mathrm{F}$$ The charge on each capacitor is the same because they are in series. Therefore, the charge is: $$Q = \frac{13 \times 60}{13} = 60 \, \mu\mathrm{C}$$

The original capacitance is given by $C_{original} = \frac{A \varepsilon_0}{d}$. The system is divided into two capacitors $C_1$ and $C_2$ in series. For $C_1$, we have: $$C_1 = \frac{A \varepsilon_0}{d/2} = \frac{2A \varepsilon_0}{d} = C.$$ For $C_2$, we have: $$C_2 = \frac{KA \varepsilon_0}{d/2} = \frac{2KA \varepsilon_0}{d} = \frac{6A \varepsilon_0}{d} = 3C.$$ Since $C_1$ and $C_2$ are in series, the new capacitance $C_{new}$ is given by: $$C_{new} = \frac{C_1 C_2}{C_1 + C_2} = \frac{C \times 3C}{C + 3C} = \frac{3C}{4}.$$ Substituting the values, we get: $$C_{new} = \frac{3}{4} \times \frac{2A \varepsilon_0}{d} = \frac{3}{2} \times \frac{A \varepsilon_0}{d}.$$ Therefore, $C_{new} = \frac{3}{2} C_{original}$. Finally, we have: $$\frac{3}{2} \times 4 = 6 \mu F.$$

Let $R$ be the radius of the combined drop and $r$ be the radius of the smaller drop. The volume will remain the same: $$\frac{4}{3} \pi R^3 = 64 \times \frac{4}{3} \pi r^3$$ Therefore, $R = 4r$. Let $Q = 64q$ where $q$ is the charge of the smaller drop and $Q$ is the charge of the combined drop. The ratio of surface charge densities is given by: $$\frac{\sigma_{bigger}}{\sigma_{smaller}} = \frac{\frac{Q}{4\pi R^2}}{\frac{q}{4\pi r^2}} = \frac{Q}{q} \cdot \frac{r^2}{R^2}$$ Substituting the values, we get: $$= 64 \cdot \frac{r^2}{16r^2} = 4$$ Thus, $$\frac{\sigma_{bigger}}{\sigma_{smaller}} = \frac{4}{1}$$

The circuit is simplified step by step. Initially, the circuit is rearranged to show symmetry. The resistors are combined in series and parallel as follows. The equivalent resistance between points A and B is calculated to be $R_{AB} = 5\,\Omega$.

Work done = MB ($\cos$ $\theta$_1 - $\cos$ $\theta$_2) $\theta_1 = 0^\circ, \theta_2 = 60^\circ$ $$= 2 \times 10^5 \times 14 \times 10^{-5} (1 - 1/2)$$ $$= 14 \, \mathrm{J}$$

Current on both the inductor is in opposite direction. Hence: $$L_{eq} = L_1 + L_2 - 2M$$

emf induced between the two ends is given by $$\frac{B_H \omega l^2}{2}$$. Substituting the values: $$\frac{0.2 \times 10^{-4} \times 5 \times 1}{2} = 0.5 \times 10^{-4} = 50 \times 10^{-6} \, \mathrm{V} = 50 \, \mu \mathrm{V}$$

From electromagnetic wave spectrum. As $\lambda$ increases, $$\lambda_{gamma-ray} < \lambda_{X-ray} < \lambda_{ultraviolet} < \lambda_{visible} < \lambda_{infrared} < \lambda_{microwave} < \lambda_{radio wave}.$$

Given $\lambda_{emitted} = 670 \, \mathrm{nm}$ and $\lambda_{obs} = 670.7 \, \mathrm{nm}$. We need to find $v$. The speed of light $c = 3 \times 10^8 \, \mathrm{m/s}$. If $v << c$, then $$\frac{\lambda_{obs} - \lambda_{emitted}}{\lambda_{emitted}} = \frac{v}{c}$$ $$\frac{670.7 - 670}{670} = \frac{v}{c}$$ Solving for $v$, we get $$v = 3.13 \times 10^5 \, \mathrm{m/s}$$

Given $\lambda = 4500 \, \mathrm{\AA}$. $B = 2 \, \mathrm{mT}$, $R = 2 \, \mathrm{mm}$. $R = \frac{\sqrt{2Km}}{qB}$. $$\frac{(qBR)^2}{2m} = K$$ $$\frac{(1.6 \times 10^{-19} \times 2 \times 10^{-3} \times 2 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} = K$$ $$\frac{(6.4)^2}{2 \times 9.1} \times \frac{10^{-50}}{10^{-31}} = K$$ $$K = 2.25 \times 10^{-19} \, \mathrm{J}$$ $$= \frac{2.25 \times 10^{-19}}{1.6 \times 10^{-19}} \, \mathrm{eV} = 1.40 \, \mathrm{eV}$$ $$E = \frac{12400}{4500} = 2.76 \, \mathrm{eV}$$ $$\phi = E - K = (2.76 - 1.40) \, \mathrm{eV} = 1.36 \, \mathrm{eV}$$

Given $\lambda_{eq} = \lambda_1 + \lambda_2$. $$\frac{\ln 2}{(t_{1/2})_{eq}} = \frac{\ln 2}{(t_{1/2})_1} + \frac{\ln 2}{(t_{1/2})_2}$$ $$\left(t_{1/2}\right)_{eq} = \frac{(t_{1/2})_1 \times (t_{1/2})_2}{(t_{1/2})_1 + (t_{1/2})_2}$$ $$= \frac{3 \times 4.5}{3 + 4.5} = \frac{3 \times 4.5}{7.5} = \frac{3 \times 3}{5} = 1.8 \, hr$$

When the amplifier connects with positive feedback, it acts as the oscillator. The feedback here is positive feedback which means some amount of voltage is given to the input.

Given $$y(t) = 40 \sin (10 \times 10^6 \pi t)$$ $$x(t) = 20 \sin (1000 \pi t)$$ $\($$\Rightarrow$ $\omega$_c = 10^7 $\pi$$\)$ $\($$\omega$_m = 10^3 $\pi$$\)$ $\($A_c = 40$\)$ $\($A_m = 20$\)$ Equation of modulated wave = $\($(A_c + A_m $\sin$ $\omega$_m t) $\sin$ $\omega$_c t$\)$ $$= A_c \left( 1 + \frac{A_m}{A_c} \sin \omega_m t \right) \sin \omega_c t$$ $$= A_c (1 + \mu \sin \omega_m t) \sin \omega_c t, \mu = \frac{A_m}{A_c}$$ $$= A_c \sin \omega_c t + \frac{\mu A_c}{2} \left[ \cos(\omega_c - \omega_m)t - \cos(\omega_c + \omega_m)t \right]$$ Amplitude of minimum frequency = $$\frac{\mu A_c}{2} = \frac{A_m}{A_c} \times \frac{A_c}{2} = \frac{A_m}{2} = 10$$

Let they meet at $t = t$. So first ball gets $t$ sec. And second gets $(t - 2)$ sec. and they will meet at the same height. $$h_1 = 50t - \frac{1}{2}gt^2$$ $$h_2 = 50(t - 2) - \frac{1}{2}g(t - 2)^2$$ $h_1 = h_2$ $$50t - \frac{1}{2}gt^2 = 50(t - 2) - \frac{1}{2}g(t - 2)^2$$ $$100 = \frac{1}{2}g \left[t^2 - (t - 2)^2\right]$$ $$100 = \frac{10}{2} \left[4t - 4\right]$$ $$5 = t - 1$$ $t = 6$ sec.

Impulse = change in momentum $$= m[v - (-v)] = 2mv$$ $$= 2 \times 0.4 \times 15 = 12 \, \mathrm{Ns}$$

The acceleration is given by $a = \frac{6mg}{10m} = \frac{6g}{10} = \frac{3g}{5}$. Taking 8, 9, 10 together, we have $T = 3 \times ma = 3m \times \frac{3g}{5} = 36 \, \mathrm{N}$.

Given $m = 2000 \, \mathrm{gm/min}$. Heat required by water per minute is $mS\Delta T$. $$= (2000) \times 4.2 \times 40 \, \mathrm{J/min}$$ $$= 336000 \, \mathrm{J/min}$$ The rate of combustion is given by $$\left( \frac{dm}{dt} \right) L = 336000 \, \mathrm{J/min}$$ $$\frac{dm}{dt} = \frac{336000}{8 \times 10^3} \, \mathrm{g/min}$$ $$= 42 \, \mathrm{gm/min}$$

Given $f_1 = f$. $f_2 = f + 4$ $f_3 = f + 2 \times 4$ $f_4 = f + 3 \times 4$ $f_{20} = f + 19 \times 4$ $f + (19 \times 4) = 2 \times f$ $f = 76 \, \mathrm{Hz}$. Frequency of last tuning forks $= 2f$ $= 152 \, \mathrm{Hz}$

It should be mentioned, 10 cm wire is part of long wire. Force experienced by unit length of wire $$= \frac{\mu_0 I_1 I_2}{2 \pi d}, \ I_1 = I_2 = 5 \, \mathrm{A}$$ Force experienced by wires of length 10 cm $$= \frac{\mu_0 I_1 I_2}{2 \pi d} \times 10 \times 10^{-2}$$ $$10^{-5} = \frac{2 \times 10^{-7} \times 5 \times 5}{d} \times 10 \times 10^{-2}$$ $$d = 50 \times 10^{-3} \, \mathrm{m}$$ $$d = 50 \times 10^{-1} \, \mathrm{cm} = 5 \, \mathrm{cm}.$$

Given $h = \sqrt{7} \, \mathrm{m}$ and $\mu = \frac{4}{3}$. The critical angle $C$ is given by: $$\tan C = \frac{r}{h}$$ Therefore, $r = h \tan C$. Since $\sin C = \frac{1}{\mu} = \frac{3}{4}$, we have: $$\tan C = \frac{3}{\sqrt{7}}$$ Substituting back, we find: $$r = \sqrt{7} \times \frac{3}{\sqrt{7}} = 3$$ The area of the surface is $\pi r^2 = 9 \pi \, \mathrm{m}^2$.

$50$ VSD $=49$ MSD $\Rightarrow 1\ \mathrm{VSD}=\dfrac{49}{50}\ \mathrm{MSD}$ Least count $=1\ \mathrm{MSD}-1\ \mathrm{VSD}$ $=\left(1-\dfrac{49}{50}\right)\mathrm{MSD}$ $=\dfrac{1}{50}\,\mathrm{MSD}$ $1\ \mathrm{MSD}=\dfrac{1}{40}\,\mathrm{cm}$ Least count $=\dfrac{1}{50\times40}\,\mathrm{cm}$ $=\dfrac{1}{2000}\,\mathrm{cm}$ $=\dfrac{1}{2}\times10^{-5}\,\mathrm{m}$ $=0.5\times10^{-5}\,\mathrm{m}$ $=5\times10^{-6}\,\mathrm{m}$

Stopping potential $V_0 = 0.42 \, \mathrm{V}$ $\lambda = 6630 \, \mathrm{\AA}$ $E = \phi + eV_0$ $E$: energy of incident photon $V_0$: Stopping potential $\phi = E - eV_0$ $$E = \frac{12400}{6630} \, \mathrm{eV} = 1.87 \, \mathrm{eV}$$ $$\phi = (1.87 - 0.42) = 1.45 \, \mathrm{eV}$$ $\phi = h\nu_0$; $\nu_0$: threshold frequency $$1.45 \times 1.6 \times 10^{-19} = 6.63 \times 10^{-34} \times \nu_0$$ $$\nu_0 = 0.35 \times 10^{15}$$ $$= 35 \times 10^{13} \, \mathrm{sec^{-1}}$$ $$= 35$$

Radial node = n - l - 1 = 4 - 2 - 1 = 1. Angular node (l) = 2.

Invertase: Cane sugar $\rightarrow$ Glucose and fructose Zymase: Glucose $\rightarrow$ Ethanol and $\mathrm{CO_2}$ Diastase: Starch $\rightarrow$ Maltose Maltase: Maltose $\rightarrow$ Glucose

Te (Metalloid) refer NCERT (Page No. 83)

Metals are discussed in NCERT (Page No. 154).

$Mg(HCO_3)_2 \xrightarrow{\mathrm{Boil}} Mg(OH)_2+2CO_2\uparrow$ $Ca(HCO_3)_2 \xrightarrow{\mathrm{Boil}} CaCO_3+H_2O+CO_2\uparrow$

Refer NCERT (Page No. 300)

NO_2 $\rightarrow$ 7 + 2 $\times$ 8 = 23 electron (odd)

3MnO_4^{2-} + 4H^+ $\rightarrow$ 2MnO_4^- + MnO_2 + 2H_2O +6 +7 +4 So, it is disproportionation reaction.

Ce $= [\mathrm{Xe}] \, 4f^1 \, 5d^1 \, 6s^2$ $\mathrm{Ce^{3+}} = [\mathrm{Xe}] \, 4f^1 \, 5d^0$ $\mathrm{Ce^{4+}} = [\mathrm{Xe}] \, 4f^0 \, 5d^0$ (Noble gas configuration)

BOD for clean water $< 5 \, ppm$

Nucleophilicity is proportional to electron density on donor atom and proportional to the size of donor atom (in gas). It is also proportional to $\($ $\frac{1}{EN of atom}$ $\)$ (for period).

The reaction involves the oxidation of toluene using $\mathrm{CrO_3}$ and acetic anhydride to form an intermediate. This intermediate undergoes hydrolysis in the presence of $\mathrm{H_2O}$ and $\mathrm{H^+}$ to yield the final product, $1$-$\mathrm{CH_3COOH}$, along with $2 \ \mathrm{CH_3COOH}$.

Electrophile will attack at ortho and para position with respect to better electron releasing group (ERG). ERG: $-\mathrm{OH} > -\mathrm{CH_3}$. Para position with respect to $-\mathrm{OH}$ (+R) group and it will be meta position with respect to $-\mathrm{CH_3}$ group.

Hinsberg's reagent reacts with primary amines to form a product that is soluble in alkali. With secondary amines, the product is insoluble in alkali. Tertiary amines do not react with Hinsberg's reagent.

Refer NCERT (Page No. 452)

Refer NCERT (Page No. 426)

Given the reaction: $\mathrm{CH_4} + 2\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O}$. Calculate the moles: $$\frac{100}{16} = 6.25$$ $$\frac{208}{32} = 6.5$$ Mole to stoichiometric coefficient ratio: $$\frac{Mole}{Stoi. Coeff.} = \frac{6.25}{1} \frac{6.5}{2} = 3.25$$ So, $\mathrm{O_2}$ is the limiting reagent. Mole-Mole analysis: $$\frac{n_{\mathrm{O_2}}}{2} = \frac{n_{\mathrm{CO_2}}}{1}$$ $$\frac{6.5}{2} = n_{\mathrm{CO_2}}$$ Mass of $\mathrm{CO_2}$: $$\frac{6.5}{2} \times 44 = 143 \, \mathrm{gm}$$

$A \rightarrow 4 - \left(2 \times \dfrac{1}{2}\right) = 3$ $B \rightarrow 12 \times \dfrac{1}{4} + 1 \times 1 = 4$ So, Compound is $\mathrm{A_3B_4}$ The value of $x$ is $3$.

Number of lone pair on central atom for $\mathrm{H_2O}$ and $\mathrm{XeF_4}$ is equal to 2.

The reaction is $\mathrm{H_2O (l) \rightarrow H_2O (g)}$. The number of moles $n$ is calculated as $n = \frac{36}{18} = 2 \, \mathrm{mol}$. The change in internal energy $\Delta U$ is given by $\Delta U = \Delta H - \Delta n_g \, RT$. Substituting the values, we have $$\Delta U = 41.1 - \frac{1 \times 8.31 \times 373}{1000} \, \mathrm{kJ/mol}$$ which simplifies to $$\Delta U = 38 \, \mathrm{kJ/mol}$$

Given $\pi = iCRT$. $$= \frac{1 \times 2}{60000 \times 0.2} \times 0.083 \times 300$$ $$= 0.00415 \, bar (\because \, 1 \, bar = 10^5 \, Pa)$$ So, $0.00415 \times 10^5 \, Pa = 415 \, Pa$

The reaction is given by: $$ \mathrm{HI} \rightleftharpoons \frac{1}{2} \mathrm{H_2} + \frac{1}{2} \mathrm{I_2} $$ Initial concentration $t_i$ is 1. At equilibrium $t_{eq}$, the concentrations are: $$ 1 - 0.4, \frac{0.4}{2}, \frac{0.4}{2} $$ The equilibrium constant $K_p$ is calculated as: $$ K_p = \frac{(0.2)^{\frac{1}{2}} (0.2)^{\frac{1}{2}}}{1 - 0.4} = \frac{0.2}{0.6} = \frac{1}{3} $$ The change in Gibbs free energy is given by: $$ \Delta G = \Delta G^\circ + RT \ln K = 0 $$ Solving for $\Delta G^\circ$: $$ \Delta G^\circ = -RT \ln K \Rightarrow -8.31 \times 300 \times 2.3 \times \log \left( \frac{1}{3} \right) $$ This results in: $$ = 2735 \, \mathrm{J/mol} $$

Cu(s) + Sn$^{2+}$(0.001 M) $\rightarrow$ Cu$^{2+}$(0.01 M) + Sn(s) $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ $= -0.14 - 0.34$ $= -0.48\ V$ $E_{cell} = E^\circ_{cell} - \frac{0.059}{2}\log\left(\frac{0.01}{0.001}\right)$ $= -0.48 - \frac{0.059}{2}\log(10)$ $= -0.5095\ V$ $\Delta G = -nFE_{cell}$ $= -2 \times 96500 \times (-0.5095)$ $= 98333.5\ J\,mol^{-1}$ $= 98.335\ kJ\,mol^{-1}$ $= 983.35 \times 10^{-1}\ kJ\,mol^{-1}$ Nearest integer $= 983$

Given $K = Ae^{-\frac{E_a}{RT}}$. $K_{cat} = Ae^{-\frac{E_a}{RT}}$, $K_{uncat.} = Ae^{-\frac{E_a}{RT}}$. $$\frac{K_{cat}}{K_{uncat.}} = e^{\frac{E_a - E_{a1}}{RT}} = e^{\frac{10 \times 1000}{8.31 \times 300}} = e^{4.009} = e^x$$ Therefore, $x = 4$.

The reaction is given by: $$[\mathrm{Co(H_2O)_6}]^{2+} + \mathrm{NH_3 (excess)} \rightarrow [\mathrm{Co(NH_3)_6}]^{3+} + 6\mathrm{H_2O}$$ The complex is diamagnetic and forms a low spin complex. For $\mathrm{Co^{3+}}$, the electron configuration is $3d^6 \, 4s^0$. This implies $t_{2g}^6 \, e_g^0$. The total number of electrons is 6.

The reaction is given by $\mathrm{CH_4} + 2\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O}$. POAC on H atom gives: $$n_{\mathrm{CH_4}} \times 4 = n_{\mathrm{H_2O}} \times 2$$ Calculating $n_{\mathrm{CH_4}}$: $$n_{\mathrm{CH_4}} = \frac{81}{18} \times 2 \times \frac{1}{4} = \frac{81}{36}$$ Thus, $n_{\mathrm{CH_4}} = 2.25$. This is equal to $225 \times 10^{-2}$. The nearest integer is $225$.