JEE Main 31 August 2021 Shift 2 question paper with solutions

Given $\alpha + \beta + \gamma = 2\pi$. $$\begin{vmatrix} 1 & \cos \gamma & \cos \beta \\ \cos \gamma & 1 & \cos \alpha \\ \cos \beta & \cos \alpha & 1 \end{vmatrix}$$ $$= 1 + 2 \cos \alpha \cdot \cos \beta \cdot \cos \gamma - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma$$ $$= \sin^2 \gamma - \cos^2 \alpha - \cos^2 \beta + (\cos(\alpha + \beta) + \cos(\alpha - \beta)) \cos \gamma$$ $$= \sin^2 \gamma - \cos^2 \alpha - \cos^2 \beta + \cos^2 \gamma + \cos(\alpha - \beta) \cos \gamma$$ $$= \sin^2 \alpha - \cos^2 \beta + \cos(\alpha - \beta) \cos(\alpha + \beta)$$ $$= \sin^2 \alpha - \cos^2 \beta + \cos^2 \alpha - \sin^2 \beta = 0$$

Suppose $\vec{r} = x \vec{a} + y \vec{b} + 2 \vec{c}$ and $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. $$\vec{a} \times \{ (\vec{r} - \vec{b}) \times \vec{a} \} + \vec{b} \times \{ (\vec{r} - \vec{c}) \times \vec{b} \} + \vec{c} \times \{ (\vec{r} - \vec{a}) \times \vec{c} \} = \vec{0}$$ $$\Rightarrow k^2 (\vec{r} - \vec{b}) - k^2 x \vec{a} + k^2 (\vec{r} - \vec{c}) - k^2 y \vec{b} + k^2 (\vec{r} - \vec{a}) - k^2 z \vec{c} = \vec{0}$$ $$\Rightarrow 3 \vec{r} - (\vec{a} + \vec{b} + \vec{c}) - \vec{r} = \vec{0}$$ $$\Rightarrow \vec{r} = \frac{\vec{a} + \vec{b} + \vec{c}}{2}$$

Given $f(x) = \sin^{-1}\left(\frac{3x^2 + x - 1}{(x-1)^2}\right) + \cos^{-1}\left(\frac{x-1}{x+1}\right)$. $$-1 \leq \frac{x-1}{x+1} \leq 1 \implies 0 \leq x < \infty \ldots (1)$$ $$-1 \leq \frac{3x^2 + x - 1}{(x-1)^2} \leq 1 \implies x \in \left[\frac{1}{4}, \frac{1}{2}\right] \cup \{0\} \ldots (2)$$ $(1) \& (2)$ $$\Rightarrow Domain = \left[\frac{1}{4}, \frac{1}{2}\right] \cup \{0\}$$

Given $g(3) = 2$, $g(1)$ can be defined in 3 ways. The number of onto functions in this condition is $3 \times 4!$. The total number of onto functions is $6!$. The required probability is $$\frac{3 \times 4!}{6!} = \frac{1}{10}.$$

Given $f(m+n) = f(m) + f(n)$. Put $m = 1$, $n = 1$. $$f(2) = 2f(1)$$ Put $m = 2$, $n = 1$. $$f(3) = f(2) + f(1) = 3f(1)$$ Put $m = 3$, $n = 3$. $$f(6) = 2f(3) \Rightarrow f(3) = 9$$ $$\Rightarrow f(1) = 3, f(2) = 6$$ $$f(2) \cdot f(3) = 6 \times 9 = 54$$

Given $P_1: 2x + 3y + 2z = 0$ which implies $\vec{n}_1 = 2\hat{i} + 3\hat{j} + 2\hat{k}$. For $P_2: x - 2y + z = 0$, we have $\vec{n}_2 = \hat{i} - 2\hat{j} + \hat{k}$. The direction vector of line $L$, which is the line of intersection of $P_1$ and $P_2$, is $\vec{r} = \vec{n}_1 \times \vec{n}_2 = 7\hat{i} - 7\hat{k}$. The direction ratios of $L$ are $(1, 0, -1)$. Therefore, the equation of $L$ is $\frac{x}{1} = \frac{y}{0} = \frac{z}{-1} = \lambda$. The direction ratios of $\overrightarrow{PQ}$ are $(\lambda + 1, -2, 2 - \lambda)$. Since $\overrightarrow{PQ} \perp \vec{r}$, we have $(\lambda + 1)(1) + (-2)(0) + (2 - \lambda)(-1) = 0$. Solving gives $\lambda = \frac{1}{2}$, which implies $Q\left(\frac{1}{2}, 0, -\frac{1}{2}\right)$. Therefore, $PQ = \frac{\sqrt{34}}{2}$.

Given $\sim (A \Rightarrow B) = A \land \sim B$. Therefore, $\sim ((p \lor r) \Rightarrow (q \lor r))$ is equal to $(p \lor r) \land (\sim q \land \sim r)$. This simplifies to $((p \lor r) \land (\sim r)) \land (\sim q)$, which further simplifies to $p \land (\sim r) \land (\sim q)$.

Given $\($ $\alpha$ = $\lim$_{x $\to$ $\frac{\pi}{4}$} $\frac{\tan^3 x - \tan x}{\cos \left( x + \frac{\pi}{4} \right)}$ $\)$; $\($ $\frac{0}{0}$ $\)$ form. Using L'Hopital's rule: $\[$ $\alpha$ = $\lim$_{x $\to$ $\frac{\pi}{4}$} $\frac{3 \tan^2 x \sec^2 x - \sec^2 x}{-\sin \left( x + \frac{\pi}{4} \right)}$ $\]$ $\($ $\Rightarrow$ $\alpha$ = -4 $\)$ $\($ $\beta$ = $\lim$_{x $\to$ 0} ($\cos$ x) $\cot$ x = e^{$\lim$_{x $\to$ 0} $\frac{(\cos x - 1)}{\tan x}$} $\)$ $\($ $\beta$ = e $\lim$_{x $\to$ 0} $\frac{-(1 - \cos x)}{x^2}$ , $\frac{x^2}{\left( \frac{\tan x}{x} \right) x}$ $\)$ $\($ $\beta$ = e $\lim$_{x $\to$ 0} $\left$( -$\frac{1}{2}$ $\right$) $\cdot$ $\frac{x}{1}$ = e^0 $\Rightarrow$ $\beta$ = 1 $\)$ $\($ $\alpha$ = -4; $\beta$ = 1 $\)$ If $\($ ax^2 + bx - 4 = 0 $\)$ are the roots then $\($ 16a - 4b - 4 = 0 $\)$ and $\($ a + b - 4 = 0 $\)$ $\($ $\Rightarrow$ a = 1 $\)$ and $\($ b = 3 $\)$

General point on $\frac{x^2}{4} + \frac{y^2}{9} = 1$ is $A(2 \cos \theta, 3 \sin \theta)$. Given $B(-3, -5)$. Midpoint $C \left( \frac{2 \cos \theta - 3}{2}, \frac{3 \sin \theta - 5}{2} \right)$. Let $h = \frac{2 \cos \theta - 3}{2}$ and $k = \frac{3 \sin \theta - 5}{2}$. Therefore, $$\left( \frac{2h + 3}{2} \right)^2 + \left( \frac{2k + 5}{3} \right)^2 = 1$$ which simplifies to $$36x^2 + 16y^2 + 108x + 80y + 145 = 0.$$

Given $\($ $\frac{dy}{dx}$ = $\frac{2^x(y+2^y)}{2^x(1+2^y \ln 2)}$ $\)$. This implies $$ \int \frac{(1+2^y) \ln 2}{(y+2^y)} \, dy = \int \, dx $$ which leads to $$ \ln |y + 2^y| = x + c $$ Using the initial condition $\($ x = 0; y = 0 $\)$, we find $\($ c = 0 $\)$. Thus, $$ x = \ln |y + 2^y| $$ At $\($ y = 1 $\)$, we have $\($ x = $\ln$ 3 $\)$. Therefore, $\($ 3 $\in$ (e, e^2) $\Rightarrow$ x $\in$ (1, 2) $\)$.

Given $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, x^2 + y^2 = ab$. Differentiating, we have: $$\frac{2x_1}{a^2} + \frac{2y_1 y_1'}{b^2} = 0$$ This implies: $$y_1' = \frac{-x_1}{a^2} \frac{b^2}{y_1} \ldots (1)$$ Also, $2x_1 + 2y_1 y_1' = 0$ gives: $$y_2' = \frac{-x_1}{y_1} \ldots (2)$$ Here $(x_1, y_1)$ is the point of intersection of both curves. Thus, $x_1^2 = \frac{a^2 b}{a+b}, y_1^2 = \frac{ab^2}{a+b}$. Therefore, $$\tan \theta = \frac{y_1' - y_2'}{1 + y_1' y_2'} = \frac{\frac{-x_1 b^2}{a^2 y_1} + \frac{x_1}{y_1}}{1 + \frac{x_1^2 b^2}{a^2 y_1^2}}$$ Simplifying, we find: $$\tan \theta = \frac{-b^2 x_1 y_1 + a^2 x_1 y_1}{a^2 y_1^2 + b^2 x_1^2}$$ Finally, $$\tan \theta = \frac{a-b}{\sqrt{ab}}$$

Let, $y = tx$ $$\frac{dy}{dx} = t + x \frac{dt}{dx}$$ Therefore, $$tx \left( t + x \frac{dt}{dx} \right) = x \left( t^2 + \frac{\varphi(t^2)}{\varphi'(t^2)} \right)$$ $$t^2 + xt \frac{dt}{dx} = t^2 + \frac{\varphi(t^2)}{\varphi'(t^2)}$$ $$\int \frac{t \varphi'(t^2)}{\varphi(t^2)} dt = \int \frac{dx}{x}$$ Let $\varphi(t^2) = p$ Therefore, $$\varphi'(t^2) 2t dt = dp$$ Thus, $$\int \frac{dy}{2p} = \int \frac{dx}{x}$$ $$\frac{1}{2} \ln \varphi(t^2) = \ln x + \ln c$$ $$\varphi(t^2) = x^2 k$$ $$\varphi \left( \frac{y^2}{x^2} \right) = kx^2, \varphi(1) = k$$ $$\varphi \left( \frac{y^2}{4} \right) = 4 \varphi(1)$$

Given the equation: $$x + 1 - 2 \log_2(3 + 2^x) + 2 \log_4(10 - 2^{-x}) = 0$$ We simplify using logarithmic properties: $$\log_2(2^{x+1}) - \log_2(3 + 2^x)^2 + \log_2(10 - 2^{-x}) = 0$$ This becomes: $$\log_2\left(\frac{2^{x+1} \cdot (10 - 2^{-x})}{(3 + 2^x)^2}\right) = 0$$ Thus: $$\frac{2(10 \cdot 2^x - 1)}{(3 + 2^x)^2} = 1$$ Solving gives: $$\Rightarrow 20 \cdot 2^x - 2 = 9 + 2^{2x} + 6 \cdot 2^x$$ Therefore: $$\therefore (2^x)^2 - 14(2^x) + 11 = 0$$ The roots are $2^{x_1}$ and $2^{x_2}$. Thus: $$\therefore 2^{x_1} \cdot 2^{x_2} = 11$$ Finally: $$x_1 + x_2 = \log_2(11)$$

Given $\frac{z-i}{z-1}$ is purely imaginary number. Let $z = x + iy$. $$\frac{x+i(y-1)}{(x-1)+iy} \times \frac{(x-1)-iy}{(x-1)-iy}$$ $$\Rightarrow \frac{x(x-1)+y(y-1)+i(-y-x+1)}{(x-1)^2+y^2}$$ is purely imaginary number. $$\Rightarrow x(x-1) + y(y-1) = 0$$ $$\Rightarrow \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{2}$$ Therefore, $|z - (3 + 3i)|_{\min} = |PC| - \frac{1}{\sqrt{2}}$ $$= \frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 2\sqrt{2}$$

Given $\frac{\frac{10}{2}(2a_1 + 9,d)}{\frac{p}{2}(2a_1 + (p-1)d)} = \frac{100}{p^2}$ $(2a_1 + 9,d),p = 10\left(2a_1 + (p-1)d\right)$ $2a_1p + 9dp = 20a_1 + 10d(p-1)$ $9dp = (20 - 2p)a_1 + 10d(p-1)$ $9p = (20 - 2p)\frac{a_1}{d} + 10(p-1)$ $\frac{a_1}{d} = \frac{10-p}{2(10-p)} = \frac{1}{2}$ Therefore, $\frac{a_{11}}{a_{10}} = \frac{a_1 + 10d}{a_1 + 9d}$ $= \frac{\frac{1}{2} + 10}{\frac{1}{2} + 9}$ $= \frac{21}{19}$

Given the determinant: $$\begin{vmatrix} 1 & 5 & 6 \\ \frac{1}{2} & 3 & 2 \\ 1 & \alpha & \beta \end{vmatrix} = 12$$ We have the equation: $$4\alpha - 2\beta = \pm 24 + 8$$ This implies: $$4\alpha - 2\beta = +24 + 8 \Rightarrow 2\alpha - \beta = 16$$ Equation (1): $$2x - y - 16 = 0 \ldots (1)$$ Also: $$4\alpha - 2\beta = -24 + 8 \Rightarrow 2\alpha - \beta = -8$$ Equation (2): $$2x - y + 8 = 0 \ldots (2)$$ The perpendicular distance of (1) from $(0, 0)$ is $$\left| \frac{0 - 0 - 16}{\sqrt{5}} \right| = \frac{16}{\sqrt{5}}$$ The perpendicular distance of (2) from $(0, 0)$ is $$\left| \frac{0 - 0 + 8}{\sqrt{5}} \right| = \frac{8}{\sqrt{5}}$$

Given $$(32)^{\tan^2 x} + (32)^{\sec^2 x} = 81$$ which implies $$(32)^{\tan^2 x} + (32)^{1 + \tan^2 x} = 81$$ leading to $$(32)^{\tan^2 x} = \frac{81}{33}.$$ In the interval $$\left[0, \frac{\pi}{4}\right]$$ there is only one solution.

Given $f(0) = 0$, $f(1) = 1$ and $f(2) = 2$. Let $h(x) = f(x) - x$ has three roots. By Rolle's theorem $h'(x) = f'(x) - 1$ has at least two roots. $h''(x) = f''(x) = 0$ has at least one root.

Given $$\pi^2 \left[ \int_0^1 \sin \frac{\pi x}{2} \, dx + \int_1^2 \sin \frac{\pi x}{2} (x-1) \, dx \right]$$ This equals $$\pi^2 \left[ -\frac{2}{\pi} \left( \cos \frac{\pi x}{2} \right) + \left( (x-1) \left( -\frac{2}{\pi} \cos \frac{\pi x}{2} \right) \right)_1^2 - \int_1^2 -\frac{2}{\pi} \cos \frac{\pi x}{2} \, dx \right]$$ This simplifies to $$\pi^2 \left[ 0 + \frac{2}{\pi} + \frac{2}{\pi} + \frac{2}{\pi} \cdot \frac{2}{\pi} \left( \sin \frac{\pi x}{2} \right)_1^2 \right]$$ Finally, we have $$= 4\pi - 4 = 4(\pi - 1)$$

Let 8, 16, $x_1, x_2, x_3, x_4, x_5$ be the observations. Now $$\frac{x_1 + x_2 + \cdots + x_5 + 14}{7} = 8$$ which implies $$\sum_{i=1}^{5} x_i = 42 \ldots (1)$$ Also $$\frac{x_1^2 + x_2^2 + \cdots + x_5^2 + 8^2 + 6^2}{7} - 64 = 16$$ which implies $$\sum_{i=1}^{5} x_i^2 = 560 - 100 = 460 \ldots (2)$$ So the variance of $x_1, x_2, \ldots, x_5$ is $$\frac{460}{5} - \left(\frac{42}{5}\right)^2 = \frac{2300 - 1764}{25} = \frac{536}{25}.$$

Given $\($ $\frac{10!}{\alpha! \beta! \gamma!}$ a^$\alpha$ (2$\,$ b)^$\beta$ $\cdot$ (4ab)^$\gamma$ $\)$ and $\($ $\frac{10!}{\alpha! \beta! \gamma!}$ a^{$\alpha$ + $\gamma$} $\cdot$ b^{$\beta$ + $\gamma$} $\cdot$ 2^$\beta$ $\cdot$ 4^$\gamma$ $\)$. $\($ $\alpha$ + $\beta$ + $\gamma$ = 10 $\)$ ...(1) $\($ $\alpha$ + $\gamma$ = 7 $\)$ ...(2) $\($ $\beta$ + $\gamma$ = 8 $\)$ ...(3) Adding (2) and (3) and subtracting (1) gives $\($ $\gamma$ = 5 $\)$. $\($ $\alpha$ = 2 $\)$ $\($ $\beta$ = 3 $\)$ So coefficients = $\($ $\frac{10!}{2!3!5!}$ $\cdot$ 2^3 $\cdot$ 2^{10} $\)$ = $\($ $\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{2 \times 3 \times 2 \times 5!}$ $\times$ 2^{13} $\)$ = $\($ 315 $\times$ 2^{16} $\Rightarrow$ k = 315 $\)$.

Point $(2, 2, -2)$ also lies on given plane. So $2 + 3 \times 2 - 2(-2) + \beta = 0$ $$\Rightarrow 2 + 6 + 4 + \beta = 0 \Rightarrow \beta = -12$$ Also $\alpha \times 1 - 5 \times 3 + 2 \times -2 = 0$ $$\Rightarrow \alpha - 15 - 4 = 0 \Rightarrow \alpha = 19$$ Therefore, $\alpha + \beta = 19 - 12 = 7$

A = 4-digit numbers divisible by 3 A = 1002, 1005, $\ldots$, 9999. 9999 = 1002 + (n - 1)3 $\Rightarrow$ (n - 1)3 = 8997 $\Rightarrow$ n = 3000 B = 4-digit numbers divisible by 7 B = 1001, 1008, $\ldots$, 9996 $\Rightarrow$ 9996 = 1001 + (n - 1)7 $\Rightarrow$ n = 1286 A $\cap$ B = 1008, 1029, $\ldots$, 9996 9996 = 1008 + (n - 1)21 $\Rightarrow$ n = 429 So, no divisible by either 3 or 7 = 3000 + 1286 - 429 = 3857 total 4-digits numbers = 9000 required numbers = 9000 - 3857 = 5143

= $\int$ $\frac{\sin x}{\cos^3 x}$ dx = $\int$ $\frac{\tan x \cdot \sec^2 x}{(\tan x + 1)(1 + \tan^2 x - \tan x)}$ dx Let $\($ $\tan$ x = t $\Rightarrow$ $\sec$^2 x $\cdot$ dx = dt $\)$ = $\int$ $\frac{t}{(t+1)(t^2-t+1)}$ dt = $\int$ $\left$( $\frac{A}{t+1}$ + $\frac{B(2t-1)}{t^2-t+1}$ + $\frac{C}{t^2-t+1}$ $\right$) dx $\Rightarrow$ A(t^2 - t + 1) + B(2t - 1)(t^2 - t + 1) + C(t + 1) = t $\Rightarrow$ t^2(A + 2B) + t(-A + B + C) + A - B + C = 1 $\therefore$ A + 2B = 0 -A + B + C = 1 A - B + C = 0 $\Rightarrow$ C = $\frac{1}{2}$ $\Rightarrow$ A - B = -$\frac{1}{2}$ $\cdots$ (4) A + 2B = 0 A - B = -$\frac{1}{2}$ $\Rightarrow$ 3B = $\frac{1}{2}$ $\Rightarrow$ B = $\frac{1}{6}$ A = -$\frac{1}{3}$ I = -$\frac{1}{3}$ $\int$ $\frac{dt}{1+t}$ + $\frac{1}{6}$ $\int$ $\frac{2t-1}{t^2-t+1}$ dt + $\frac{1}{2}$ $\int$ $\frac{dt}{t^2-t+1}$ = -$\frac{1}{3}$ $\ln$ |(1 + $\tan$ x)| + $\frac{1}{6}$ $\ell$ $\tan$^2 x - $\tan$ x + 1| + $\frac{1}{2}$ $\cdot$ $\frac{2}{\sqrt{3}}$ $\tan$^{-1} $\left$( $\frac{\tan x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}$ $\right$) = -$\frac{1}{3}$ $\ln$ |(1 + $\tan$ x)| + $\frac{1}{6}$ $\ell$ $\tan$^2 x - $\tan$ x + 1| + $\frac{1}{\sqrt{3}}$ $\tan$^{-1} $\left$( $\frac{2 \tan x - 1}{\sqrt{3}}$ $\right$) + C $\alpha$ = -$\frac{1}{3}$, $\beta$ = $\frac{1}{6}$, $\gamma$ = $\frac{1}{\sqrt{3}}$ 18 $\left$( $\alpha$ + $\beta$ + $\gamma$^2 $\right$) = 18 $\left$( -$\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{3}$ $\right$) = 3

The tangent of $y^2 = 8x$ is $y = mx + \frac{2}{m}$. At point $P(2, -4)$, we have $$-4 = 2m + \frac{2}{m}$$ which implies $$m + \frac{1}{m} = -2 \implies m = -1.$$ Therefore, the tangent is $y = -x - 2$. This gives $$x + y + 2 = 0 \ldots (1)$$ Equation (1) is also tangent to $x^2 + y^2 = a$. So $$\frac{2}{\sqrt{2}} = \sqrt{a} \implies \sqrt{a} = \sqrt{2}$$ Thus, $a = 2$.

Given $$S = \frac{7}{5} + \frac{9}{5^2} + \frac{13}{5^3} + \frac{19}{5^4} + \cdots$$ $$\frac{1}{5} S = \frac{7}{5^2} + \frac{9}{5^3} + \frac{13}{5^4} + \cdots$$ On subtracting $$\frac{4}{5} S = \frac{7}{5} + \frac{2}{5^2} + \frac{4}{5^3} + \frac{6}{5^4} + \cdots$$ $$S = \frac{7}{4} + \frac{1}{10} \left( 1 + \frac{2}{5} + \frac{3}{5^2} + \cdots \right)$$ $$S = \frac{7}{4} + \frac{1}{10} \left( 1 - \frac{1}{5} \right)^{-2}$$ $$= \frac{7}{4} + \frac{1}{10} \times \frac{25}{16} = \frac{61}{32}$$ $$\Rightarrow 160 \ S = 5 \times 61 = 305$$

$(I - A)^3 = I^3 - A^3 - 3A(I - A) = I - A^3$ $\Rightarrow 3A(I - A) = 0$ or $A^2 = A$ $\Rightarrow \begin{bmatrix} a^2 & ab + bd \\ 0 & d^2 \end{bmatrix} = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix}$ $\Rightarrow a^2 = a, \quad b(a + d - 1) = 0, \quad d^2 = d$ If $b \neq 0$, $a + d = 1 \Rightarrow 4$ ways If $b = 0$, $a = 0, 1$ & $d = 0, 1 \Rightarrow 4$ ways $\Rightarrow$ Total $8$ matrices

\text{Total area} $=\int_{0}^{3/2}(1+4x-x^{2})\,dx$ $=\left(x+2x^{2}-\frac{x^{3}}{3}\right)\Bigg|_{0}^{3/2}$ $=\frac{39}{8}$ and $\frac{39}{16}=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{3}{2}\cdot m$ $\Rightarrow 3m=\frac{13}{2}$ $\Rightarrow 12m=26$

Given $\tan \theta = \frac{3}{2}$. The ratio of the area of $\triangle APQ$ to the area of $\triangle BPQ$ is given by $$\frac{Area \triangle APQ}{Area \triangle BPQ} = \frac{AR}{RB} = \frac{3 \sin \theta}{2 \cos \theta} = \frac{9}{4}.$$ Therefore, $$8 \left( \frac{Area \triangle APQ}{Area \triangle BPQ} \right) = 18.$$

Given $$F'(x) = a(x-1)(x+3)$$ Then $$F''(x) = 6a(x+1)$$ Also, $$F'(x) = 3a(x+1)^2 + b$$ At $x = 1$, $$F'(1) = 0 \Rightarrow b = -12a$$ Now, $$F(x) = a(x+1)^3 - 12ax + c$$ Simplifying, $$= (x+1)^3 - 12x - 6$$ Finally, $$F(3) = 64 - 36 - 6 = 22$$

Force on each column $=\frac{mg}{4}$ Strain $=\frac{mg}{4AY}$ $=\frac{50 \times 10^3 \times 9.8}{4 \times \pi (1 - 0.25) \times 2 \times 10^{11}}$ $=2.6 \times 10^{-7}$

Given the triangle, the magnetic field $B$ is calculated as follows: $$B = 3 \left[ \frac{\mu_0 i}{4 \pi} (\sin 60^\circ + \sin 60^\circ) \right]$$ The tangent of $60^\circ$ is given by: $$\tan 60^\circ = \frac{\ell/2}{r}$$ Where $r = \frac{9 \times 10^{-2}}{2 \sqrt{3}} \, \mathrm{M}$. Therefore, the magnetic field is: $$\therefore \, B = 3 \times 10^{-5} \, \mathrm{T}$$ The current is flowing in a clockwise direction, so $\vec{B}$ is inside the plane of the triangle by the right-hand rule.

Given $M = 1.5 \, \mathrm{kg}$, $r = 0.5 \, \mathrm{m}$, $d = \frac{5}{2} \, \mathrm{m}$. The moment of inertia $I$ is given by $$I = 2 \left( \frac{2}{5} M r^2 + M d^2 \right)$$ Calculating, $$I = 19.05 \, \mathrm{kg \, m^2}$$

Given $\vec{A} = \vec{P} + \vec{Q}$ and $\vec{B} = \vec{P} - \vec{Q}$ with $\vec{P} \perp \vec{Q}$. The magnitudes are $|\vec{A}| = |\vec{B}| = \sqrt{P^2 + Q^2}$. The magnitude of $\vec{A} + \vec{B}$ is $|\vec{A} + \vec{B}| = \sqrt{2 \left(P^2 + Q^2\right)(1 + \cos \theta)}$. For $|\vec{A} + \vec{B}| = \sqrt{3 \left(P^2 + Q^2\right)}$, $\theta_1 = 60^\circ$. For $|\vec{A} + \vec{B}| = \sqrt{2 \left(P^2 + Q^2\right)}$, $\theta_2 = 90^\circ$.

For every large distance P.E. = 0 and total energy = 2.6 + 0 = 2.6 eV. Finally in first excited state of H atom total energy = -3.4 eV. Loss in total energy = 2.6 - (-3.4) = 6 eV. It is emitted as photon $$\lambda = \frac{1240}{6} = 206 \, nm$$ $$f = \frac{3 \times 10^8}{206 \times 10^{-9}} = 1.45 \times 10^{15} \, Hz$$ $$= 1.45 \times 10^9 \, Hz$$

Thermal resistance of spherical sheet of thickness $dr$ and radius $r$ is $$dR = \frac{dr}{K(4\pi r^2)}$$ $$R = \int_{r_1}^{r_2} \frac{dr}{K(4\pi r^2)}$$ $$R = \frac{1}{4\pi K} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) = \frac{1}{4\pi K} \left( \frac{r_2 - r_1}{r_1 r_2} \right)$$ Thermal current $(i) = \frac{\theta_2 - \theta_1}{R}$ $$i = \frac{4\pi K r_1 r_2}{r_2 - r_1} (\theta_2 - \theta_1)$$

Given $V_A = 5 \, \mathrm{V} \Rightarrow A = 1$, $V_A = 0 \, \mathrm{V} \Rightarrow A = 0$, $V_B = 5 \, \mathrm{V} \Rightarrow B = 1$, $V_B = 0 \, \mathrm{V} \Rightarrow B = 0$. If $A = B = 0$, there is no potential anywhere here $V_0 = 0$. If $A = 1$, $B = 0$, Diode $D_1$ is forward biased, here $V_0 = 5 \, \mathrm{V}$. If $A = 0$, $B = 1$, Diode $D_2$ is forward biased hence $V_0 = 5 \, \mathrm{V}$. If $A = 1$, $B = 1$, both diodes are forward biased hence $V_0 = 5 \, \mathrm{V}$. Truth table for 1st circuit: $$\begin{array}{ccc} A & B & Output \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}$$ Thus, the given circuit is an OR gate. For the 2nd circuit, $V_B = 5 \, \mathrm{V}$, $A = 1$ and $V_B = 0 \, \mathrm{V}$, $A = 0$. When $A = 0$, the E-B junction is unbiased, there is no current through it, therefore $V_0 = 1$. When $A = 1$, the E-B junction is forward biased, $V_0 = 0$. Thus, this circuit is not a gate.

Given the uncertainty principle: $$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$$ We have: $$\Delta x = \frac{h}{4\pi m \Delta v}$$ and $$v = \sqrt{\frac{3KT}{m}}$$ Therefore, $$\frac{\Delta x_e}{\Delta x_p} = \sqrt{\frac{m_p}{m_e}}$$

Given $T = 2\pi \sqrt{\ell/g}$. When bob is immersed in liquid, $mg_{eff} = mg - Buoyant force$. $mg_{eff} = mg - v \rho g$ ($\sigma$ = density of liquid$). $= mg - v \frac{\rho}{4} g$ $= mg - \frac{mg}{4} = \frac{3mg}{4}$ Therefore, $g_{eff} = \frac{3g}{4}$. $$T_1 = 2\pi \sqrt{\frac{\ell_1}{g_{eff}}}$$ $$\ell_1 = \ell + \frac{\ell}{3} = \frac{4\ell}{3}, \ell_{eff} = \frac{3g}{4}$$ By solving, $$T_1 = \frac{4}{3} 2\pi \sqrt{\ell/g}$$ $$T_1 = \frac{4}{3} T$$

Here $\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0$ $$\vec{F}_1 + \vec{F}_2 = -\vec{F}_3$$ Since $\vec{F}_{nt} = 0$ (equilibrium) Both statements correct

Given $[M] = K[F]^a[T]^b[V]^c$. $$[M^1] = \left[M^1 L^1 T^{-2}\right]^a \left[T^1\right]^b \left[L^1 T^{-1}\right]^c$$ a = 1, b = 1, c = -1 Therefore, $[M] = \left[FTV^{-1}\right]$

Given $\vec{F} = q(\vec{V} \times \vec{B})$. $$\vec{F}_1 = 4\pi \left[ 0.5 \hat{i} \times B_0 \left( \frac{\hat{i} + \hat{j}}{2} \right) \cos \left( K \cdot \frac{\pi}{K} - 0 \right) \right]$$ $$\vec{F}_2 = 2\pi \left[ 0.5c \hat{i} \times B_0 \left( \frac{\hat{i} + \hat{j}}{2} \right) \cos \left( K \cdot \frac{3\pi}{K} - 0 \right) \right]$$ $\cos \pi = -1$, $\cos 3\pi = -1$. Therefore, $\frac{F_1}{F_2} = 2$.

The circuit is simplified by combining resistors in parallel and series. First, the two pairs of $2 \, \Omega$ resistors are combined in parallel, resulting in $2 \, \Omega$. Next, the $3 \, \Omega$ resistors are combined in parallel, resulting in $1.5 \, \Omega$. These results are then combined in series with the $2 \, \Omega$ resistor. Finally, the equivalent resistance is calculated as follows: $$R_{eq} = \frac{3 \times 3/2}{3 + 3/2} = \frac{9/2}{9/2} = 1 \, \Omega.$$

Since $\phi = \vec{E} \cdot \vec{A} = EA \cos \theta$. $\theta = 90^\circ$ Therefore, $\phi = 0$

Given the equation $PV = nRT$, we have: $$400 \times 10^3 \times 500 \times 10^{-6} = n \left( \frac{25}{3} \right) (300)$$ Solving for $n$, we get: $$n = \frac{2}{25}$$ We also have $n = n_1 + n_2$: $$\frac{2}{25} = \frac{M_1}{2} + \frac{M_2}{32}$$ Additionally, $M_1 + M_2 = 0.76 \, \mathrm{gm}$. The ratio is given by: $$\frac{M_2}{M_1} = \frac{16}{3}$$

Given the equation $3MV_0 = 2MV_2 + MV_1$. We have $3 V_0 = 2 V_2 + V_1$. Solving for $V_2$, we get $120 = 2 V_2 + 60 \Rightarrow V_2 = 30 \, \mathrm{m/s}$. The change in kinetic energy is given by $$\frac{\Delta K.E.}{K.E} = \frac{\frac{1}{2}MV_1^2 + \frac{1}{2}2MV_2^2 - \frac{1}{2}3MV_0^2}{\frac{1}{2}3MV_0^2}$$ which simplifies to $$= \frac{V_1^2 + 2 V_2^2 - 3 V_0^2}{3 V_0^2}$$ $$= \frac{3600 + 1800 - 4800}{4800}$$ $$= \frac{1}{8}$$

The vector $\vec{B}$ must not be parallel to the plane of the coil for non-zero flux, and according to Lenz's law, if $\vec{B}$ is outward, it should be decreasing for anticlockwise induced current.

In S.H.M. total mechanical energy remains constant and also $\langle K.E. \rangle = \langle P.E. \rangle = \frac{1}{4} K A^2$ (for 1 time period)

To convert pulsating dc into steady dc both of mentioned method are correct.

Given $$g_{up} = \frac{g}{\left(1 + \frac{r}{R}\right)^2}$$ $$g_{down} = g \left(1 - \frac{r}{R}\right)$$ The ratio is $$\frac{g_{down}}{g_{up}} = \left(1 - \frac{r}{R}\right) \left(1 + \frac{r}{R}\right)^2$$ Expanding the expression, we have $$= \left(1 - \frac{r}{R}\right) \left(1 + \frac{2r}{R} + \frac{r^2}{R^2}\right)$$ Simplifying further, $$= 1 + \frac{r}{R} - \frac{r^2}{R^2} - \frac{r^3}{R^3}$$

Signal bandwidth = 2fm = 12$\mathrm{kHz}$ Therefore, N = $\frac{6\,\mathrm{MHz}}{12\,\mathrm{kHz}}$ = $\frac{6 \times 10^6}{12 \times 10^3}$ = 500

Given $\Delta U = \frac{1}{2}(\Lambda C)V^2$. $$\Delta U = \frac{1}{2}(KC - C)V^2$$ $$\Delta U = \frac{1}{2}(2 - 1)CV^2$$ $$\Delta U = \frac{1}{2} \times 200 \times 10^{-6} \times 200 \times 200$$ $$\Delta U = 4 \, \mathrm{J}$$

$\frac{\Delta M}{M}$ $=\frac{\Delta p_{x}}{p_{x}}$ $=\frac{250}{500}$ $=\frac{1}{2}$ $\frac{1}{2}$ $=\frac{x}{495}$ $\Rightarrow x=250$

Given $$y = mx + C$$ $$v^2 = \frac{20}{10}x + 20$$ $$v^2 = 2x + 20$$ $$2v \frac{dv}{dx} = 2$$ Therefore, $$a = v \frac{dv}{dx} = 1$$

Given $8 \beta = 2.4 \, \mathrm{cm}$. $$\frac{8 \lambda \Delta}{d} = 2.4 \, \mathrm{cm}$$ Substituting the values, $$\frac{8 \times 1.5 \times c}{0.3 \times 10^{-3} \times f} = 2.4 \times 10^{-2}$$ Solving for $f$, we find $$f = 5 \times 10^{14} \, \mathrm{Hz}$$

\[ \begin{aligned} 9\mathrm{MSD} &= 10\mathrm{VSD} \\ 9 \times 1\,\mathrm{mm} &= 10\mathrm{VSD} \\ \therefore\quad 1\mathrm{VSD} &= 0.9\,\mathrm{mm} \\ \mathrm{LC} &= 1\mathrm{MSD} - 1\mathrm{VSD} = 0.1\,\mathrm{mm} \\ \text{Reading} &= \mathrm{MSR} + \mathrm{VSR} \times \mathrm{LC} \\ &= 10 + 8 \times 0.1 = 10.8\,\mathrm{mm} \\ \text{Actual reading} &= 10.8 - 0.4 = 10.4\,\mathrm{mm} \\ \text{Radius} &= \frac{d}{2} = \frac{10.4}{2} = 5.2\,\mathrm{mm} \\ &= 52 \times 10^{-2}\,\mathrm{cm} \end{aligned} \]

Given $X_L = 2 \pi f L$. $f$ is very large. Therefore, $X_L$ is very large, hence open circuit. $X_C = \frac{1}{2 \pi f C}$. $f$ is very large. Therefore, $X_C$ is very small, hence short circuit. Final circuit $Z_{eq} = 1 + \frac{2 \times 2}{2 + 2} = 2$

Given $i = A = 60^\circ$. $\delta_{\min} = 2i - A$ $$= 2 \times 60^\circ - 60^\circ = 60^\circ$$ $$\mu = \frac{\sin^{-1}\left(\frac{\delta_{\min} + A}{2}\right)}{\sin'\left(\frac{A}{2}\right)}$$ $$= \sqrt{3}$$ $$V_{prism} = \frac{3 \times 10^8}{\sqrt{3}}$$ $$AP = 10 \times 10^{-2} \times \frac{\sqrt{3}}{2}$$ $$time = \frac{5 \times 10^{-2}}{3 \times 10^8} \times \sqrt{3} \times \sqrt{3}$$ $$= 5 \times 10^{-10} \, sec$$ Ans = 5

Given $E = i^2 R t$. $$192 = 16(R)(1)$$ Therefore, $R = 12 \, \Omega$. Now, $E^1 = (8)^2 (12)(5)$. Thus, $E = 3840 \, \mathrm{J}$.

More stable less potential energy. Stability order: I > III > IV > II. So potential energy: II > IV > III > I.

$\mathrm{Eu}\rightarrow[\mathrm{Xe}]\,4f^7\,6s^2$ $\mathrm{Eu}^{2+}\rightarrow[\mathrm{Xe}]\,4f^7$

The reaction begins with the formation of an acylium ion in the presence of $AlCl_3$. This ion undergoes electrophilic aromatic substitution (E.A.S.) with benzene to form compound (A), which is $C_6H_5CH_2CH_2COOH$. Further reduction with $Zn-Hg$ and $HCl$ leads to compound (B), which is $Ph-CH_2-CH_2-CH_2-COOH$.

No option contains all species that show disproportionation reaction. $\mathrm{MnO_4^-}$ Mn is in $+7$ oxidation state (highest) hence cannot be simultaneously oxidized or reduced.

Cell constant \[ \left(\frac{\ell}{A}\right) \Rightarrow \text{Units}=\mathrm{m^{-1}} \] Molar conductivity $(\Lambda_m)$ \[ \Rightarrow \text{Units}=\mathrm{S\,m^2\,mol^{-1}} \] Conductivity $(\kappa)$ \[ \Rightarrow \text{Units}=\mathrm{S\,m^{-1}} \] Degree of dissociation $(\alpha)$ \[ \Rightarrow \text{Dimensionless} \] Therefore, \[ (a)\text{--}(iii),\qquad (b)\text{--}(i),\qquad (c)\text{--}(iv),\qquad (d)\text{--}(ii) \]

The reaction begins with the nucleophilic attack of the amine group $\mathrm{NH_2}$ on the carbonyl carbon of the ester $\mathrm{H_3C-C(=O)-O-C(=O)-CH_3}$. This forms an intermediate $\mathrm{Ph-NH_2^+-O-C-CH_3}$. The intermediate rearranges to form $\mathrm{Ph-NH-C-CH_3}$ and $\mathrm{O=C-CH_3}$. The proton transfer results in $\mathrm{Ph-NH-C-CH_3}$ (A). Upon treatment with $\mathrm{Br_2}$ and $\mathrm{CH_3COOH}$, the compound is transformed into $\mathrm{NH-C-CH_3}$ with a bromine substituent on the benzene ring, resulting in the major product (B).

Keratin, collagen and myosin are examples of fibrous protein.

Oxides of nitrogen and sulphur are acidic and settle down on ground as dry deposition. Ammonium salts in rain drops result in wet deposition.

The reaction involves the conversion of an aldehyde group to a carboxylate ion using $\mathrm{NaOH}$ and heat ($\Delta$). The reaction proceeds as follows: The starting compound is a benzene ring with an aldehyde group ($\mathrm{CHO}$) and a hydroxymethyl group ($\mathrm{CH_2OH}$). Upon treatment with $\mathrm{NaOH}$ and heat, the aldehyde group is converted to a carboxylate ion ($\mathrm{COO^-Na^+}$), resulting in the formation of a benzene ring with a hydroxymethyl group and a sodium carboxylate group. The reaction is then treated with $\mathrm{H_3O^+}$, which converts the sodium carboxylate group to a carboxylic acid group ($\mathrm{COOH}$). The final products are a benzene ring with two hydroxymethyl groups and a benzene ring with a hydroxymethyl group and a carboxylic acid group.

One unpaired electron. Spin only magnetic moment is given by $$\sqrt{3} \, \mathrm{B.M.} = 1.73 \, \mathrm{BM}$$

Lithium salts are hydrated due to high hydration energy of $\mathrm{Li^+}$. $\mathrm{Li^+}$ due to smallest size in IA group has highest polarizing power.

Option (2) is incorrect $$\Delta G^\circ = -RT \ln K$$ $$\Delta H^\circ - T \Delta S^\circ = -RT \ln K$$ $$\ln K = -\left[ \frac{\Delta H^\circ - \Delta S^\circ}{RT} \right]$$

Atomic hydrogen is produced at high temperature in an electric arc or under ultraviolet radiations. The dissociation of dihydrogen at 2000 K is only 0.081%. H-H bond dissociation enthalpy is highest for a single bond for any diatomic molecule. Dihydrogen can be produced on reacting Zn with dil. HCl as well as NaOH (aq.).

Novalac is a linear polymer of $[\mathrm{Ph} - \mathrm{OH} + \mathrm{HCHO}]$. So ester linkage not present. So novalac is not a polyester.

Stability of oxides of Halogens is $$\mathrm{I} > \mathrm{Cl} > \mathrm{Br}$$

$\text{Mn}^{2+} \rightarrow \text{III group}$ $\text{As}^{3+} \rightarrow \text{II B group}$ $\text{Cu}^{2+} \rightarrow \text{II A group}$ $\text{Al}^{3+} \rightarrow \text{IV group}$

NaOH + EtOH is known as alcoholic NaOH, so it gives $\mathrm{E^2}$ reaction with given alkyl halide.

The reaction sequence begins with benzene reacting with $\mathrm{Br_2}$ and $\mathrm{Fe}$ under heat $\Delta$ to form bromobenzene. Bromobenzene then reacts with $\mathrm{Mg}$ in dry ether to form phenylmagnesium bromide ($\mathrm{PhMgBr}$). This Grignard reagent reacts with $\mathrm{CH_3OH}$ to form a compound with $\mathrm{Br}$ and $\mathrm{MgOCH_3}$ groups.

The reaction sequence starts with benzene. It undergoes nitration with $\mathrm{HNO_3}$ and $\mathrm{H_2SO_4}$ under heat $\Delta$ to form nitrobenzene $[A]$. Then, $[A]$ reacts with $\mathrm{Cl_2}$ in the presence of anhydrous $\mathrm{AlCl_3}$ to form $[B]$, which is chloronitrobenzene. Finally, $[B]$ is reduced with $\mathrm{Fe/HCl}$ to form $[C]$, which is chloroaniline.

Sulphurous acid has 1 S=O bond. Peroxodisulphuric acid has 4 S=O bonds. Pyrosulphuric acid has 4 S=O bonds.

We know $\frac{x}{m} = KP^{1/n}$; using $(x \propto V)$ $\Rightarrow \frac{10}{1} = K \times (100)^{1/n} \ldots (1)$ $\frac{15}{1} = K \times (200)^{1/n} \ldots (2)$ $\frac{V}{1} = K \times (300)^{1/n} \ldots (3)$ Divide $(2)/(1)$ $\frac{15}{10} = 2^{1/n}$ $\log\left(\frac{3}{2}\right) = \frac{1}{n}\log 2$ $\frac{1}{n} = \frac{\log 3 - \log 2}{\log 2} = \frac{0.4771 - 0.3010}{0.3010}$ $\frac{1}{n} = 0.585$ Divide $(3)/(1)$ $\frac{V}{10} = 3^{1/n}$ $\log\left(\frac{V}{10}\right) = \frac{1}{n}\log 3$ $\log\left(\frac{V}{10}\right) = 0.585 \times 0.4771 = 0.2791$ $\frac{V}{10} = 10^{0.2791}$ $\Rightarrow V = 10 \times 10^{0.2791}$ $\Rightarrow V = 10^{1.279} = 10^x$ $\Rightarrow x = 1.279$ $\Rightarrow x = 128 \times 10^{-2}$ (Nearest integer)

With benzene as solvent $$\Delta T_b = i K_b m$$ $$\Delta T_b = \frac{1}{2} \times 2.6 \times \frac{1.22/M_w}{100/1000} \ldots (1)$$ With Acetone as solvent $$\Delta T_b = i K_b m$$ $$0.17 = 1 \times 1.7 \times \frac{1.22/M_w}{100/1000} \ldots (2)$$ (1) / (2) $$\frac{\Delta T_b}{0.17} = \frac{\frac{1}{2} \times 2.6 + \frac{1.22/M_w}{100/1000}}{1 \times 1.7 \times \frac{1.22/M_w}{100/1000}}$$ $$\Delta T_b = \frac{0.26}{2}$$ $$\Delta T_b = 13 \times 10^2$$ $$\Rightarrow x = 13$$

The electron configuration for $\mathrm{Zn}^+$ is $1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^1$. The outermost electron is in the $4s$ subshell. Therefore, $m = 0$.

Anions from CCP or FCC ($A^-$) $= 4A^-$ per unit cell. Cations occupy all octahedral voids ($B^+$) $= 4B^+$ per unit cell. Cell formula $\rightarrow A_4 B_4$ Empirical formula $\rightarrow AB$ $\rightarrow (x = 1)$

Anode mud contains Sb, Se, Te, Ag, Au and Pt.

The reaction is given by: $$\mathrm{HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(\ell)}$$ At $t = 0$, the volumes and concentrations are $50 \, \mathrm{ml}, 1 \, \mathrm{M}$ for HCl and $30 \, \mathrm{ml}, 1 \, \mathrm{M}$ for NaOH. At $t = \infty$, the volumes are $20 \, \mathrm{mm}$ for HCl and none for NaOH. The concentration of HCl is calculated as: $$[\mathrm{HCl}] = \frac{20}{80} = \frac{1}{4} \, \mathrm{M} = 2.5 \times 10^{-1} \, \mathrm{M}$$ The pH is calculated as: $$\mathrm{pH} = -\log 2.15 \times 10^{-1} = 1 - 0.3979 = 0.6021$$ Finally, the pH is: $$\mathrm{pH} = 6021 \times 10^{-4}$$

Given $\log K = 20.35 - \frac{2.47 \times 10^3}{T}$. We know $\log K = \log A - \frac{E_a}{2.303RT}$. Therefore, $$\frac{E_a}{2.303RT} = 2.47 \times 10^3$$ $$E_a = 2.47 \times 10^3 \times 2.303 \times \frac{8.314}{1000} KJ/mole$$ $$= 47.29 = 47 (Nearest integer)$$

$\mathrm{Na_2O+H_2O\rightarrow2NaOH}$ $\frac{20}{62}\ \text{moles}$ Moles of $\mathrm{NaOH}$ formed $=\frac{20}{62}\times2$ $[\mathrm{NaOH}]=\frac{\frac{40}{62}}{\frac{500}{1000}}=1.29\,\mathrm{M}=13\times10^{-1}\,\mathrm{M}$ (Nearest integer)

Molecular orbital configuration of $\mathrm{O_2^{2-}}$ is $$\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} (\pi 2p_x^2 = \pi 2p_y^2) (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*2})$$ Zero unpaired electron

Given the reaction: $$\mathrm{C_2H_7N} + \left(2x + \frac{y}{2}\right) \mathrm{CuO} \rightarrow x \mathrm{CO_2} + \frac{y}{2} \mathrm{H_2O} + \frac{z}{2} \mathrm{N_2} + \left(2x + \frac{y}{2}\right) \mathrm{Cu}$$ On balancing $$\mathrm{C_2H_7N} + \frac{15}{2} \mathrm{CuO} \rightarrow 2 \mathrm{CO_2} + \frac{7}{2} \mathrm{H_2O} + \frac{1}{2} \mathrm{N_2} + \frac{15}{2} \mathrm{Cu}$$ On comparing $$y = 7$$