JEE Main 29 January 2024 Shift 2 question paper with solutions

Given $|P^{-1}AP - 2I| = |P^{-1}AP - 2P^{-1}P|$. This equals $|P^{-1}(A - 2I)P|$. Using the property of determinants, this becomes $|P^{-1}||A - 2I||P|$. Since $|P^{-1}| = 1/|P|$, it simplifies to $|A - 2I|$. Calculating the determinant: $$\begin{vmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{vmatrix} = 69$$ So, the prime factors of 69 are 3 and 23. Thus, the sum is 26.

3 Shelf empty: $(8, 0, 0, 0) \rightarrow 1 way$ $(7, 1, 0, 0)$ $(6, 2, 0, 0)$ 2 shelf empty: $(5, 3, 0, 0)$ $(4, 4, 0, 0) \rightarrow 4 ways$ 1 shelf empty: $(6, 1, 1, 0)$ $(4, 2, 1, 0)$ $(1, 2, 3, 2)$ $(2, 2, 2, 2)$ 0 Shelf empty: $(3, 3, 2, 0)$ $(4, 2, 2, 0) \rightarrow 5 ways$ $(5, 1, 1, 1)$ $(3, 3, 1, 1)$ $(4, 2, 1, 1) \rightarrow 5 ways$ Total $= 15 ways$

Direction ratio of PR = (4, 1, -1). Direction ratio of PQ = (1, 4, -1). Now, $\($ $\cos$ $\theta$ = $\left$| $\frac{4 + 4 + 1}{\sqrt{18} \cdot \sqrt{18}}$ $\right$| $\)$. $\($ $\theta$ = $\frac{\pi}{3}$ $\)$.

Given $\bar{X} = \frac{24}{5}$; $\sigma^2 = \frac{194}{25}$. Let first four observations be $x_1, x_2, x_3, x_4$. Here, $$\frac{x_1 + x_2 + x_3 + x_4 + x_5}{5} = \frac{24}{5} \cdots$$ Also, $$\frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{7}{2}$$ which implies $$x_1 + x_2 + x_3 + x_4 = 14.$$ Now from equation 1, $x_5 = 10$. Now, $\sigma^2 = \frac{194}{25}$. $$\frac{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2}{5} - \frac{576}{25} = \frac{194}{25}$$ which implies $$x_1^2 + x_2^2 + x_3^2 + x_4^2 = 54.$$ Now, variance of first 4 observations $$Var = \frac{\sum_{i=1}^{4} x_i^2}{4} - \left(\frac{\sum_{i=1}^{4} x_i}{4}\right)^2$$ $$= \frac{54}{4} - \frac{49}{4} = \frac{5}{4}.$$

Given $f(x) = 2x + 3(x)^{\frac{2}{3}}$. The derivative is $f'(x) = 2 + 2x^{-\frac{1}{3}}$. This can be rewritten as: $$= 2 \left( 1 + \frac{1}{x^{\frac{1}{3}}} \right)$$ $$= 2 \left( \frac{x^{\frac{1}{3}} + 1}{x^{\frac{1}{3}}} \right)$$ The sign chart shows: - Positive for $x 0$ So, maxima (M) at $x = -1$ and minima (m) at $x = 0$.

Given $z = 2 - i \left( 2 \tan \frac{5\pi}{8} \right) = x + iy$ (let). $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1} \frac{y}{x}$. $$r = \sqrt{(2)^2 + \left( 2 \tan \frac{5\pi}{8} \right)^2}$$ $$= \left| 2 \sec \frac{5\pi}{8} \right| = \left| 2 \sec \left( \pi - \frac{3\pi}{8} \right) \right|$$ $$= 2 \sec \frac{3\pi}{8}$$ and $\theta = \tan^{-1} \left( \frac{-2 \tan \frac{5\pi}{8}}{2} \right)$. $$= \tan^{-1} \left( \tan \left( \pi - \frac{5\pi}{8} \right) \right)$$ $$= \frac{3\pi}{8}$$

Given $\($ $\frac{3 \cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x}$ = x^3 - x^2 + 6 $\)$. This implies $$ \frac{\cos 2x \left( 3 + \cos^2 2x \right)}{\cos 2x \left( 1 - \sin^2 x \cos^2 x \right)} = x^3 - x^2 + 6 $$ which simplifies to $$ \frac{4 \left( 3 + \cos^2 2x \right)}{\left( 4 - \sin^2 2x \right)} = x^3 - x^2 + 6 $$ Further simplifying gives $$ \frac{4 \left( 3 + \cos^2 2x \right)}{\left( 3 + \cos^2 2x \right)} = x^3 - x^2 + 6 $$ Thus, $$ x^3 - x^2 + 2 = 0 \Rightarrow (x + 1) \left( x^2 - 2x + 2 \right) = 0 $$ So, the sum of real solutions is $\($-1$\)$.

Area of parallelogram, $S = |\vec{a} \times \vec{b}|$. Area of quadrilateral = Area($\triangle OAB$) + Area($\triangle OBC$). $$= \frac{1}{2} \{ |\vec{a} \times (12 \vec{a} + 4 \vec{b})| + |\vec{b} \times (12 \vec{a} + 4 \vec{b})| \}$$ $$= 8 |(\vec{a} \times \vec{b})|$$ Ratio $= \frac{8 |(\vec{a} \times \vec{b})|}{|(\vec{a} \times \vec{b})|} = 8$

Given $\log_e a$, $\log_e b$, $\log_e c$ are in A.P. Therefore, $b^2 = ac \ldots (i)$ Also $\log_e \left( \frac{a}{2b} \right)$, $\log_e \left( \frac{2b}{3c} \right)$, $\log_e \left( \frac{3c}{a} \right)$ are in A.P. $$\left( \frac{2b}{3c} \right)^2 = \frac{a}{2b} \times \frac{3c}{a}$$ $$\frac{b}{c} = \frac{3}{2}$$ Putting in eq. (i) $b^2 = a \times \frac{2b}{3}$ $$\frac{a}{b} = \frac{3}{2}$$ $a : b : c = 9 : 6 : 4$

Given $$\int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^3 x \cos^3 x \sin(x - \theta)}} \, dx$$ Let $$I = \int \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sqrt{\sin^3 x \cos^3 x (\sin x \cos \theta - \cos x \sin \theta)}} \, dx$$ This can be rewritten as: $$= \int \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x \cos^2 x \sqrt{\tan x \cos \theta - \sin \theta}} \, dx + \int \frac{\cos^{\frac{3}{2}} x}{\sin^2 x \cos^{\frac{3}{2}} x \sqrt{\cos \theta - \cot x \sin \theta}} \, dx$$ Let $$I = I_1 + I_2$$ For $I_1$, let $\tan x \cos \theta - \sin \theta = t^2$ Then $$\sec^2 x \, dx = \frac{2t \, dt}{\cos \theta}$$ For $I_2$, let $\cos \theta - \cot x \sin \theta = z^2$ Then $$\csc^2 x \, dx = \frac{2z \, dz}{\sin \theta}$$ Thus, $$I = I_1 + I_2$$ This becomes: $$= \int \frac{2t \, dt}{\cos \theta} + \int \frac{2z \, dz}{\sin \theta}$$ Simplifying gives: $$= \frac{2t}{\cos \theta} + \frac{2z}{\sin \theta}$$ Finally, $$= 2 \sec \theta \sqrt{\tan x \cos \theta - \sin \theta} + 2 \csc \theta \sqrt{\cos \theta - \cot x \sin \theta}$$ Comparing, we have $$AB = 8 \csc 2\theta$$

Writing P in terms of parametric coordinates $2 + r \cos \theta, 3 + r \sin \theta$ as $\tan \theta = \sqrt{3}$ $$P \left( 2 + \frac{r}{2}, 3 + \frac{\sqrt{3}r}{2} \right)$$ P must satisfy $2x - 3y + 28 = 0$ So, $2 \left( 2 + \frac{r}{2} \right) - 3 \left( 3 + \frac{\sqrt{3}r}{2} \right) + 28 = 0$ We find $r = 4 + 6\sqrt{3}$

Differential equation: $$x \cos \frac{y}{x} \frac{dy}{dx} = y \cos \frac{y}{x} + x$$ $$\cos \frac{y}{x} \left[ x \frac{dy}{dx} - y \right] = x$$ Divide both sides by $x^2$ $$\cos \frac{y}{x} \left( \frac{x \frac{dy}{dx} - y}{x^2} \right) = \frac{1}{x}$$ Let $\frac{y}{x} = t$ $$\cos t \left( \frac{dt}{dx} \right) = \frac{1}{x}$$ $$\cos t \, dt = \frac{1}{x} \, dx$$ Integrating both sides $$\sin t = \ln |x| + c$$ $$\sin \frac{y}{x} = \ln |x| + c$$ Using $y(1) = \frac{\pi}{3}$, we get $c = \frac{\sqrt{3}}{2}$ So, $\alpha = \sqrt{3} \Rightarrow \alpha^2 = 3$

Let $r$'th term of the GP be $ar^{n-1}$. Given, $$2a_r = a_{r+1} + a_{r+2}$$ $$2ar^{n-1} = ar^n + ar^{n+1}$$ $$\frac{2}{r} = 1 + r$$ $$r^2 + r - 2 = 0$$ Hence, we get $r = -2$ (as $r \neq 1$). So, $S_{20} - S_{18} = (\text{Sum up to 20 terms}) - (\text{Sum up to 18 terms}) = T_{19} + T_{20}$ $$T_{19} + T_{20} = ar^{18}(1+r)$$ Putting the values $a = \dfrac{1}{8}$ and $r = -2$: $$T_{19} + T_{20} = -2^{15}$$

Solving lines $L_1(3x + 2y = 14)$ and $L_2(5x - y = 6)$ to get $A(2, 4)$ and solving lines $L_3(4x + 3y = 8)$ and $L_4(6x + y = 5)$ to get $B\left(\frac{1}{2}, 2\right)$. Finding Eqn. of $AB : 4x - 3y + 4 = 0$. Calculate distance $PM$ $$\Rightarrow \left| \frac{4(5) - 3(-2) + 4}{5} \right| = 6$$

Assume $\sin^{-1} x = \theta$. $$\cos(2\theta) = \frac{1}{9}$$ $$\sin \theta = \pm \frac{2}{3}$$ As $m$ and $n$ are co-prime natural numbers, $$x = \frac{2}{3}$$ i.e. $m = 2$, $n = 3$. So, the quadratic equation becomes $2x^2 - 3x + 1 = 0$ whose roots are $\alpha = 1$, $\beta = \frac{1}{2} \left(1, \frac{1}{2}\right)$ lies on $$5x + 8y = 9$$

Given $$f(x) = \frac{x}{x^2 - 6x - 16}$$ Now, $$f'(x) = \frac{-(x^2 + 16)}{(x^2 - 6x - 16)^2}$$ $$f'(x) < 0$$ Thus $f(x)$ is decreasing in $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$

Given $$y = \log_e \left( \frac{1-x^2}{1+x^2} \right)$$ Differentiating, $$\frac{dy}{dx} = y' = \frac{-4x}{1-x^4}$$ Again, $$\frac{d^2y}{dx^2} = y'' = \frac{-4(1+3x^4)}{(1-x^4)^2}$$ Again $$y' - y'' = \frac{-4x}{1-x^4} + \frac{4(1+3x^4)}{(1-x^4)^2}$$ At $x = \frac{1}{2}$, $$y' - y'' = \frac{736}{225}$$ Thus $$225 \left( y' - y'' \right) = 225 \times \frac{736}{225} = 736$$

Given set = {1, 2, 3, $\ldots$, 50$\}$ P(A) = Probability that number is multiple of 4 $P(B)$ = Probability that number is multiple of 6 $P(C)$ = Probability that number is multiple of 7 Now, $$P(A) = \frac{12}{50}, \; P(B) = \frac{8}{50}, \; P(C) = \frac{7}{50}$$ again $$P(A \cap B) = \frac{4}{50}, \; P(B \cap C) = \frac{1}{50}, \; P(A \cap C) = \frac{1}{50}$$ $$P(A \cap B \cap C) = 0$$ Thus $$P(A \cup B \cup C) = \frac{12}{50} + \frac{8}{50} + \frac{7}{50} - \frac{4}{50} - \frac{1}{50} - \frac{1}{50} + 0$$ $$= \frac{21}{50}$$

Unit vector $\mathbf{\hat{u}} = x \mathbf{\hat{i}} + y \mathbf{\hat{j}} + z \mathbf{\hat{k}}$. $$\mathbf{P}_1 = \frac{1}{\sqrt{2}} \mathbf{\hat{i}} + \frac{1}{\sqrt{2}} \mathbf{\hat{k}}, \mathbf{P}_2 = \frac{1}{\sqrt{2}} \mathbf{\hat{j}} + \frac{1}{\sqrt{2}} \mathbf{\hat{k}}$$ $$\mathbf{P}_3 = \frac{1}{\sqrt{2}} \mathbf{\hat{i}} + \frac{1}{\sqrt{2}} \mathbf{\hat{j}}$$ Now angle between $\mathbf{\hat{u}}$ and $\mathbf{P}_1 = \frac{\pi}{2}$. $$\mathbf{\hat{u}} \cdot \mathbf{P}_1 = 0 \Rightarrow \frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0$$ $$\Rightarrow x + z = 0 \ldots (i)$$ Angle between $\mathbf{\hat{u}}$ and $\mathbf{P}_2 = \frac{\pi}{3}$. $$\mathbf{\hat{u}} \cdot \mathbf{P}_2 = |\mathbf{\hat{u}}| \cdot |\mathbf{P}_2| \cos \frac{\pi}{3}$$ $$\Rightarrow \frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \Rightarrow y + z = \frac{1}{\sqrt{2}} \ldots (ii)$$ Angle between $\mathbf{\hat{u}}$ and $\mathbf{P}_3 = \frac{2\pi}{3}$. $$\mathbf{\hat{u}} \cdot \mathbf{P}_3 = |\mathbf{\hat{u}}| \cdot |\mathbf{P}_3| \cos \frac{2\pi}{3}$$ $$\Rightarrow \frac{x}{\sqrt{2}} + \frac{4}{\sqrt{2}} = -\frac{1}{2} \Rightarrow x + y = \frac{-1}{\sqrt{2}} \ldots (iii)$$ From equation (i), (ii) and (iii) we get $$x = -\frac{1}{\sqrt{2}}, y = 0, z = \frac{1}{\sqrt{2}}$$ Thus $\mathbf{\hat{u}} - \mathbf{\hat{v}} = -\frac{1}{\sqrt{2}} \mathbf{\hat{i}} + \frac{1}{\sqrt{2}} \mathbf{\hat{k}} - \frac{1}{\sqrt{2}} \mathbf{\hat{i}} - \frac{1}{\sqrt{2}} \mathbf{\hat{j}} - \frac{1}{\sqrt{2}} \mathbf{\hat{k}}$ $$\mathbf{\hat{u}} - \mathbf{\hat{v}} = \frac{-2}{\sqrt{2}} \mathbf{\hat{i}} - \frac{1}{\sqrt{2}} \mathbf{\hat{j}}$$ $$\therefore |\mathbf{\hat{u}} - \mathbf{\hat{v}}|^2 = \left( \sqrt{\frac{4}{2} + \frac{1}{2}} \right)^2 = \frac{5}{2}$$

Given the equation $x^2 - \sqrt{6}x + 6 = 0$, we have roots $\alpha$ and $\beta$. The roots are given by $$x = \frac{\sqrt{6} \pm i \sqrt{6}}{2} = \frac{\sqrt{6}}{2} (1 \pm i).$$ Thus, $$\alpha = \sqrt{3} \left( e^{i \frac{\pi}{4}} \right), \beta = \sqrt{3} \left( e^{-i \frac{\pi}{4}} \right).$$ Therefore, $$\frac{\alpha^{99}}{\beta} + \alpha^{98} = \alpha^{98} \left( \frac{\alpha}{\beta} + 1 \right).$$ This simplifies to $$\frac{\alpha^{98}(\alpha + \beta)}{\beta} = 3^{49} \left( e^{i 99 \frac{\pi}{4}} \right) \times \sqrt{2}.$$ Thus, $$3^{49} (-1 + i) = 3^n (a + ib).$$ So, $n = 49$, $a = -1$, $b = 1$. Finally, $n + a + b = 49 - 1 + 1 = 49.$

Given $a, b, c$ are in A.P. Therefore, $2b = a + c \Rightarrow a - 2b + c = 0$. Thus, $ax + by + c$ passes through the fixed point $(1, -2)$. Therefore, $P = (1, -2)$. For infinite solution, $D = D1 = D2 = D3 = 0$. $$D : \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 0$$ This implies $\alpha = 8$. $$D_1 : \begin{vmatrix} 6 & 1 & 1 \\ \beta & 5 & \alpha \\ 4 & 2 & 3 \end{vmatrix} = 0 \Rightarrow \beta = 6$$ Therefore, $Q = (8, 6)$. Thus, $PQ^2 = 113$.

Parabola is $x^2 = 8y$. Chord with mid point $(x_1, y_1)$ is $T = S_1$. Therefore, $xx_1 - 4(y + y_1) = x_1^2 - 8y_1$. Thus, $(x_1, y_1) = \left(1, \frac{5}{4}\right)$. Therefore, $x - 4\left(y + \frac{5}{4}\right) = 1 - 8 \times \frac{5}{4} = -9$. Therefore, $x - 4y + 4 = 0 \ldots (i)$. $(\alpha, \beta)$ lies on (i) and also on $y^2 = 4x$. Therefore, $\alpha - 4\beta + 4 = 0 \ldots (ii)$ and $\beta^2 = 4\alpha \ldots (iii)$. Solving (ii) and (iii), $\beta^2 = 4(4\beta - 4) \Rightarrow \beta^2 - 16\beta + 16 = 0$. Therefore, $\beta = 8 \pm 4\sqrt{3}$ and $\alpha = 4\beta - 4 = 28 \pm 16\sqrt{3}$. Thus, $(\alpha, \beta) = (28 + 16\sqrt{3}, 8 + 4\sqrt{3})$ and $(28 - 16\sqrt{3}, 8 - 4\sqrt{3})$. Therefore, $(\alpha - 28)(\beta - 8) = (\pm 16\sqrt{3})(\pm 4\sqrt{3}) = 192$.

The integral from $\frac{\pi}{6}$ to $\frac{\pi}{3}$ of $\sqrt{1 - \sin 2x} \, dx$ is evaluated as follows: $$= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin x - \cos x \, dx$$ This can be split into two integrals: $$= \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx$$ Evaluating these integrals gives: $$= -1 + 2\sqrt{2} - \sqrt{3}$$ This can be expressed as: $$= \alpha + \beta \sqrt{2} + \gamma \sqrt{3}$$ Where $\alpha = -1$, $\beta = 2$, $\gamma = -1$. Finally, we have: $$3\alpha + 4\beta - \gamma = 6$$

The area is given by the integral: $$A = \int_{0}^{2} (x^2 + 2) \, dx + \int_{2}^{3} (2x + 2) \, dx$$ Calculating the integrals, we find: $$A = \frac{41}{3}$$ Therefore, multiplying by 12 gives: $$12A = 41 \times 4 = 164$$

Given $$L_1 : \frac{x - 5}{4} = \frac{y - 4}{1} = \frac{z - 5}{3} = \lambda$$ $$drs(4, 1, 3) = \mathbf{b_1}$$ $$M(4\lambda + 5, \lambda + 4, 3\lambda + 5)$$ $$L_2 : \frac{x + 8}{12} = \frac{y + 2}{5} = \frac{z + 11}{9} = \mu$$ $$N(12\mu - 8, 5\mu - 2, 9\mu - 11)$$ $$\overrightarrow{MN} = (4\lambda - 12\mu + 13, \lambda - 5\mu + 6, 3\lambda - 9\mu + 16) \ldots (1)$$ Now $$\overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 3 \\ 12 & 5 & 9 \end{vmatrix} = -6\hat{i} + 8\hat{k} \ldots (2)$$ Equation (1) and (2) $$\therefore \frac{4\lambda - 12\mu + 13}{-6} = \frac{\lambda - 5\mu + 6}{0} = \frac{3\lambda - 9\mu + 16}{8}$$ I and II $$\lambda - 5\mu + 6 = 0 \ldots (3)$$ I and III $$\lambda - 3\mu + 4 = 0 \ldots (4)$$ Solve (3) and (4) we get $$\lambda = -1, \mu = 1$$ $$\therefore M(1, 3, 2)$$ $$N(4, 3, -2)$$ $$\therefore \overrightarrow{OM} \cdot \overrightarrow{ON} = 4 + 9 - 4 = 9$$

Given $f(1) = 1$, $f(a) = 0$. $$f^2(x) = \lim_{r \to x} \left( \frac{2r^2 \left( f^2(r) - f(x)f(r) \right)}{r^2 - x^2} - r^3 e^{\frac{f(x)}{r}} \right)$$ $$= \lim_{r \to x} \left( \frac{2r^2 f(r)}{r + x} \frac{(f(r) - f(x))}{r - x} - r^3 e^{\frac{f(x)}{r}} \right)$$ $$f^2(x) = \frac{2x^2 f(x)}{2x} f'(x) - x^3 e^{\frac{f(x)}{x}}$$ $$y^2 = xy \frac{dy}{dx} - x^3 e^{\frac{y}{x}}$$ $$\frac{y}{x} \frac{dy}{dx} - \frac{x^2}{y} e^{\frac{y}{x}}$$ Put $y = vx$ then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. $$v = v + x \frac{dv}{dx} - \frac{x}{v} e^v$$ $$\frac{dv}{dx} = \frac{e^v}{v} \Rightarrow e^{-v} v dv = dx$$ Integrating both sides: $$e^v (x + c) + 1 + v = 0$$ $f(1) = 1 \Rightarrow x = 1, y = 1$ $$\Rightarrow c = -1 - \frac{2}{e}$$ $$e^v \left( -1 - \frac{2}{e} + x \right) + 1 + v = 0$$ $$e^{\frac{y}{x}} \left( -1 - \frac{2}{e} + x \right) + 1 + \frac{y}{x} = 0$$ $x = a, y = 0 \Rightarrow a = \frac{2}{e}$ $$ae = 2$$

Let $32^{32} = t$. $$64^{32^{32}} = 64^t = 8^{2t} = (9 - 1)^{2t}$$ $$= 9k + 1$$ Hence remainder = 1.

Given the equation $x^2 - 2^y = 2023$. Solving for $x$ and $y$, we find $x = 45$ and $y = 1$. Therefore, the sum $$\sum_{(x,y)=C} (x+y) = 46.$$

According to the question, $$27r_1 + \frac{9r_2}{2} = -9$$ $$\lim_{x \to 3} \frac{\int_3^x 8t^2 \, dt}{\frac{3r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3x}$$ $$= \lim_{x \to 3} \frac{8x^2}{\frac{3r_2}{2} - 2r_2 x - 3r_1 x^2 - 3} (using LH' Rule)$$ $$= \frac{72}{\frac{3r_2}{2} - 6r_2 - 27r_1 - 3}$$ $$= \frac{72}{-\frac{9r_2}{2} - 27r_1 - 3}$$ $$= \frac{72}{9 - 3} = 12$$

Given $\($ $\mathbf{n_1}$ $\times$ $\frac{hc}{\lambda_1}$ = 200 $\)$ and $\($ $\mathbf{n_2}$ $\times$ $\frac{hc}{\lambda_2}$ = 200 $\)$. Therefore, $\($ $\frac{\mathbf{n_1}}{\mathbf{n_2}}$ = $\frac{\lambda_1}{\lambda_2}$ = $\frac{300}{500}$ $\)$. Simplifying gives $\($ $\frac{\mathbf{n_1}}{\mathbf{n_2}}$ = $\frac{3}{5}$ $\)$.

The expression for the output is given by $$Y = A \cdot \overline{B} + \overline{A} \cdot B.$$ This can be rewritten as $$= (A + \overline{A}) \cdot B.$$ Since $A + \overline{A} = 1$, we have $$Y = 1 \cdot B.$$ Therefore, the final output is $$Y = B.$$

Given $$Q = \frac{a^4 b^3}{c^2}$$ The relative error in $Q$ is given by $$\frac{\Delta Q}{Q} = 4 \frac{\Delta a}{a} + 3 \frac{\Delta b}{b} + 2 \frac{\Delta c}{c}$$ Multiplying by 100 to find the percentage error, we have $$\frac{\Delta Q}{Q} \times 100 = 4 \left( \frac{\Delta a}{a} \times 100 \right) + 3 \left( \frac{\Delta b}{b} \times 100 \right) + 2 \left( \frac{\Delta c}{c} \times 100 \right)$$ Substituting the given percentage errors, % error in $Q = 4 \times 3\% + 3 \times 4\% + 2 \times 5\%$ $$= 12\% + 12\% + 10\%$$ $$= 34\%$$

Given $P_{avg} = V_{rms} I_{rms} \cos(\Delta \phi)$. $$= \frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \cos\left(\frac{\pi}{3}\right)$$ $$= \frac{10^4}{2} \times \frac{1}{2} \times 10^{-3}$$ $$= \frac{10}{4} = 2.5 \, W$$

Given the equation $PV = nRT$. We can express $PV$ as $PV = \frac{N}{N_A} RT$. Here, $N$ is the total number of molecules. The pressure $P$ is given by $P = \frac{N}{V} kT$. Substituting the values, we have: $$1.38 \times 1.01 \times 10^5 = 2 \times 10^{25} \times 1.38 \times 10^{-23} \times T$$ Simplifying, we get: $$1.01 \times 10^5 = 2 \times 10^2 \times T$$ Solving for $T$: $$T = \frac{1.01 \times 10^3}{2} \approx 500 \, \mathrm{K}$$

Given that $$m = 900 \mathrm{gm} = \frac{900}{1000} \mathrm{kg} = \frac{9}{10} \mathrm{kg}$$ $$r = 1 \, \mathrm{m}$$ $$\omega = \frac{2 \pi N}{60} = \frac{2 \pi (10)}{60} = \frac{\pi}{3} \, \mathrm{rad/sec}$$ $$T - mg = mr \omega^2$$ $$T = mg + mr \omega^2$$ $$= \frac{9}{10} \times 9.8 + \frac{9}{10} \times 1 \left( \frac{\pi}{3} \right)^2$$ $$= 8.82 + \frac{9}{10} \times \frac{\pi^2}{9}$$ $$= 8.82 + 0.98$$ $$= 9.80 \, \mathrm{N}$$

Given $\ell = 10 \, \mathrm{m}$. Initial energy $= mg\ell$. So, $\($ $\frac{9}{10}$ mg$\ell$ = $\frac{1}{2}$ mv^2 $\)$. Therefore, $\($ $\frac{9}{10}$ $\times$ 10 $\times$ 10 = $\frac{1}{2}$ v^2 $\)$. Solving for $v^2$, we get $v^2 = 180$. Thus, $v = \sqrt{180} = 6\sqrt{5} \, \mathrm{m/s}$.

Given the magnification $m = 2$, we have: $$m = \frac{-v}{u}$$ Substituting $v = 15 - u$, we get: $$2 = \frac{-(15 - u)}{-u}$$ Simplifying, we find: $$2u = 15 - u$$ $$3u = 15 \Rightarrow u = 5 \, \mathrm{cm}$$ Now, substituting back to find $v$: $$v = 15 - u = 15 - 5 = 10 \, \mathrm{cm}$$ Using the lens formula: $$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ Substitute the values: $$\frac{1}{f} = \frac{1}{10} + \frac{1}{(-5)} = \frac{1 - 2}{10} = \frac{-1}{10}$$ Thus, the focal length is: $$f = -10 \, \mathrm{cm}$$

Given $$R = \frac{mv}{qB} = \frac{p}{qB} = \frac{\sqrt{2\, m (\mathrm{KE})}}{qB} = \frac{\sqrt{2mqV}}{qB}$$ We have $$R \propto \sqrt{m}$$ And $$m \propto R^2$$ Thus, $$\frac{m_1}{m_2} = \left(\frac{R_1}{R_2}\right)^2$$

Given $\Delta x = \frac{7\lambda}{4}$. $$\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \times \frac{7\lambda}{4} = \frac{7\pi}{2}$$ $I = I_{\max} \cos^2 \left( \frac{\phi}{2} \right)$ $$\frac{I}{I_{\max}} = \cos^2 \left( \frac{\phi}{2} \right) = \cos^2 \left( \frac{7\pi}{2 \times 2} \right) = \cos^2 \left( \frac{7\pi}{4} \right)$$ $$= \cos^2 \left( 2\pi - \frac{\pi}{4} \right)$$ $$= \cos^2 \frac{\pi}{4}$$ $$= \frac{1}{2}$$

Volume constant $$\frac{4}{3} \pi R^3 = 27 \times \frac{4}{3} \times \pi r^3$$ $$R^3 = 27r^3$$ $$R = 3r$$ $$r = \frac{R}{3}$$ $$r^2 = \frac{R^2}{9}$$ Work done $= T \cdot \Delta A$ $$= 27 \, T \left(4 \pi r^2\right) - T 4 \pi R^2$$ $$= 27 \, T 4 \pi \frac{R^2}{9} - 4 \pi R^2 \, T$$ $$= 8 \pi R^2 \, T$$

Apply energy conservation between A and B $$\frac{1}{2} m V_L^2 = \frac{1}{2} m V_H^2 + mg(2L)$$ Therefore, $V_L = \sqrt{5gL}$ So, $V_H = \sqrt{gL}$ $$\frac{(\mathrm{K.E})_A}{(\mathrm{K.E})_B} = \frac{\frac{1}{2} m (\sqrt{5gL})^2}{\frac{1}{2} m (\sqrt{gL})^2} = \frac{5}{1}$$

Given $$Y = \frac{stress}{strain}$$ $$Y = \frac{\frac{F}{\ell/2}}{\frac{\ell}{L}}$$ $$F = Y \pi r^2 \times \frac{\ell}{L} \cdots (i)$$ $$Y = \frac{\frac{F r/2}{\Delta \ell/2}}{L}$$ $$F = Y \frac{\Delta \ell}{L} \times 2 \times \frac{\pi r^2}{4}$$ From (i) $$Y \pi r^2 \frac{\ell}{L} = Y \frac{\Delta \ell}{L} \frac{\pi r^2}{2}$$ $$\Delta \ell = 2 \ell$$

Given $T^2 \propto r^3$. Therefore, $\($ $\frac{T_1^2}{r_1^3}$ = $\frac{T_2^2}{r_2^3}$ $\)$. Substituting the given values, $\($ $\frac{(200)^2}{r^3}$ = $\frac{T_2^2}{\left(\frac{r}{4}\right)^3}$ $\)$. Simplifying, $\($ $\frac{200 \times 200}{4 \times 4 \times 4}$ = T_2^2 $\)$. Solving for $T_2$, we get $\($ T_2 = $\frac{200}{4 \times 2}$ $\)$. Therefore, $T_2 = 25$ days.

Given $\($ $\frac{E}{B}$ = C $\)$ and $\($ $\frac{E}{B}$ = 3 $\times$ 10^8 $\)$. Therefore, $\($ B = $\frac{E}{3 \times 10^8}$ = $\frac{9.6}{3 \times 10^8}$ $\)$. Thus, $\($ B = 3.2 $\times$ 10^{-8} $\mathrm{T}$ $\)$. The direction of $\($ $\mathbf{\hat{B}}$ = $\mathbf{\hat{v}}$ $\times$ $\mathbf{\hat{E}}$ $\)$ is $\($ $\mathbf{\hat{i}}$ $\times$ $\mathbf{\hat{j}}$ = $\mathbf{\hat{k}}$ $\)$. So, $\($ $\mathbf{B}$ = 3.2 $\times$ 10^{-8} $\mathbf{\hat{k}}$ $\mathrm{T}$ $\)$.

The equivalent resistance is given by $R_{eq} = 2\,\Omega + 2\,\Omega + 1\,\Omega = 5\,\Omega$. The current $i$ is calculated as $i = \frac{V}{R_{eq}} = \frac{10}{5} = 2\,A$. Current in resistance $R_3$ is calculated as follows: $$R_3 = 2 \times \left( \frac{4}{4 + 4} \right)$$ $$= 2 \times \frac{4}{8}$$ $$= 1\,A$$

Given $x = t^3 - 6t^2 + 20t + 15$. Differentiate with respect to $t$: $$\frac{dx}{dt} = v = 3t^2 - 12t + 20$$ Differentiate again to find acceleration: $$\frac{dv}{dt} = a = 6t - 12$$ When $a = 0$, solve for $t$: $$6t - 12 = 0; \ t = 2 sec$$ At $t = 2 sec$, calculate $v$: $$v = 3(2)^2 - 12(2) + 20$$ $$v = 8 \, \mathrm{m/s}$$

Given the equation for $f_{eq}$: $$f_{eq} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$$ For diatomic gas, $f_{eq} = 5$. $$5 = \frac{(N)(6) + (2)(3)}{N + 2}$$ Simplifying, we have: $$5N + 10 = 6N + 6$$ Solving for $N$: $$N = 4$$

According to Rutherford atomic model, most of mass of atom and all its positive charge is concentrated in tiny nucleus and electron revolve around it. According to Thomson atomic model, atom is spherical cloud of positive charge with electron embedded in it. Hence, Statement I is true but statement II false.

Given $\vec{E} = 6\hat{i} + 5\hat{j} + 3\hat{k}$ and $\vec{A} = 30\hat{i}$. The flux $\phi$ is given by the dot product $\vec{E} \cdot \vec{A}$. Therefore, $$\phi = (6\hat{i} + 5\hat{j} + 3\hat{k}) \cdot (30\hat{i})$$ $$\phi = 6 \times 30 = 180$$

Given $$Y = \frac{Stress}{Strain} = \frac{F/A}{\Delta \ell / \ell} = \frac{F \ell}{A \Delta \ell}$$ $$\Delta \ell = \frac{F \ell}{A Y}$$ $$V = A \ell \Rightarrow \ell = \frac{V}{A}$$ $$\Delta \ell = \frac{F V}{A^2 Y}$$ $Y$ and $V$ are the same for both the wires. $$\Delta \ell \propto \frac{F}{A^2}$$ $$\frac{\Delta \ell_1}{\Delta \ell_2} = \frac{F_1}{A_1^2} \times \frac{A_2^2}{F_2}$$ $$\Delta \ell_1 = \Delta \ell_2$$ $$F_1 A_2^2 = F_2 A_1^2$$ $$\frac{F_1}{F_2} = \frac{A_1^2}{A_2^2} = \left( \frac{4}{1} \right)^2 = 16$$

Given $B_H = 0.60 \times 10^{-4} \, \mathrm{Wb/m^2}$. Induced emf $e = B_H v \ell$. $$e = 0.60 \times 10^{-4} \times 10 \times 5$$ $$= 3 \times 10^{-3} \, \mathrm{V}$$

For minimum potential difference electron has to make transition from $n = 3$ to $n = 2$ state but first electron has to reach to $n = 3$ state from ground state. So, energy of bombarding electron should be equal to energy difference of $n = 3$ and $n = 1$ state. $$\Delta E = 13.6 \left[ 1 - \frac{1}{3^2} \right] e = \mathrm{eV}$$ $$\frac{13.6 \times 8}{9} = V$$ $$V = 12.09 \, \mathrm{V} \approx 12.1 \, \mathrm{V}$$ So, $\alpha = 121$

Given $q = 4 \, \mu \mathrm{C}$, $\vec{v} = 4 \times 10^6 \hat{\jmath} \, \mathrm{m/s}$ and $\vec{B} = 2 \hat{k} \, \mathrm{T}$. The force $\vec{F}$ is given by $q(\vec{v} \times \vec{B})$. $$\vec{F} = 4 \times 10^{-6} \left( 4 \times 10^6 \hat{\jmath} \times 2 \hat{k} \right)$$ $$= 4 \times 10^{-6} \times 8 \times 10^6 \hat{\imath}$$ $$\vec{F} = 32 \hat{\imath} \, \mathrm{N}$$ Therefore, $x = 32$.

From phasor diagram particle has to move from P to Q in a circle of radius equal to amplitude of SHM. $$\cos \phi = \frac{\sqrt{3} A}{2 A} = \frac{\sqrt{3}}{2}$$ $$\phi = \frac{\pi}{6}$$ Now, $$\frac{\pi}{6} = \omega t$$ $$\frac{\pi}{6} = \frac{2\pi}{T} t$$ $$\frac{\pi}{6} = \frac{2\pi}{6\pi} t$$ $$t = \frac{\pi}{2}$$ So, $$x = 2$$

At steady state, the capacitor behaves as an open circuit and current flows in the circuit as shown in the diagram. $$R_{eq} = 9 \, \Omega$$ $$i = \frac{9 \, V}{9 \, \Omega} = 1 \, A$$ $$\Delta V_{6\Omega} = 1 \times 6 = 6 \, V$$ $$V_A = 3 \, V$$ So, the potential difference across $6 \, \mu F$ is $6 \, V$. Hence $$Q = C \Delta V$$ $$= 6 \times 6 \times 10^{-6} \, C$$ $$= 36 \, \mu C$$

For $n^{th}$ minima $$b \sin \theta = n \lambda$$ ($\lambda$ is small so $\sin \theta$ is small, hence $\sin \theta \simeq \tan \theta$) $$b \tan \theta = n \lambda$$ $$\frac{b}{D} y = n \lambda$$ $$\Rightarrow \; y_n = \frac{n \lambda D}{b} \; (Position of $n^{th}$ minima)$$ $B \rightarrow 1^{st}$ minima, $A \rightarrow 3^{rd}$ minima $$y_3 = \frac{3 \lambda D}{b}, \; y_1 = \frac{\lambda D}{b}$$ $$\Delta y = y_3 - y_1 = \frac{2 \lambda D}{b}$$ $$3 \times 10^{-3} = \frac{2 \times 6000 \times 10^{-10} \times 0.5}{b}$$ $$b = \frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}}$$ $$b = 2 \times 10^{-4} \, \mathrm{m}$$ $$x = 2$$

Given, $i_1 = 0.3 \, \mathrm{A}$, $i_1 + i_2 + i_3 = 0.9 \, \mathrm{A}$. So, $V_{AB} = i_1 \times 20\, \Omega = 20 \times 0.3 \, \mathrm{V} = 6 \, \mathrm{V}$. $$i_2 = \frac{6 \, \mathrm{V}}{15\, \Omega} = \frac{2}{5} \, \mathrm{A}$$ $$i_1 + i_2 + i_3 = \frac{9}{10} \, \mathrm{A}$$ $$\frac{3}{10} + \frac{2}{5} + i_3 = \frac{9}{10}$$ $$\frac{7}{10} + i_3 = \frac{9}{10}$$ $$i_3 = 0.2 \, \mathrm{A}$$ So, $i_3 \times R_1 = 6 \, \mathrm{V}$. $$(0.2)R_1 = 6$$ $$R_1 = \frac{6}{0.2} = 30\, \Omega$$

Given $|\vec{a}_c| = |\vec{a}_t|$. $$\frac{v^2}{r} = \frac{dv}{dt}$$ This implies $$\int_4^v \frac{dv}{v^2} = \int_0^t \frac{dt}{r}$$ Therefore, $$\left[ -\frac{1}{v} \right]_4^v = \frac{t}{r}$$ This gives $$-\frac{1}{v} + \frac{1}{4} = 2t$$ Thus, $$v = \frac{4}{1 - 8t} = \frac{ds}{dt}$$ Integrating, $$4 \int_0^t \frac{dt}{1 - 8t} = \int_0^s ds$$ Given $r = 0.5 \, \mathrm{m}$ and $s = 2\pi r = \pi$, we have $$4 \times \left[ \ln(1 - 8t) \right]_0^t = \pi$$ This simplifies to $$\frac{\ln(1 - 8t)}{-8} = -2\pi$$ Therefore, $$1 - 8t = e^{-2\pi}$$ Solving for $t$, $$t = \left(1 - e^{-2\pi}\right) \frac{1}{8} s$$ So, $\alpha = 8$

Given the equation of the line: $$y - x - 4 = 0$$ $d_1$ is the perpendicular distance of the given line from the origin. $$d_1 = \left| \frac{-4}{\sqrt{1^2 + 1^2}} \right| \Rightarrow 2\sqrt{2} \, \mathrm{m}$$ So, $$|\vec{L}| = m v d_1 = 5 \times 3\sqrt{2} \times 2\sqrt{2} \, \mathrm{kg \, m^2/s}$$ $$= 60 \, \mathrm{kg \, m^2/s}$$

The stability of the conjugate base is proportional to acidic strength. The order is $\mathrm{C} < \mathrm{D} < \mathrm{B} < \mathrm{A}$.

A-II, B-III, C-I, D-IV Fact based.

Ethanol $\rightarrow 15.9$ Phenol $\rightarrow 10$ M-Nitrophenol $\rightarrow 8.3$ P-Nitrophenol $\rightarrow 7.1$

The reaction involves the addition of HI to the alkene. According to Markovnikov's rule, the iodine (I) will add to the more substituted carbon atom, resulting in the formation of the product shown.

Cyclohex-2-en-1-ol

Given $\mathrm{K_2MnO_4}$. $2 + x - 8 = 0$ Therefore, $x = +6$. The oxidation state of Mn is $+6$. IUPAC Name = Potassium tetraoxidomanganate(VI)

The reaction is given by: $$\mathrm{Ni^{2+} + 2dmg^- \rightarrow [Ni(dmg)_2]}$$ This results in a rosy red or bright red precipitate.

It is Reimer Tiemann Reaction

A - II, B - IV, C - III, D - I Fact based.

Nessler's Reagent Reaction: $$2 \, \mathrm{K_2HgI_4} + \mathrm{NH_3} + 3\mathrm{KOH} \rightarrow \mathrm{HgO.Hg(NH_2)I} + 7\mathrm{KI} + 2\mathrm{H_2O}$$ (Nessler’s Reagent) (iodine of Millon’s base Brown precipitate)

The reaction involves the conversion of aniline to chlorobenzene. First, aniline ($\mathrm{C_6H_5NH_2}$) is treated with $\mathrm{NaNO_2}$ and $\mathrm{HCl}$ at $0^\circ \mathrm{C}$ to form a diazonium salt ($\mathrm{C_6H_5N_2^+Cl^-}$). This is followed by a Sandmeyer reaction using $\mathrm{Cu_2Cl_2}$ to replace the diazonium group with a chlorine atom, resulting in chlorobenzene ($\mathrm{C_6H_5Cl}$).

The reaction sequence involves three main steps. First, the ester group is selectively reduced to an aldehyde using DIBAL-H. This is followed by an intramolecular aldol condensation facilitated by alcoholic NaOH, leading to the formation of a cyclic compound with an aldehyde group. Finally, Clemmensen reduction using Zn/HCl removes the carbonyl group, resulting in the formation of the final product.

For $\mathrm{Eu}^{+2}$, the electron configuration is $[\mathrm{Xe}]4f^7 6s^0$. When $\mathrm{Eu}^{+3}$ is formed, it loses one electron: $[\mathrm{Xe}]4f^7 6s^0 \rightarrow \mathrm{Eu}^{+3} + 1e^-$. The resulting configuration is $[\mathrm{Xe}]4f^6 6s^0$.

Q11 (A) $\mathrm{Zn}$, $\mathrm{Cd}$, $\mathrm{Hg}$ exhibit lowest enthalpy of atomization in respective transition series. (C) Compounds of $\mathrm{Zn}$, $\mathrm{Cd}$ and $\mathrm{Hg}$ are diamagnetic in nature.

Al < Si < C < N; IE_1 order.

Covalent character of $\mathrm{AgCN}$.

Due to unsymmetrical.

Statement-1 is false because chlorine has most negative electron gain enthalpy in its group.

Fact Based.

Antibonding molecular orbital from $2s = 1$ Antibonding molecular orbital from $2p = 3$ Total $= 4$

The complex is $[\mathrm{Fe(H_2O)_5(NO)]^{2+}}$. The oxidation number of $\mathrm{Fe}$ is $+1$.

Given the equilibrium constant expression: $$K_C = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$$ Substitute the given concentrations: $$K_C = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2}) \times (3 \times 10^{-2})^3}$$ Calculate the value: $$K_C = 417$$

Given the formula for molality: $$m = \frac{M \times 1000}{d_{sol} \times 1000 - M \times Molar mass_{solute}}$$ The calculated molality is $815 \times 10^{-3} \, m$.

Equivalent of Oxalic acid = Equivalents of NaOH $$50 \times 0.5 \times 2 = 25 \times M \times 1$$ $$M_{NaOH} = 2\, M$$ $$W_{NaOH} in 50\, ml$ $ = 2 \times 50 \times 40 \times 10^{-3}\, g = 4\, g$$

The molecule shown has a total of 22 bonds.

Half life of bromine-82 = 36 hours $$t_{1/2} = \frac{0.693}{K}$$ $$K = \frac{0.693}{36} = 0.01925 \, \mathrm{hr}^{-1}$$ First order reaction kinetic equation $$t = \frac{2.303}{K} \log \frac{a}{a-x}$$ $$\log \frac{a}{a-x} = \frac{t \times K}{2.303} (t = 1 \, \mathrm{day} = 24 \, \mathrm{hr})$$ $$\log \frac{a}{a-x} = \frac{24 \, \mathrm{hr} \times 0.01925 \, \mathrm{hr}^{-1}}{2.303}$$ $$\log \frac{a}{a-x} = 0.2006$$ $$\frac{a}{a-x} = anti log(0.2006)$$ $$\frac{a}{a-x} = 1.587$$ If $a = 1$ $$\frac{1}{1-x} = 1.587 \Rightarrow 1-x = 0.6301 = Fraction remain after one day$$

Given $\Delta H^0_{vap} \ \mathrm{CCl_4} = 30.5 \, \mathrm{kJ/mol}$. Mass of $\mathrm{CCl_4} = 284 \, \mathrm{gm}$. Molar mass of $\mathrm{CCl_4} = 154 \, \mathrm{g/mol}$. Moles of $\mathrm{CCl_4} = \frac{284}{154} = 1.844 \, \mathrm{mol}$. $\Delta H^0_{vap}$ for 1 mole $= 30.5 \, \mathrm{kJ/mol}$. $\Delta H^0_{vap}$ for 1.844 mol $= 30.5 \times 1.844 = 56.242 \, \mathrm{kJ}$.

Given $\frac{W}{E}=\frac{\text{charge}}{1\,\mathrm{F}}$. $1.314\times\frac{197}{3}=\frac{Q}{1\,\mathrm{F}}$ $Q=2\times10^{-2}\,\mathrm{F}$

Molecules with zero dipole moment are $\mathrm{CO_2}$, $\mathrm{CH_4}$, $\mathrm{BF_3}$.