JEE Main 1 February 2024 Shift 2 question paper with solutions

Given $f(x) = |2x^2 + 5|x| - 3|$. The graph of $y = 12x^2 + 5x - 3$ is shown. The graph of $f(x)$ is also shown. The number of points of discontinuity is $0 = m$. The number of points of non-differentiability is $3 = n$.

Given $px^2 + qx - r = 0$ with roots $\alpha$ and $\beta$. Let $p = A$, $q = AR$, $r = AR^2$. Then the equation becomes $Ax^2 + ARx - AR^2 = 0$. Simplifying, we have $x^2 + Rx - R^2 = 0$ with roots $\alpha$ and $\beta$. Therefore, $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}$. Thus, $\frac{\alpha + \beta}{\alpha \beta} = \frac{3}{4} \Rightarrow \frac{-R}{-R^2} = \frac{3}{4} \Rightarrow R = \frac{4}{3}$. Now, $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = R^2 - 4 \left(-R^2\right) = 5 \left(\frac{16}{9}\right) = \frac{80}{9}$.

Given $4 \sin^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0; \; x \in [-2\pi, 2\pi]$. Rearranging, we have: $$4 - 4 \cos^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0$$ Simplifying further: $$4 \cos^3 x + 4 \cos^2 x + 4 \cos x - 13 = 0$$ This implies: $$4 \cos^3 x + 4 \cos^2 x + 4 \cos x = 13$$ However, the left-hand side $\leq 12$ can't be equal to 13.

Given $$I = \int_0^1 \left(2x^3 - 3x^2 - x + 1\right)^{\frac{1}{3}} \, dx$$ Using $\int_0^{2a} f(x) \, dx$ where $f(2a - x) = -f(x)$, here $$f(1-x) = f(x)$$ Therefore, $I = 0$

Given $h = 3 \cos \theta$ and $k = \frac{18}{7} \sin \theta$. Therefore, the locus is $$\frac{x^2}{9} + \frac{49y^2}{324} = 1.$$ The eccentricity $e$ is calculated as $$e = \sqrt{1 - \frac{324}{49 \times 9}} = \frac{\sqrt{117}}{21} = \frac{\sqrt{13}}{7}.$$

$$\left(\frac{x^{\frac{1}{3}}}{3} + \frac{x^{\frac{-2}{3}}}{18}\right)^{18}$$ $$t_7 = {}^{18}c_6\left(\frac{x^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{x^{\frac{-2}{3}}}{2}\right)^{6} = {}^{18}c_6\frac{1}{(3)^{12}} \cdot \frac{1}{2^6}$$ $$t_{13} = {}^{18}c_{12}\left(\frac{x^{\frac{1}{3}}}{3}\right)^{6}\left(\frac{x^{\frac{-2}{3}}}{2}\right)^{12} = {}^{18}c_{12}\frac{1}{(3)^{6}} \cdot \frac{1}{2^{12}} \cdot x^{-6}$$ $$m = {}^{18}c_6 \cdot 3^{-12} \cdot 2^{-6} \quad : \quad n = {}^{18}c_{12} \cdot 2^{-12} \cdot 3^{-6}$$ $$\left(\frac{n}{m}\right)^{\frac{1}{3}} = \left(\frac{2^{-12} \cdot 3^{-6}}{3^{-12} \cdot 2^{-6}}\right)^{\frac{1}{3}} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4}$$

Given $f(0) = 2$, $\lim_{x \to -\infty} f(x) = 1$. The equation is $f'(x) - x \cdot f(x) = 3$. The integrating factor is $I.F = e^{-\alpha x}$. Integrating, we have: $$y \left( e^{-\alpha x} \right) = \int 3 \cdot e^{-\alpha x} \, dx$$ This gives: $$f(x) \cdot \left( e^{-\alpha x} \right) = \frac{3 e^{-\alpha x}}{-\alpha} + c$$ At $x = 0$, we have $2 = \frac{-3}{\alpha} + c \implies \frac{3}{\alpha} = c - 2 \ldots (1)$ Thus: $$f(x) = \frac{-3}{\alpha} + c \cdot e^{\alpha x}$$ As $x \to -\infty$, $1 = \frac{-3}{\alpha} + c(0)$ Solving, $\alpha = -3$, therefore $c = 1$. At $f(-\ln 2) = \frac{-3}{\alpha} + c \cdot e^{\alpha x}$: $$1 + e^{3 \ln 2} = 9$$ (But $\alpha$ should be greater than 0 for finite value of $c$)

Given $P(8\lambda - 3, 2\lambda + 4, 2\lambda - 1)$. PR = 6. $$(8\lambda - 4)^2 + (2\lambda + 2)^2 + (2\lambda - 4)^2 = 36$$ $\lambda = 0, 1$. Hence $P(-3, 4, -1)$ and $Q(5, 6, 1)$. Centroid of $\triangle PQR = (1, 4, 1) \equiv (\alpha, \beta, \gamma)$. $$\alpha^2 + \beta^2 + \gamma^2 = 18$$

A(1,3,2); B(-2,8,0); C(3,6,7); $\overrightarrow{AC}=2\hat{i}+3\hat{j}+5\hat{k}$ $AB=\sqrt{9+25+4}=\sqrt{38}$ $AC=\sqrt{4+9+25}=\sqrt{38}$ $\overrightarrow{AD}=\frac{1}{2}\hat{i}-4\hat{j}-\frac{3}{2}\hat{k} =\frac{1}{2}(\hat{i}-8\hat{j}-3\hat{k})$ Length of projection of $\overrightarrow{AD}$ on $\overrightarrow{AC}$ $=\left|\frac{\overrightarrow{AD}\cdot\overrightarrow{AC}}{|\overrightarrow{AC}|}\right| =\frac{37}{2\sqrt{38}}$

Given $S_{10} = 390$. $$\frac{10}{2} [2a + (10 - 1)d] = 390$$ This implies $2a + 9d = 78 \ldots (1)$. Given $\frac{t_{10}}{t_5} = \frac{15}{7}$, we have $\frac{a + 9d}{a + 4d} = \frac{15}{7}$, which implies $8a = 3d \ldots (2)$. From (1) and (2), $a = 3$ and $d = 8$. Now, $S_{15} - S_{5} = \frac{15}{2} (6 + 14 \times 8) - \frac{5}{2} (6 + 4 \times 8)$. $$= \frac{15 \times 118 - 5 \times 38}{2} = 790$$

The integral to solve is $$\int_0^{\pi/3} \cos^4 x \, dx$$. This can be rewritten as $$\int_0^{\pi/3} \left( \frac{1 + \cos 2x}{2} \right)^2 \, dx$$. Simplifying, we have $$= \frac{1}{4} \int_0^{\pi/3} (1 + 2 \cos 2x + \cos^2 2x) \, dx$$. This becomes $$= \frac{1}{4} \left[ \int_0^{\pi/3} dx + 2 \int_0^{\pi/3} \cos 2x \, dx + \int_0^{\pi/3} \frac{1 + \cos 4x}{2} \, dx \right]$$. Evaluating each integral, we get $$= \frac{1}{4} \left[ \frac{\pi}{3} + (\sin 2x)_0^{\pi/3} + \frac{1}{2} \left( \frac{\pi}{3} \right) + \frac{1}{8} (\sin 4x)_0^{\pi/3} \right]$$. This simplifies to $$= \frac{1}{4} \left[ \frac{\pi}{3} + (\sin 2x)_0^{\pi/3} + \frac{1}{2} \left( \frac{\pi}{3} \right) + \frac{1}{8} (\sin 4x)_0^{\pi/3} \right]$$. Further simplification gives $$= \frac{1}{4} \left[ \frac{\pi}{2} + \frac{\sqrt{3}}{2} + \frac{1}{8} \times \left( -\frac{\sqrt{3}}{2} \right) \right]$$. Finally, we have $$= \frac{\pi}{2} + \frac{7\sqrt{3}}{64}$$. Thus, $a = \frac{1}{8}$ and $b = \frac{7}{64}$. Therefore, $9a + 8b = \frac{9}{8} + \frac{7}{8} = 2$.

Given $|z| \geq 1$. The minimum value of $|z + \frac{3}{2} + 2i|$ is actually zero.

Given $$f(x) = \frac{\sqrt{x^2 - 25}}{4 - x^2} + \log_{10}(x^2 + 2x - 15)$$ Domain: $x^2 - 25 \geq 0$ implies $x \in (-\infty, -5] \cup [5, \infty)$. $4 - x^2 \neq 0$ implies $x \neq \{-2, 2\}$. $x^2 + 2x - 15 > 0$ implies $(x + 5)(x - 3) > 0$. This implies $x \in (-\infty, -5) \cup (3, \infty)$. Therefore, $x \in (-\infty, -5) \cup [5, \infty)$. Let $\alpha = -5$ and $\beta = 5$. Therefore, $\alpha^2 + \beta^3 = 150$.

Is $\overrightarrow{\mathrm{PN}} \cdot \mathbf{b} = 0$? $3(3\lambda - 2) + 2(2\lambda - 5) + (\lambda - 7) = 0$ $14\lambda = 23 \Rightarrow \lambda = \frac{23}{14}$ $\mathrm{N} \left( \frac{83}{14}, \frac{32}{14}, \frac{51}{14} \right)$ Therefore, $\frac{\alpha + 3}{2} = \frac{83}{14} \Rightarrow \alpha = \frac{62}{7}$ $\frac{\beta + 4}{2} = \frac{32}{14} \Rightarrow \beta = \frac{4}{7}$ $\frac{\gamma + 9}{2} = \frac{51}{14} \Rightarrow \gamma = \frac{-12}{7}$ Ans. $14(\alpha + \beta + \gamma) = 108$

Given $$f(x) = \begin{cases} x - 1; & x = even \\ 2x; & x = odd \end{cases}$$ $$f(f(f(a))) = 21$$ C-1: If $a = even$ $$f(a) = a - 1 = odd$$ $$f(f(a)) = 2(a - 1) = even$$ $$f(f(f(a))) = 2a - 3 = 21 \Rightarrow a = 12$$ C-2: If $a = odd$ $$f(a) = 2a = even$$ $$f(f(a)) = 2a - 1 = odd$$ $$f(f(f(a))) = 4a - 2 = 21 (Not possible)$$ Hence $a = 12$ Now $$\lim_{x \to 12^-} \left( \frac{|x|^3}{2} - \left[ \frac{x}{12} \right] \right)$$ $$= \lim_{x \to 12^-} \frac{|x|^3}{12} - \lim_{x \to 12^-} \left[ \frac{x}{12} \right]$$ $$= 144 - 0 = 144.$$

Given the equations: $$x + 2y + 3z = 5$$ $$2x + 3y + z = 9$$ $$4x + 3y + \lambda z = \mu$$ For infinite solutions, the following must hold: $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$. The determinant $\Delta$ is: $$\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda \end{vmatrix} = 0 \Rightarrow \lambda = -13$$ The determinant $\Delta_1$ is: $$\begin{vmatrix} 5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13 \end{vmatrix} = 0 \Rightarrow \mu = 15$$ The determinant $\Delta_2$ is: $$\begin{vmatrix} 1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13 \end{vmatrix} = 0$$ The determinant $\Delta_3$ is: $$\begin{vmatrix} 1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & 15 \end{vmatrix} = 0$$ For $\lambda = -13$, $\mu = 15$, the system of equations has infinite solutions. Hence $\lambda + 2\mu = 17$.

Given $x_1, x_2, \ldots, x_{10}$, we have $$\sum_{i=1}^{10} (x_i - \alpha) = 2 \Rightarrow \sum_{i=1}^{10} x_i - 10\alpha = 2$$ Therefore, the mean $\mu = \frac{6}{5} = \frac{\sum x_i}{10}$. Thus, $$\sum x_i = 12$$ $$10\alpha + 2 = 12 \Rightarrow \alpha = 1$$ Now, $$\sum_{i=1}^{10} (x_i - \beta)^2 = 40$$ Let $y_i = x_i - \beta$. Therefore, $$\sigma_y^2 = \frac{1}{10} \sum y_i^2 - (\bar{y})^2$$ $$\sigma_x^2 = \frac{1}{10} \sum (x_i - \beta)^2 - \left(\frac{\sum_{i=1}^{10} (x_i - \beta)}{10}\right)^2$$ $$\frac{84}{25} = 4 - \left(\frac{12 - 10\beta}{10}\right)^2$$ Therefore, $$\left(\frac{6 - 5\beta}{5}\right)^2 = 4 - \frac{84}{25} = \frac{16}{25}$$ $$6 - 5\beta = \pm 4 \Rightarrow \beta = \frac{2}{5} (not possible) or \beta = 2$$ Hence, $\frac{\beta}{\alpha} = 2$

Given $\overline{P(A)} = \frac{2}{7} = p$. $P(A \cap V) = \frac{1}{5} = q$. $P(A) = \frac{5}{7}$. Ans. $P(A \cap \overline{V}) = \frac{18}{35}$.

Given $mOM \cdot mCM = -1$. Therefore, $\($ $\frac{k}{h}$ $\cdot$ $\frac{k-1}{h}$ = -1 $\)$. Thus, the locus is $x^2 + y(y-1) = 0$. Simplifying, we get $x^2 + y^2 - y = 0$. For $p = \left| \frac{1/2}{\sqrt{2}} \right|$, we have $p = \frac{1}{2\sqrt{2}}$. The length $PQ = 2\sqrt{r^2 - p^2}$. Calculating, $PQ = 2\sqrt{\frac{1}{4} - \frac{1}{8}} = \frac{1}{\sqrt{2}}$.

Given $a,\ ar,\ ar^2$ forming a geometric progression. Sum of any two sides is greater than the third side: $a + ar > ar^2$,\quad $a + ar^2 > ar$,\quad $ar + ar^2 > a$ $r^2 - r - 1 0$ is always true. $r^2 + r - 1 > 0$ $$r \in \left(-\infty,\ \frac{-1-\sqrt{5}}{2}\right) \cup \left(\frac{-1+\sqrt{5}}{2},\ \infty\right) \hfill \ldots(2)$$ Taking the intersection of (1) and (2): $$r \in \left(\frac{-1+\sqrt{5}}{2},\ \frac{1+\sqrt{5}}{2}\right)$$ As $r > 1$: $$r \in \left(1,\ \frac{1+\sqrt{5}}{2}\right)$$ $[r] = 1$,\quad $[-r] = -2$ $3[r] + [-r] = 1$

Given $A = I_2 - 2MM^T$. $$A^2 = \left(I_2 - 2MM^T\right) \left(I_2 - 2MM^T\right)$$ $$= I_2 - 2MM^T - 2MM^T + 4MM^TM^T$$ $$= I_2 - 4MM^T + 4MM^T$$ $$= I_2$$ $AX = \lambda X$ $A^2X = \lambda AX$ $X = \lambda (\lambda X)$ $X = \lambda^2 X$ $X (\lambda^2 - 1) = 0$ $\lambda^2 = 1$ $\lambda = \pm 1$ Sum of square of all possible values $= 2$

Given $F^1(x) = x f(x)$ and $F(x^2) = x^4 + x^5$, let $x^2 = t$. Then $F(t) = t^2 + t^{5/2}$. Differentiating, we have $F'(t) = 2t + \frac{5}{2}t^{3/2}$. Therefore, $t \cdot f(t) = 2t + \frac{5}{2}t^{3/2}$, which implies $f(t) = 2 + \frac{5}{2}r^{1/2}$. Now, we calculate the sum: $$\sum_{r=1}^{12} f(r^2) = \sum_{r=1}^{12} \left(2 + \frac{5}{2}r\right)$$ $$= 24 + \frac{5}{2} \left[ \frac{12(13)}{2} \right]$$ $$= 219$$

Given $$y = \frac{(\sqrt{x} + 1) \left( x^2 - \sqrt{x} \right)}{x \sqrt{x} + x + \sqrt{x}} + \frac{1}{15} \left( 3 \cos^2 x - 5 \right) \cos^3 x$$ Simplifying, we have $$y = \frac{(\sqrt{x} + 1)(\sqrt{x}) \left( (\sqrt{x})^3 - 1 \right)}{(\sqrt{x}) \left( (\sqrt{x})^2 + (\sqrt{x}) + 1 \right)} + \frac{1}{5} \cos^5 x - \frac{1}{3} \cos^3 x$$ Further simplifying, $$y = (\sqrt{x} + 1)(\sqrt{x} - 1) + \frac{1}{5} \cos^5 x - \frac{1}{3} \cos^3 x$$ Differentiating, $$y' = 1 - \cos^4 x \cdot (\sin x) + \cos^2 x (\sin x)$$ Evaluating at $x = \frac{\pi}{6}$, $$y' \left( \frac{\pi}{6} \right) = 1 - \frac{9}{16} \times \frac{1}{2} + \frac{3}{4} \times \frac{1}{2}$$ Simplifying, $$= \frac{32 - 9 + 12}{32} = \frac{35}{32}$$ Finally, $$= 96 y' \left( \frac{\pi}{6} \right) = 105$$

Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} + 8\hat{j} + 2\hat{k}$, and $\vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}$. We know $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$. Therefore, $\left( \vec{b} - \vec{c} \right) \times \vec{a} = 0$. This implies $\vec{b} - \vec{c} = \lambda \vec{\alpha}$, where $\vec{b} = \vec{c} + \lambda \vec{\alpha}$. Substituting, $-\hat{i} - 8\hat{j} + 2\hat{k} = \left( 4\hat{i} + c_2\hat{j} + c_3\hat{k} \right) + \lambda (\hat{i} + \hat{j} + \hat{k})$. Solving, $\lambda + 4 = -1 \Rightarrow \lambda = -5$. $\lambda + c_2 = -8 \Rightarrow c_2 = -3$. $\lambda + c_3 = 2 \Rightarrow c_3 = 7$. Thus, $\vec{c} = 4\hat{i} - 3\hat{j} + 7\hat{k}$. Now, $\cos \theta = \frac{12 - 12 + 7}{\sqrt{26} \cdot \sqrt{74}} = \frac{7}{\sqrt{26} \cdot \sqrt{74}} = \frac{7}{2\sqrt{481}}$. $\tan^2 \theta = \frac{625 \times 3}{49}$. Therefore, $[\tan^2 \theta] = 38$.

Given $L_1$, $L_3$, $L_5$, $-$ $L_{19}$ are parallel. $L_2$, $L_4$, $L_6$, $-$ $L_{20}$ are concurrent. Total points of intersection are calculated as follows: $$\binom{20}{2} - \binom{10}{2} - \binom{10}{2} + 1$$ which equals 101.

Given the points $P(a, a^2)$ and $Q(-b, b^2)$, the area $S_2$ is given by the determinant: $$S_2 = \frac{1}{2} \left| \begin{array}{ccc} 0 & 0 & 1 \\ a & a^2 & 1 \\ -b & b^2 & 1 \end{array} \right| = \frac{1}{2}(ab^2 + a^2b).$$ The line $PQ$ has the equation: $$PQ: -y - a^2 = \frac{a^2 - b^2}{a + b}(x - a).$$ Simplifying, we have: $$y - a^2 = (a - b)x - (a - b)a$$ $$y = (a - b)x + ab.$$ The area $S_1$ is: $$S_1 = \int_{-b}^{a} ((a - b)x + ab - x^2) \, dx$$ Evaluating the integral: $$= (a - b) \frac{x^2}{2} + (ab)x - \frac{x^3}{3} \bigg|_{-b}^{a}$$ $$= \frac{(a-b)^2(a+b)}{2} + ab(a + b) - \frac{(a^3+b^3)}{3}.$$ The ratio $\frac{S_1}{S_2}$ is: $$\frac{S_1}{S_2} = \frac{(a-b)^2(a+b)}{2} + ab(a+b) - \frac{(a^3+b^3)}{3} \cdot \frac{2}{ab^2 + a^2b}$$ $$= \frac{3(a-b)^2 + 6ab - 2(a^2+b^2-ab)}{3ab}$$ $$= \frac{1}{3} \left[ a + \frac{b}{a} + 2 \right].$$ Given $\frac{4}{3} = \frac{m}{n}$, we find $m + n = 7.$

Given $ky^2 = 2(y-x)$ and $2y^2 = kx$. Point of intersection leads to $$ky^2 = \left( \frac{y - 2y^2}{k} \right)$$ For $y = 0$, $ky = 2 \left( 1 - \frac{2y}{k} \right)$. Solving $ky + \frac{4y}{k} = 2$, we find $y = \frac{2}{k+\frac{4}{k}} = \frac{2k}{k^2 + 4}$. The area $A$ is given by $$A = \int_0^{\frac{2k}{k^2+4}} \left( \left( y - \frac{ky^2}{2} \right) - \left( \frac{2y^2}{k} \right) \right) \cdot dy$$ This simplifies to $$= \frac{y^2}{2} - \left( \frac{k}{2} + \frac{2}{k} \right) \cdot \frac{y^3}{3} \bigg|_0^{\frac{2k}{k^2+4}}$$ Evaluating the integral, we have $$= \left( \frac{2k}{k^2+4} \right)^2 \left[ \frac{1}{2} - \frac{k^2 + 4}{2k} \times \frac{1}{3} \times \frac{2k}{k^2+4} \right]$$ This results in $$= \frac{1}{6} \times 4 \times \left( \frac{1}{2} \cdot \frac{k + \frac{4}{k}}{2} \right) \geq 2$$ Thus, $k + \frac{4}{k} \geq 4$. The area is maximum when $k = \frac{4}{k}$. Solving gives $k = 2, -2$.

$\dfrac{dx}{dy}-\dfrac{x}{y}=\dfrac{1-y^2}{y}$ Integrating factor $=e^{\int -\frac{1}{y}\,dy}$ $=\dfrac{1}{y}$ $\dfrac{x}{y}=\int \dfrac{1-y^2}{y^2}\,dy$ $\dfrac{x}{y}=-\dfrac{1}{y}-y+c$ $x=-1-y^2+cy$ $x(1)=1$ $1=-1-1+c \Rightarrow c=3$ $x=-1-y^2+3y$ $5x(2)=5(-1-4+6)$ $=5$

Using the sine rule, we have $\($ $\frac{c}{\sin 30^\circ}$ = $\frac{4\sqrt{3}}{\sin 120^\circ}$ $\)$. Solving for $\($ c $\)$, we get $\($ 2c = 8 $\Rightarrow$ c = 4 $\)$. The length $\($ AB = |(b + 1)| = 4 $\)$. Given $\($ b = 3 $\)$, the slope $\($ m_{AB} = 0 $\)$. The slope $\($ m_{BC} = $\frac{-1}{\sqrt{3}}$ $\)$. For line $\($ BC $\)$, we have $\($ -y = $\frac{-1}{\sqrt{3}}$(x - 3) $\)$ which simplifies to $\($ $\sqrt{3}$y + x = 3 $\)$. The point of intersection is given by the equations $\($ y = x + 3 $\)$ and $\($ $\sqrt{3}$y + x = 3 $\)$. Solving these, $\($ ($\sqrt{3}$ + 1)y = 6 $\)$ gives $\($ y = $\frac{6}{\sqrt{3} + 1}$ $\)$. For $\($ x $\)$, $\($ x = $\frac{6}{\sqrt{3} + 1}$ - 3 = $\frac{6 - 3\sqrt{3} - 3}{\sqrt{3} + 1}$ = 3$\left$($\frac{1 - \sqrt{3}}{1 + \sqrt{3}}$$\right$) = $\frac{-6}{(1 + \sqrt{3})^2}$ $\)$. Finally, $\($ $\frac{\beta^4}{\alpha^2}$ = 36 $\)$.

Given the circuit, we have the equation $I_S S = I_g G$. Substituting the given values, we get $$\frac{95}{100} I S = \frac{5I}{100} G.$$ Solving for $S$, we find $$S = \frac{G}{19}.$$ The resistance $R_A$ is given by $$R_A = \frac{S G}{S + G} = \frac{\frac{G^2}{19}}{\frac{20G}{19}}.$$ Simplifying, we get $$R_A = \frac{G}{20}.$$

For no deflection $\frac{0.8}{1} = \frac{R}{3}$. Therefore, $R = 2.4 \, \mathrm{m\Omega}$. Temperature fall in 10 s $= 20^\circ \mathrm{C}$. $$\Delta R = R \alpha \Delta t$$ $$\alpha = \frac{\Delta R}{R \Delta t} = \frac{-0.6}{3 \times 20}$$ $$= -10^{-2} \, \mathrm{C^{-1}}$$

Part of Theory

Given $\gamma = 1 + \frac{2}{f} = 1.4 \Rightarrow \frac{2}{f} = 0.4$. Therefore, $f = 5$. The work done $W = nR\Delta T = 200 \, \mathrm{J}$. The heat $Q$ is given by $$Q = \left(\frac{f + 2}{2}\right) nR\Delta T$$ $$= \frac{7}{2} \times 200 = 700 \, \mathrm{J}$$

Only the translational kinetic energy of the disc changes into gravitational potential energy. And rotational KE remains unchanged as there is no friction. $$\frac{1}{2}mv^2 = mgh$$ $$h = \frac{v^2}{2g}$$

Given $E_g = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} J$. $$= \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9} \times 1.6 \times 10^{-19}} eV$$ $$= \frac{15}{8} eV$$ So $x = 15$

Let's say the radius of small droplets is $r$ and that of the big drop is $R$. $$\frac{4}{3} \pi R^3 = 1000 \frac{4}{3} \pi r^3$$ $$R = 10r$$ $$U_i = 1000 \left(4 \pi r^2 S\right)$$ $$U_f = 4 \pi R^2 S$$ $$= 100 \left(4 \pi r^2 S\right)$$ $$U_f = \frac{1}{10} U_i$$

Given the formula for wavelength, we have: $$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{60 \times 10^6} = 5 \, \mathrm{m}$$

The average force is given by the change in momentum over the change in time. $$F_{av} = \frac{\Delta p}{\Delta t}$$ Substituting the given values: $$= \frac{0.12 \times 25}{0.1} = 30 \, \mathrm{N}$$

Given $P = nh\nu$. $$n = \frac{P}{h\nu} = \frac{2 \times 10^{-3}}{6.63 \times 10^{-34} \times 6 \times 10^{14}}$$ $$= 5 \times 10^{15}$$

For first minima $a \sin \theta = \lambda$ $$\sin \theta = \frac{\lambda}{a} = \frac{1}{2}$$ $$\theta = 30^\circ$$ Angular spread $= 60^\circ$

Given $\($ $\phi$_{smaller cube} = $\frac{2Q}{\epsilon_0}$ $\)$ and $\($ $\phi$_{bigger cube} = $\frac{5Q}{\epsilon_0}$ $\)$. Therefore, $\($ $\frac{\phi_{smaller cube}}{\phi_{bigger cube}}$ = $\frac{2}{5}$ $\)$.

The root mean square velocity $V_{rms}$ is given by $$V_{rms} = \sqrt{\frac{3RT}{M}}.$$ The ratio of velocities is $$\frac{V_1}{V_2} = \sqrt{\frac{M_2}{M_1}} \Rightarrow \frac{2}{V_2} = \sqrt{\frac{32}{2}}.$$ Solving for $V_2$, we find $$V_2 = 0.5 \, km/s.$$

Given $V_A = 20 \, \mathrm{m/s}$ and $V_B = -30 \, \mathrm{m/s}$. Velocity of B with respect to A is given by $$V_{B/A} = -50 \, \mathrm{m/s}$$ Velocity of ground with respect to B is $$V_{G/B} = 30 \, \mathrm{m/s}$$

In steady state, Req = 12$\,$ $\Omega$. I = $\frac{6}{12}$ = 0.5$\,$ A. P.D across C_1 = 3$\,$ V. P.D across C_2 = 4$\,$ V. q_1 = C_1 V_1 = 12$\,$ $\muC$. q_2 = C_2 V_2 = 24$\,$ $\muC$. $\frac{q_1}{q_2}$ = $\frac{1}{2}$.

First case $\frac{2}{40} = \frac{X}{60} \implies X = 3\, \Omega$. In second case $X' = \frac{2 \times 3}{2 + 3} = 1.2\, \Omega$. $$\frac{2}{\ell} = \frac{1.2}{100 - \ell}$$ $$200 - 2\ell = 1.2\ell$$ $$\ell = \frac{200}{3.2} = 62.5\, cm$$ Balance length changes by $22.5\, cm$.

Theoretical

Efficiency is given by $$\frac{E_s I_s}{E_p I_p}$$. Given $$0.8 = \frac{240 I_s}{4000}$$, we have $$3200 = 240 I_s$$. Solving for $I_s$, we get $$I_s = \frac{3200}{240} = 13.33 \, \mathrm{A}$$.

Given the equation for force, $$F = \frac{GMm}{R^{3/2}} = m\omega^2 R$$. We have $$\omega^2 \propto \frac{1}{R^{5/2}}$$, therefore, $$T = \frac{2\pi}{\omega}$$ so $$T^2 \propto R^{5/2}$$.

Net force $= 8\hat{i} + 4\hat{j} + 4\hat{k}$ $$\vec{a} = \frac{\vec{F}}{m} = 2\hat{i} + \hat{j} + \hat{k}$$

Given the equations for $f_1$ and $f_2$: $$f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$$ $$f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{9m}}$$ We find the ratio: $$\frac{f_1}{f_2} = \sqrt{\frac{9}{1}} = \frac{3}{1}$$

Given $V = 4\sqrt{x}$. Acceleration $a$ is given by $a = V \frac{dv}{dx}$. Substituting the values, we have: $$a = 4\sqrt{x} \times 4 \times \frac{1}{2} x^{-1/2} = 8 \, \mathrm{m/s^2}$$

Given $\tau = \mathrm{BINA} \sin \theta$. $C \theta = \mathrm{BINA} \sin 90^\circ$. $C = \frac{\mathrm{BINA}}{\theta} = \frac{0.01 \times 10 \times 10^{-3} \times 100 \times 2 \times 10^{-4}}{0.05} = 4 \times 10^{-5} \, \mathrm{N} - \mathrm{m/rad}$. $x = 4$

The change in length $\Delta L$ is given by the formula: $$\Delta L = \frac{FL}{AY}$$ where $F$ is the force, $L$ is the original length, $A$ is the cross-sectional area, and $Y$ is Young's modulus. The ratio of change in length to the original length is: $$\frac{\Delta L}{L} = \frac{F}{AY}$$ For two different forces $F_1$ and $F_2$, the ratio of their changes in length is: $$\frac{\Delta L_1}{L_1} \Big/ \frac{\Delta L_2}{L_2} = \frac{F_1}{F_2} = \frac{30}{10} = 3$$

Given the energy equation: $$E = -13.6 z^2 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right)$$ We have $$E = C \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$ Thus, $$h\nu = C \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$$ The ratio of frequencies is given by: $$\frac{\nu_1}{\nu_2} = \frac{\left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]_{2 \to 1}}{\left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]_{3 \to 1}}$$ Substituting the values, we get: $$= \frac{\left[ \frac{1}{1} - \frac{1}{4} \right]}{\left[ \frac{1}{1} - \frac{1}{9} \right]} = \frac{3/4}{8/9}$$ Simplifying further: $$= \frac{3}{4} \times \frac{9}{8}$$ Thus, $$\frac{\nu_1}{\nu_2} = \frac{27}{32}$$ Therefore, $$\nu_2 = \frac{32}{27} \nu_1 = \frac{32}{27} \times 3 \times 10^{15} \, \mathrm{Hz} = \frac{32}{9} \times 10^{15} \, \mathrm{Hz}$$

In steady state there will be no current in branch of capacitor, so no voltage drop across $R_2 = 5 \Omega$. $I_2 = 0$ $I_1 = I_3 = \frac{10}{4 + 6} = 1 \, \mathrm{A}$ $V_{R_3} = V_c + V_{R_2}$, $V_{R_2} = 0$ $I_3 R_3 = V_c$ $V_c = 1 \times 6 = 6 \, \mathrm{volt}$ $q_c = CV_c = 10 \times 6 = 60 \, \mu \mathrm{C}$

Given $\phi = NAB \cos(\omega t)$. The electromotive force $\varepsilon$ is given by $$\varepsilon = -\frac{d\phi}{dt} = NAB \omega \sin(\omega t).$$ The maximum electromotive force $\varepsilon_{max}$ is $$\varepsilon_{max} = NAB \omega$$ $$= 200 \times 0.2 \times 0.01 \times \pi$$ $$= \frac{4\pi}{10} = \frac{2\pi}{5} volt.$$

Let intensity of light on screen due to each slit is $I_0$. So intensity at centre of screen is $4I_0$. Intensity at distance $y$ from centre: $$I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi$$ $$I_{max} = 4I_0$$ $$\frac{I_{max}}{2} = 2I_0 = 2I_0 + 2I_0 \cos \phi$$ $$\cos \phi = 0$$ $$\phi = \frac{\pi}{2}$$ $$K \Delta x = \frac{\pi}{2}$$ $$\frac{2\pi}{\lambda} d \sin \theta = \frac{\pi}{2}$$ $$\frac{2}{\lambda} d \times \frac{y}{D} = \frac{1}{2}$$ $$y = \frac{\lambda D}{4d} = \frac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}}$$ $$= 125 \times 10^{-6}$$ $$= 125$$

Impulse $J = 0.2 \, \mathrm{N} - \mathrm{S}$ $J = \int F \, dt = 0.2 \, \mathrm{N} - \mathrm{s}$ Angular impulse $\vec{M}$ $$\vec{M}_c = \int \tau \, dt$$ $$= \int F \frac{L}{2} \, dt$$ $$= \frac{L}{2} \int F \, dt = \frac{L}{2} \times J$$ $$= \frac{0.3}{2} \times 0.2$$ $$= 0.03$$ $$I_{cm} = \frac{L^2}{12} = \frac{2 \times (0.3)^2}{12} = \frac{0.09}{6}$$ $$M = I_{cm} (\omega_f - \omega_i)$$ $$0.03 = \frac{0.09}{6} (\omega_f)$$ $$\omega_f = 2 \, \mathrm{rad/s}$$ $$\theta = \omega t$$ $$t = \frac{\theta}{\omega} = \frac{\pi}{2 \times 2} = \frac{\pi}{4} \, \mathrm{sec.}$$ $X = 4$

Given $\sin \theta = \frac{10}{20} = \frac{1}{2}$. Therefore, $\theta = 30^\circ$. We have $\tan \theta = \frac{qE}{mg}$. Substituting the values, $\tan 30^\circ = \frac{q \times 2 \times 10^4}{1 \times 10^{-3} \times 10}$. This simplifies to $\frac{1}{\sqrt{3}} = q \times 10^6$. Solving for $q$, we get $q = \frac{1}{\sqrt{3}} \times 10^{-6} \, \mathrm{C}$. Therefore, $x = 3$.

3rd Ionisation energy: [NCERT Data] V: $2833 \, \mathrm{KJ/mol}$ Cr: $2990 \, \mathrm{KJ/mol}$ Mn: $3260 \, \mathrm{KJ/mol}$ Fe: $2962 \, \mathrm{KJ/mol}$ Alternative Mn: $3d^5 4s^2$ Fe: $3d^6 4s^2$ Cr: $3d^5 4s^1$ V: $3d^3 4s^2$ So Mn has highest 3rd IE among all the given elements due to $d^5$ configuration.

A $\pi$ bonding molecular orbital has higher electron density above and below the inter nuclear axis.

$\mathrm{Cr}^{2+}$ is reducing as its configuration changes from $d^4$ to $d^3$ due to formation of $\mathrm{Cr}^{3+}$, which has half-filled $t_{2g}$ level. On the other hand, the change $\mathrm{Mn}^{3+}$ to $\mathrm{Mn}^{2+}$ results in half-filled $d^5$ configuration, which has extra stability.

The given reactions involve phenol as the starting compound. 1. When phenol is treated with zinc dust and heated, it forms benzene. 2. When phenol is treated with chloroform and sodium hydroxide, followed by acidification with hydrochloric acid, it forms salicylaldehyde. 3. When phenol is treated with carbon dioxide and sodium hydroxide, followed by acidification with hydrochloric acid, it forms salicylic acid. 4. When phenol is treated with concentrated nitric acid, it forms picric acid.

I. In p-Block both metals and non-metals are present but in d-Block only metals are present. II. EN and IE of non-metals are greater than that of metals. I - False, II - True

Strongest reducing agent: $\mathrm{BiH_3}$, explained by its low bond dissociation energy.

$\mathrm{CuSO_4\cdot 5H_2O}$ $\mathrm{Cu}^{2+} : 3d^9\,4s^0$ Unpaired electron is present, so it shows colour due to $d$-$d$ transition.

All are $-M$, hence meta directing groups.

For $\mathrm{H_2O}$, $\mathrm{NH_3}$, $\mathrm{C_2H_5OH}$, there is intermolecular H-bonding. The given structure shows intramolecular H-bonding.

Lassaigne's test is used for detection of all elements N, S, P, X.

Due to H-bonding, the boiling point of alcohol is high.

The reaction of 2-pentyne with hydrogen in the presence of Pd/C catalyst results in the formation of a fully saturated alkane. The reaction of 2-pentyne with sodium in liquid ammonia results in the formation of trans-2-butene.

For 3p: $n = 3$, $\ell = 1$. Number of radial nodes $= n - \ell - 1$. $$= 3 - 1 - 1 = 1$$

$CCl_4$ used in fire extinguisher. $CH_2Cl_2$ used as paint remover. Freons used in refrigerator and AC. DDT used as non Biodegradable insecticide.

The group shown exhibits a $-R$ effect, while the remaining three groups show a $+R$ effect via lone pair.

For $[\mathrm{Co(NH_3)_6}]^{3+}$, $\mathrm{Co^{3+}}$ (strong field ligand) $\Rightarrow 3\, d^6 \left(t_{2g}^6, e_g^0\right)$, Hybridisation: $d^2sp^3$. Inner orbital complex (spin paired complex). Pairing will take place. For $[\mathrm{CoF_6}]^{3-}$, $\mathrm{Co^{3+}}$ (weak field ligand) $\Rightarrow 3\, d^6 \left(t_{2g}^4, e_g^2\right)$. Hybridisation: $sp^3d^2$. Outer orbital complex (spin free complex) no pairing will take place.

SiO$_2$ and GeO$_2$ are acidic and SnO, PbO are amphoteric. Carbon does not have d-orbitals so can not form p$\pi$ - d$\pi$ bond with itself. Due to properties of catenation and p$\pi$ - p$\pi$ bond formation, carbon is able to show allotropic forms.

The reaction sequence starts with $\mathrm{C_2H_5Br}$ treated with alcoholic $\mathrm{KOH}$ to form $\mathrm{CH_2=CH_2}$ (A). Next, $\mathrm{Br_2}$ in $\mathrm{CCl_4}$ is added to form $\mathrm{CH_2Br-CH_2Br}$ (B). This compound is then treated with excess $\mathrm{KCN}$ to form $\mathrm{CH_2CN-CH_2CN}$ (C). Finally, hydrolysis with $\mathrm{H_3O^+}$ converts it to succinic acid, $\mathrm{CH_2(COOH)-CH_2(COOH)}$ (D).

Given the solubility equation: $$S = \frac{W \times 10}{M}$$ The dissociation of $\mathrm{Ca_3(PO_4)_2(s)}$ is given by: $$\mathrm{Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+}(aq.) + 2PO_4^{3-}(aq.)}$$ Let $3s$ and $2s$ be the concentrations of $\mathrm{Ca^{2+}}$ and $\mathrm{PO_4^{3-}}$ respectively. The solubility $S$ is: $$S = \frac{W \times 1000}{M \times 100} = \frac{W \times 10}{M}$$ The solubility product $K_{sp}$ is: $$K_{sp} = (3s)^3 (2s)^2$$ $$= 108 \, s^5$$ Substituting $s$: $$= 108 \times 10^5 \times \left(\frac{W}{M}\right)^5$$ $$= 1.08 \times 10^7 \left(\frac{W}{M}\right)^5$$

The reaction of $\mathrm{Ni^{2+}}$ with $\mathrm{NH_4OH}$ and dmg forms a complex with a five-membered ring structure. The structure shown is a coordination complex where $\mathrm{Ni^{2+}}$ is coordinated with dmg ligands forming a stable chelate. This complex is characterized by the presence of two five-membered rings.

The question shows four different chlorinated compounds. The correct answer is determined by identifying the compound with the correct structural feature or position of chlorine.

For the reaction $\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$, at $t = 0$, the concentration of $\mathrm{A}$ is $0.1$. At $t = 115 \, \mathrm{sec}$, the concentration of $\mathrm{A}$ is $0.1 - x$ and the concentration of $\mathrm{B}$ is $2x$. Given $0.1 + 2x = 0.28$, we solve for $x$: $$2x = 0.18$$ $$x = 0.09$$ The rate constant $K$ is calculated as: $$K = \frac{1}{115} \ln \frac{0.1}{0.1 - 0.09}$$ $$= 0.0200 \, \mathrm{sec^{-1}}$$ $$= 2 \times 10^{-2} \, \mathrm{sec^{-1}}$$

Let 3 different amino acids be A, B, C. Then the following combinations of tripeptides can be formed: ABC, ACB, BAC, BCA, CAB, CBA.

Given $\Delta T_f = i K_f \times molality$. $$24 = (1) \times 1.86 \times \frac{W}{62 \times 18.6}$$ $W = 14880 \, gm$ $= 14.880 \, kg$

Given the reaction: $\mathrm{H_2SO_4} + 2\mathrm{NH_3} \rightarrow (\mathrm{NH_4})_2\mathrm{SO_4}$. Millimole of $\mathrm{H_2SO_4} \rightarrow 10 \times 2$. So millimole of $\mathrm{NH_3} = 20 \times 2 = 40$. Organic compound $\rightarrow \mathrm{NH_3}$, 40 millimole. Therefore, mole of $N = \frac{40}{1000}$. Weight of $N = \frac{40}{1000} \times 14$. Percentage composition of $N$ in organic compound: $$\frac{40 \times 14}{1000 \times 1} \times 100$$ $$= 56\%$$

The reaction is given by $2\mathrm{H_2O} \rightarrow \mathrm{O_2} + 4\mathrm{H^+} + 4e^-$. The equation for work done is $\frac{W}{E} = \frac{Q}{96500}$. The moles multiplied by the n-factor is equal to $\frac{Q}{96500}$. Therefore, $1 \times 2 = \frac{Q}{96500}$. Solving for $Q$, we get $Q = 2 \times 96500\, \mathrm{C} = 1.93 \times 10^5\, \mathrm{C}$.

Given $\Delta G^\circ = -RT \ln K$. $$= -8.314 \times 300 \ln(10)$$ $$= 5744.14 \, \mathrm{J/mole}$$ $$= 57.44 \times 10^{-1} \, \mathrm{kJ/mole}$$

CellRx; $\mathrm{MnO_4^- + H_2C_2O_4 \rightarrow Mn^{2+} + CO_2}$ $E^\circ_{cell} = E^\circ_{op}$ of anode $+ E^\circ_{RP}$ of cathode $$= 0.49 + 1.51 = 2.00 \, \mathrm{V}$$ At equilibrium $E_{cell} = 0,$ $E^\circ_{cell} = \frac{0.059}{n} \log K$ (As per NCERT $\frac{RT}{F} = 0.059$ But $\frac{RT}{F} = 0.0591$ can also be taken.) $$2 = \frac{0.059}{10} \log K$$ $$\log K = 338.98$$

The reaction is given as: $$\mathrm{CxHy} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}$$ The balanced equation is: $$\mathrm{CxHy} + \left( x + \frac{y}{4} \right) \mathrm{O_2} \rightarrow x\mathrm{CO_2} + \frac{y}{2}\mathrm{H_2O}$$ Given: $$10x = 40$$ Solving for $x$ gives: $$x = 4$$ Next, solve for $y$: $$5y = 50$$ This gives: $$y = 10$$ Thus, the compound is $\mathrm{C_4H_{10}}$.