JEE Main 7 April 2025 Shift 2 question paper with solutions

Solve line PQ and QR. Point Q $\left( \frac{1-c}{m-1}, \frac{1-c}{m-1} + 1 \right)$. $$m_{2H} = \frac{1-c}{m-1} + 2 = \frac{1-c+2m-2}{1-c-3m+3} = \frac{-1}{4}$$ $$m_{2H} = \frac{m-1}{1-c} + 2 = \frac{1-c+2m-2}{1-c-3m+3} = \frac{-1}{4} \cdots (1)$$ Therefore, $m_{PH} = \frac{5}{0} \to \infty$. Thus, the slope of line QR $(m) = 0$. Put the value of $m$ in equation (1): $$\frac{1-c-2}{1-c+3} = \frac{-1}{4} \Rightarrow c = 0$$ So $m - c = 0$. Ans.

Apply componendo and dividendo $$\frac{2|\bar{a} + \bar{b}|}{2|\bar{a} - \bar{b}|} = \sqrt{2} + 2$$ $$\Rightarrow \frac{|\bar{a} + \bar{b}|}{|\bar{a} - \bar{b}|} = \sqrt{2}$$ $$\Rightarrow |\bar{a} + \bar{b}| = (1 + \sqrt{2})|\bar{a} - \bar{b}|$$ $$\Rightarrow |\bar{a} + \bar{b}|^2 = (3 + 2\sqrt{2})|\bar{a} - \bar{b}|^2$$ $$\Rightarrow 2|\bar{a}|^2 + 2\bar{a} \cdot \bar{b} = (3 + 2\sqrt{2}) \left( 2|\bar{a}|^2 - 2\bar{a} \cdot \bar{b} \right)$$ $$\Rightarrow 2|\bar{a}|^2(2 + 2\sqrt{2}) = 2\bar{a} \cdot \bar{b}(4 + 2\sqrt{2})$$ $$\Rightarrow \frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^2} = \frac{2 + 2\sqrt{2}}{4 + 2\sqrt{2}} = \frac{1}{\sqrt{2}}$$ Now $$\frac{|\bar{a} + \bar{b}|^2}{|\bar{a}|^2} = 1 + \frac{|\bar{b}|^2}{|\bar{a}|^2} + 2\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^2}$$ $$= 1 + 1 + 2 \left( \frac{1}{\sqrt{2}} \right) = 2 + \sqrt{2}$$

A: $|x - 1| \leq 4$ and $|y - 5| \leq 6$. (1) $\Rightarrow -4 \leq x - 1 \leq 4 \Rightarrow -6 \leq y - 5 \leq 6$ $\Rightarrow -3 \leq x \leq 5 \Rightarrow -1 \leq y \leq 11$ B: $16(x - 2)^2 + 9(y - 6)^2 \leq 144$ B: $$\frac{(x - 2)^2}{9} + \frac{(y - 6)^2}{16} \leq 1$$ From Diagram $B \subset A$

Given $y = \frac{5-x}{x^2 - 3x + 2}$. $yx^2 - 3xy + 2y + x - 5 = 0$ $yz^2 + (-3y + 1)x + (2y - 5) = 0$ Case I: If $y = 0$ (Accepted) $$\Rightarrow x = 5$$ Case II: If $y \neq 0$ $D \geq 0$ $$(-3y + 1)^2 - 4(y)(2y - 5) \geq 0$$ $$9y^2 + 1 - 6y - 8y^2 + 20y \geq 0$$ $$y^2 + 14y + 1 \geq 0$$ $$(y + 7)^2 - 48 \geq 0$$ $$|y + 7| \geq 4\sqrt{3}$$ $$\Rightarrow y + 7 \geq 4\sqrt{3} or y + 7 \leq -4\sqrt{3}$$ $$\Rightarrow y \geq 4\sqrt{3} - 7 or y \leq -4\sqrt{3} - 7$$ From Case I and Case II $$y \in (-\infty, -4\sqrt{3} - 7] \cup [4\sqrt{3} - 7, \infty)$$ So $\alpha = -4\sqrt{3} - 7$ $\beta = 4\sqrt{3} - 7$ $$\Rightarrow a^2 + b^2 = (-4\sqrt{3} - 7)^2 + (4\sqrt{3} - 7)^2$$ $$= 2(48 + 49)$$ $$= 194$$

The probability of getting a head, $P(H)$, is calculated as follows: $$P(H) = \frac{19}{20} \times \frac{1}{2} + \frac{1}{20} \times 1$$ This accounts for the selection of an unbiased coin and the occurrence of a head, as well as the selection of a biased coin where a head always occurs. The required probability is: $$\frac{19}{20} \times \frac{1}{2} + \frac{1}{20} \times 1 = \frac{19}{21}$$ Thus, $m = 19$ and $n = 21$. Therefore, $$n^2 - m^2 = 441 - 361 = 80$$

Given $2p + 2q = \frac{1}{2}$, we have $p + q$. The expected value $E(x^2) = \sum_{i=0}^{3} x_i^2 p(x_i) = 0 \cdot p + 1 \cdot p + 4 \cdot q + 9q = p + 13q$. The expected value $E(x) = \sum_{i=0}^{3} x_i p(x_i) = 0 \cdot p + 1 \cdot p + 2q + 3q = p + 5q$. Thus, $p + 13q = 2(p + 5q)$. Solving gives $p = 3q$. So, $q = \frac{1}{8}$ and $p = \frac{3}{8}$. Therefore, $8p - 1 = 2$. Option (2).

The area $A$ is given by the integral: $$A = \int_{-2}^{1} (x + 7 - x^2 - 1) \, dx + \int_{1}^{2} (11 + 3x - x^2 - 1) \, dx$$ Evaluating the first integral: $$= \left[ \frac{x^2}{2} + 6x - \frac{x^3}{3} \right]_{-2}^{1}$$ Evaluating the second integral: $$+ \left[ 10x - \frac{3x^2}{2} - \frac{x^3}{3} \right]_{1}^{2}$$ Simplifying gives: $$= \frac{50}{3} \implies 3A = 50$$ Therefore, the correct option is Option (1).

Given \[ \lim_{x \to 0} \frac{f(x)}{x^2} = 5 \] and \[ \lim_{x \to 0} \frac{ax^4 + bx^3 + cx^2 + dx + e}{x^2} = 5. \] We have $c = 5$ and $d = e = 0$. Therefore, \[ f(x) = ax^4 + bx^3 + 5x^2. \] The derivative is \[ f'(x) = 4ax^3 + 3bx^2 + 10x = x(4ax^2 + 3bx + 10). \] This function has extrema at $x = 4$ and $x = 5$. Hence, \[ f'(4) = 0 \quad \text{and} \quad f'(5) = 0. \] Solving these equations gives \[ a = \frac{1}{8}, \qquad b = -\frac{3}{2}. \] Therefore, \[ f(2) = \frac{1}{8}\times 2^4 - \frac{3}{2}\times 2^3 + 5\times 2^2. \] Calculating, \[ f(2) = 2 - 12 + 20 = 10.

Given $\cos 2\theta \cos \frac{\theta}{2} + \cos \frac{5\theta}{2} = 2 \cos^3 \frac{5\theta}{2}$. $$\frac{1}{2} \left( 2 \cos 2\theta \cos \frac{\theta}{2} \right) + \cos \frac{5\theta}{2}$$ $$= \frac{1}{2} \left( \cos \frac{15\theta}{2} + 3 \cos \frac{5\theta}{2} \right)$$ or solving $$\cos \frac{3\theta}{2} = \cos \frac{15\theta}{2}$$ $$\cos \frac{15\theta}{2} - \cos \frac{3\theta}{2} = 0$$ $$2 \sin 30 \sin \frac{9\theta}{2} = 0$$ $$3\theta = n\pi or \frac{90}{2} = m\pi$$ $$\theta = \frac{n\pi}{3} \theta = \frac{2m\pi}{9}$$ $$\theta = \left\{ -\frac{\pi}{2}, \frac{\pi}{3}, 0 \right\}$$ $$\theta = \left\{ -\frac{4\pi}{9}, -\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{2\pi}{9} \right\}$$

Given $S_n = 700 = \frac{n}{2} [2a + (n-1)d]$...(i) $a_6 = 7 \Rightarrow a + 5d = 7$...(ii) $S_7 = 7 \Rightarrow \frac{7}{2} (2a + 6d) = 7$ $a + 3d = 1$...(iii) Solve (ii) and (iii) $$\frac{n}{2} (-16 + 3n - 3) = 700 \Rightarrow 3n^2 - 19n - 1400 = 0$$ $$(3n + 56)(n - 25) = 0$$ Therefore, $a_{25} = a + 24d = -8 + 24 \times 3$ $= -8 + 72$ $= 64$

Given $\($ $\Re$ $\left$( $\frac{z-1}{2z+i}$ $\right$) + $\Re$ $\left$( $\frac{\bar{z}-1}{2\bar{z}-i}$ $\right$) = 2 $\)$. Here, $\[$ $\frac{z-1}{2z+i}$ = $\left$( $\frac{\bar{z}-1}{2\bar{z}-i}$ $\right$) = 2 $\]$ $\[$ $\Rightarrow$ $\Re$ $\left$( $\frac{z-1}{2z+i}$ $\right$) + $\Re$ $\left$( $\frac{z-1}{2z+i}$ $\right$) = 2 $\]$ $\[$ = 2 $\Re$ $\left$( $\frac{z-1}{2z+1}$ $\right$) = 2 $\Rightarrow$ $\Re$ $\left$( $\frac{z-1}{2z+i}$ $\right$) = 1 $\]$ Let $\($ z = x + iy $\)$ $\[$ $\Re$ $\left$( $\frac{(x-1) + iy}{2x + i(2y+1)}$ $\right$) = 1 $\Rightarrow$ $\Re$ $\left$[ $\frac{((x-1) + iy)(2x-i(y+1))}{(2x + i(2y+1))(2x-i(2y+1))}$ $\right$] = 1 $\]$ $\[$ $\Rightarrow$ 2x(x-1) + y(2y+1) = 1 $\]$ $\[$ $\Rightarrow$ 4x^2 + (2y+1)^2 = 1 $\]$ $\[$ $\Rightarrow$ 2x^2 - 2x + 2y^2 + y = 4x^2 + 4y^2 + 1 + 4y $\]$ $\[$ $\Rightarrow$ 2x^2 + 2y^2 + 3y + 2x + 1 = 0 $\]$ $\[$ $\Rightarrow$ x^2 + y^2 + x + $\frac{3}{2}$ y + $\frac{1}{2}$ = 0 $\]$ $\[$ centre = $\left$( -$\frac{1}{2}$, -$\frac{3}{4}$ $\right$), $\ $r = $\sqrt{\frac{1}{4} + \frac{9}{16} - \frac{1}{2}}$ = $\frac{\sqrt{5}}{4}$ $\]$ $\[$ a = -$\frac{1}{2}$, $\ $b = -$\frac{3}{4}$, $\ $r^2 = $\frac{5}{16}$ $\]$ $\[$ 15 $\frac{ab}{r^2}$ = 15 $\times$ $\left$( -$\frac{1}{2}$ $\right$) $\times$ $\left$( -$\frac{3}{4}$ $\right$) $\times$ $\frac{16}{5}$ = 18 $\]$

Length of LR = $\frac{2b^2}{a}$ = 10 $\Rightarrow$ 5a = b^2 $\ldots$ (1) f(t) = t^2 + t + $\frac{11}{12}$ $\frac{df(t)}{dt}$ = 2t + 1 = 0 $\Rightarrow$ t = -$\frac{1}{2}$ Min value of f(t) = $\left$(-$\frac{1}{2}$$\right$)^2 + $\frac{11}{12}$ = $\frac{1}{4}$ - $\frac{1}{2}$ + $\frac{11}{12}$ = $\frac{3 - 6 + 11}{12}$ = $\frac{8}{12}$ = $\frac{2}{3}$ = e e^2 = $\frac{1 - b^2}{a^2}$ $\Rightarrow$ $\frac{4}{9}$ = $\frac{1 - b^2}{a^2}$ $\Rightarrow$ $\frac{b^2}{a^2}$ = $\frac{1 - 4}{a}$ = $\frac{5}{a}$ $\Rightarrow$ b^2 = $\frac{5a^2}{a}$ $\ldots$ (2) From (1) $\&$ (2) 5a = $\frac{5a^2}{a}$ $\Rightarrow$ a = 9, b = $\sqrt{45}$ = 3$\sqrt{5}$ $\therefore$ a^2 + b^2 = 81 + 45 = 126

Given $(x^2 + 1) \frac{dy}{dx} - 2xy = (x^4 + 2x^2 + 1) \cos x$. Rewriting, we have: $$\frac{dy}{dx} - \left( \frac{2x}{x^2 + 1} \right) y = \frac{(x^2 + 1)^2 \cos x}{cx^2 + 1} = (x^2 + 1) \cos x$$ This is a linear differential equation. Let $P = \frac{-2x}{x^2 + 1}$ and $Q = (x^2 + 1) \cos x$. The integrating factor (I.F.) is given by: $$I.F. = e^{\int P dx} = e^{\int \frac{-2x}{x^2 + 1} dx} = \frac{1}{x^2 + 1}$$ Thus, we have: $$y \cdot \frac{1}{x^2 + 1} = \int (x^2 + 1) \cos x \cdot \frac{1}{x^2 + 1} dx$$ This simplifies to: $$y = \sin x + c \implies y \cos = 1 \implies c = 1$$ Therefore, $x^2 + 1$ gives: $$y = (x^2 + 1) (\sin x + 1)$$ Now, integrating from $-3$ to $3$: $$\int_{-3}^{3} y dx = \int_{-3}^{3} (x^2 + 1) (\sin x + 1) dx$$ This becomes: $$dx = \int_{-3}^{3} x^2 \sin x + x^2 \sin x + 1 dx$$ Which simplifies to: $$\implies \int_{-3}^{3} x^2 \sin x dx + \int_{-3}^{3} x^2 dx + \int_{-3}^{3} \sin x dx + \int_{-3}^{3} 1 dx$$ Evaluating these integrals gives: $$= 0 + 18 + 0 + 6 = 24$$

Line is $\perp$ to 2 line $\Rightarrow$ line will be parallel to $\left( i + a \hat{j} + b \hat{k} \right) \times \left( -b \hat{i} + a \hat{j} + 5 \hat{k} \right)$. Parallel vector along the required line is $\hat{i} (5a - ab) - \hat{j} \left( b^2 + 5 \right) + \hat{k} (a + ab)$. Dr's of required line $\alpha (5a - ab), - \left( b^2 + 5 \right), (a + ab)$. Also Dr's of required line $\alpha - 2, d, -4$. $$\frac{5a - ab}{-2} = \frac{-\left( b^2 + 5 \right)}{d} = \frac{a + ab}{-4} \ldots (1)$$ Also point $\left( 0, -\frac{1}{2}, 0 \right)$ will lie on $\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}$. $$\frac{0 - 1}{-2} = \frac{-\frac{1}{2} + 4}{d} = \frac{0 - c}{-4} \Rightarrow d = 7, c = 2$$ From (1) $\frac{5a - ab}{-2} = \frac{-b^2 - 5}{7} = \frac{a + ab}{-4}$. $$\frac{5a - ab}{-2} = \frac{a + ab}{-4}; \frac{-b^2 - 5}{7} = \frac{a + ab}{-4}$$ $$-20a + 4ab = -2a - 2ab$$ $$4b^2 + 20 = 70 + 7ab$$ $$18a = 6ab$$ $$b = 3$$ $$36 + 20 = 70 + 21a$$ $$56 = 28a \Rightarrow a = 2$$ $a + b + c + d = 2 + 3 + 2 + 7 = 14$

Total triangles $\Rightarrow \binom{h}{3}$. Total quadrilaterals $= \binom{h}{4} = q$. $$\binom{n}{3} + \binom{n}{4} = 126 \Rightarrow \binom{n+1}{4} = 126$$ $$\Rightarrow n + 1 = 9 \Rightarrow n = 8$$ $$\frac{x^2}{16} + \frac{y^2}{n} = 1 \Rightarrow \frac{x^2}{16} + \frac{y^2}{8} = 1$$ $$e = \sqrt{1 - \frac{8}{16}} = \sqrt{\frac{8}{16}} = \frac{1}{\sqrt{2}}$$

Given the line $L_1$: $\($ $\frac{x-1}{1}$ = $\frac{y-2}{1}$ = $\frac{z-0}{2}$ $\)$. Let $Q(\lambda + 1, \lambda + 2, \lambda)$. The vector $\overrightarrow{PQ} = (\lambda - 4, \lambda - 1, \lambda + 3)$. The dot product $\overrightarrow{PQ} \cdot \mathbf{m} = 0$. Solving $\lambda - 4 + \lambda + 1 + \lambda + 3 = 0$ gives $3\lambda = 0$, so $\lambda = 0$. Therefore, $Q(1, 2, 0)$. For line $L_2$: $\($ $\frac{x-2}{1}$ = $\frac{y-0}{1}$ = $\frac{z-1}{2}$ $\)$. Let $R(\mu + 2, \mu, \mu + 1)$. The vector $\overrightarrow{PR} = (\mu - 3, \mu - 1, \mu + 4)$. The dot product $\overrightarrow{PR} \cdot \mathbf{n} = 0$. Solving $\mu - 3 + \mu - 1 + \mu + 4 = 0$ gives $\mu = 0$. Therefore, $R(2, 0, 1)$. The area of $\triangle PQR(A)$ is $\frac{1}{2} |\overrightarrow{PQ} \times \overrightarrow{PR}|$. Calculating $A = \frac{1}{2} |(-4\mathbf{i} + \mathbf{j} + 3\mathbf{k}) \times (-3\mathbf{i} + \mathbf{j} + 4\mathbf{k})|$. The cross product is $\frac{1}{2} |7(\mathbf{i} + \mathbf{j} + \mathbf{k})|$. The determinant is: $$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 1 & 3 \\ -3 & 1 & 4 \end{vmatrix} = 7\mathbf{i} + 7\mathbf{j} + 7\mathbf{k}$$ Thus, $4A^2 = 49 \times 3 = 147$.

For (I) $x < 2$: $$-x^2 + 2x - 3x + 9 + 1 = 0$$ This simplifies to: $$x^2 + x - 10 = 0$$ The solutions are: $$x = \frac{-1 + \sqrt{41}}{2}, \frac{-1 - \sqrt{41}}{2}$$ This case is not valid. For (II) $2 \leq x < 3$: $$x^2 - 2x - 3x + 9 + 1 = 0$$ This simplifies to: $$x^2 - 5x + 10 = 0$$ The discriminant $D < 0$, so there are no real roots. For (III) $x \geq 3$: $$x^2 - 2x + 3x - 9 + 2 = 0$$ This simplifies to: $$x^2 + x - 8 = 0$$ The solutions are: $$x = \frac{-1 + \sqrt{32}}{2}, \frac{-1 - \sqrt{32}}{2}$$ This case is also not valid. There is 1 real root.

Given $e_1^2 = 1 - \frac{b^2}{25}$ and $e_2^2 = 1 - \frac{b^2}{16}$. Therefore, $e_1^2 e_2^2 = 1$. $$\left(1 - \frac{b^2}{25}\right) \left(1 + \frac{b^2}{16}\right) = 1$$ This implies: $$\Rightarrow 2 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1$$ $$\Rightarrow \frac{9\, b^2}{400} = \frac{b^4}{400}$$ Thus, $b^2 = 9$. The equations are: $$\frac{x^2}{9} + \frac{y^2}{25} = 1$$ $$\frac{x^2}{16} - \frac{y^2}{9} = 0$$ Calculate $e_1$: $$e_1 \sqrt{1 - \frac{9}{25}}$$ $$e_1 = \frac{4}{5}$$ The foci are: $(0, \pm 4)$ and $(\pm 5, 0)$. Ellipse passing through all four foci: $$\frac{x^2}{25} + \frac{y^2}{16} = 1$$ Calculate $e$: $$e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$$

For infinitely many solutions, $\($ $\Delta$ = 0 $\)$. $\[$ $\begin{vmatrix}$ 1 & 5 & -1 $\\$ 4 & 3 & -3 $\\$ 24 & 1 & $\lambda$ $\\$ $\end{vmatrix}$ = 0 $\]$ $\[$ $\Rightarrow$ 1(3$\lambda$ + 3) - 5(4$\lambda$ + 72) - 1(4 - 72) = 0 $\]$ $\[$ $\Rightarrow$ -17$\lambda$ + 3 - 4 $\times$ 72 - 4 = 0 $\]$ $\[$ $\Rightarrow$ 17$\lambda$ = -289 $\]$ $\[$ $\Rightarrow$ $\lambda$ = -17 $\]$ $\($ $\Delta$_1 = 0 $\)$ $\[$ $\begin{vmatrix}$ 1 & 5 & -1 $\\$ 7 & 3 & -3 $\\$ $\mu$ & 1 & -17 $\\$ $\end{vmatrix}$ = 0 $\]$ $\[$ $\Rightarrow$ 1(-51 + 3$\mu$) - 5(-119 + 3$\mu$) - 1(7 - 3$\mu$) = 0 $\]$ $\[$ $\Rightarrow$ -48 + 595 - 15$\mu$ - 7 + 3$\mu$ = 0 $\]$ $\[$ $\Rightarrow$ 12$\mu$ = 540 $\]$ $\[$ x + 5y = z = 1 $\]$ $\[$ 4x + 3y - 3z = 7 $\]$ $\[$ 24x + y - 17z = 45 $\]$ Let $\($ z = 1 $\)$ $\[$ x + 5y = 1 + $\lambda$ $\times$ 4 $\]$ $\[$ 4x + 3y = 7 + 3$\lambda$ $\]$ $\[$ 4x + 20y = 4 + 4$\lambda$ $\]$ $\[$ -17y = 3 - $\lambda$ $\]$ $\[$ y = $\frac{\lambda - 3}{17}$, x = 1 + $\lambda$ - $\frac{5\lambda - 15}{17}$ $\]$ $\[$ = $\frac{32 - 12\lambda}{17}$ $\]$ $\[$ 7 $\leq$ $\frac{\lambda - 3}{17}$ + $\frac{32 + 12\lambda}{17}$ + $\lambda$ $\leq$ 77 $\]$ $\[$ 7 $\leq$ $\frac{30\lambda + 29}{17}$ $\leq$ 77 $\]$ $\[$ 3 $\leq$ $\lambda$ $\leq$ 42 $\]$ $\[$ $\lambda$ = 3, 20, 37 $\]$

Given $ar + ar^3 + ar^5 = 21$, $ar^7 + ar^9 + ar^{11} = 15309$. Therefore, $ar(1 + r^2 + r^4) = 21$, $ar^7(1 + r^2 + r^4) = 15309$. Dividing equation (2) by equation (1), we get $$\frac{a \cdot r^7}{ar} = \frac{15309}{21}$$ which implies $r^6 = 729$. Therefore, $r = 9$. Now, $$\frac{a \cdot (r^9 - 1)}{r - 1} = \frac{7}{91} (19683 - 1) = \frac{7 \times 19682}{91 \times 2}$$ which simplifies to $$\frac{9841}{13} = 757.$$

The limit is given by $$\lim_{x \to 0} \frac{\tan(\tan x) - \tan x}{\tan^3 x} + \frac{\tan x - \sin x}{x^3} + \frac{\sin(\sin x) - \sin x}{\sin^3 x} + \frac{\sin x}{x^3}$$ which simplifies to $$\frac{1}{3} + \frac{1}{2} + \frac{1}{6} = 2.$$

Given $$\int \left( \frac{1}{x^2} + \frac{1}{x^4} \right) \left( \frac{3}{x} + \frac{1}{x^3} \right)^{\frac{1}{23}} \, dx$$ using $$t = \frac{3}{x} + \frac{1}{x^3} \Rightarrow dt = -3 \left( \frac{1}{x^2} + \frac{1}{x^4} \right) \, dx$$ $$\int t^{1/23} \, dt = \frac{t^{24/23}}{24/23} + C$$ $$= \left( \frac{24}{23} \right)(-3)$$ $$\Rightarrow \alpha = 23, \beta = -1, \gamma = -3$$ $$\alpha + \beta + \gamma = 19$$

Given $a + b = \lim_{t \to -1^+} (\alpha + \beta) = \lim_{t \to -1^+} \left( (t+2)^{\frac{1}{6}} - 1 \right)$. Let $t + 2 = y$. Then, $$a + b = \lim_{y \to 1^+} \frac{y^{1/6} - 1}{y^{1/7} - 1} = \frac{7}{6}.$$ Now, $$72(a + b)^2 = 72 \times \frac{49}{36} = 98.$$

Equation of hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Directrix: $x = \frac{-9}{5}$ and corresponding foci $(-5, 0)$. $$\Rightarrow -\frac{a}{e} = -\frac{9}{5} and -ae = -5$$ $$\Rightarrow \frac{9e^2}{5} = 5 \Rightarrow e = \sqrt{\frac{25}{9}} = \frac{5}{3} \Rightarrow a = 3$$ $$\therefore b^2 = a^2 (e^2 - 1) = 9 \left( \frac{25}{9} - 1 \right) = 16$$ Hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$. $(\alpha, 2\sqrt{5})$ lie on it. $$\Rightarrow \frac{\alpha^2}{9} - \frac{20}{16} = 1 \Rightarrow \alpha^2 = \frac{36}{16} \times 9 = \frac{81}{4}$$ Product for distance of $(x_1, y_1)$ from the two foci $$= (ex_1 + a) |ex_1 - a|$$ $$= e^2 x_1^2 - a^2$$ For $(\alpha, 2\sqrt{5}) \Rightarrow P = \frac{25}{9} \cdot \frac{81}{4} - 9 = \frac{189}{4}$$ $$4P = 189$$

Given $ (1-x)^{20} = \binom{20}{0} - \binom{20}{1}x + \binom{20}{2}x^2 - \cdots + \binom{20}{20}x^{20} $. $$ \frac{(1-x)^{20}}{x^2} = \frac{\binom{20}{0}}{x^2} - \frac{\binom{20}{1}}{x} + \binom{20}{2} - \binom{20}{3}x + \binom{20}{4}x^2 \ldots $$ Differentiate twice and put $ x = 1 $: $$ = 6 - \binom{20}{1}(2) + A $$ $$ A = 40 - 6 = 34 $$

The electric field of an outside charge is zero inside a conductor.

Given $\rho = \frac{M}{V} = \frac{m_n \times A}{\frac{4}{3} \pi R^3} = \frac{m_n \times A}{\frac{4}{3} \pi A R_0^3}$. So $\rho$ is almost constant. $R = R_0 A^{1/3}$ $R \propto A^{1/3}$

Given $$I = \frac{1}{2} \varepsilon_0 E_0^2 \times C$$ We have $$E_0 = \sqrt{\frac{2I}{\varepsilon_0 C}}$$ $E_0$: electric field N/C

Given $L = \mu_0 N A / \ell$ and $C = \frac{A \epsilon_0}{d}$. Therefore, $\frac{L}{C} \propto \frac{\mu_0}{\epsilon_0}$. Thus, $\sqrt{\frac{\mu_0}{\epsilon_0}} \propto \sqrt{\frac{L}{C}}$. We have $\frac{L}{C} = \frac{\tau R}{(\tau/R)} = R^2$. Therefore, $\sqrt{\frac{\mu_0}{\epsilon_0}} = R$.

Given $K.E = E - W$. $\lambda_e = \frac{h}{\sqrt{2mK \cdot E}}$, $E = \frac{hc}{\lambda_i}$, $W = \frac{hc}{\lambda_0}$. $$\frac{h^2}{2m\lambda_e^2} = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0}$$ $$\lambda_e = \sqrt{\frac{h}{2mc \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right)}}$$

Given $\($ $\frac{dA}{dt}$ = $\frac{L}{2m}$ $\)$. Due to central force, torque is zero and angular momentum is constant.

Given $K.E = \frac{f}{2} KT$. For He and Ar, $f = 3$. Therefore, $K.E_{He} = 1$ and $K.E_{Ar} = 1$. Hence, the answer is the same for both.

Given the equation for height $h$: $$h = \frac{2T \cos \theta}{\rho g r} = \frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}}$$ This simplifies to: $$h = \frac{100}{7} \, cm$$ Using the sine relation: $$\sin 60^\circ = \frac{h}{\ell}$$ Solving for $\ell$: $$\ell = \frac{h \times 2}{\sqrt{3}}$$ Substitute $h$: $$\ell = \frac{100}{7} \times \frac{2}{\sqrt{3}}$$ Simplifying further: $$\ell = \frac{200}{7 \times \sqrt{3}}$$ Finally: $$\ell = 16.49 \, cm$$

Given $m = -\frac{v}{u} = -\left(\frac{v}{-u}\right) = \frac{v}{u}$. $$\frac{1}{4} = \frac{v}{u} \implies u = 4v$$ $v + u = 40$ $5v = 40$ $v = 8 \, \mathrm{cm}$ $u = 32 \, \mathrm{cm}$ $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ $$\frac{1}{8} - \frac{1}{32} = \frac{1}{f}$$ $$\frac{4 - 1}{32} = \frac{1}{f}$$ $$\frac{3}{32} = \frac{1}{f}$$ $f = \frac{32}{3} = 10.7 \, \mathrm{cm}$

Given $E = \frac{V}{d} = \frac{5}{5 \times 10^{-4}} = 10^4 \, \mathrm{V/m}$. $\tau = PE \sin \theta$ Where $P = qa = 2 \times 10^{-6} \times 5 \times 10^{-7}$ $= 1 \times 10^{-12} \, \mathrm{C \cdot m}$ $\tau = 1 \times 10^{-12} \times 10^4 \times \frac{1}{2} = 5 \times 10^{-9} \, \mathrm{N \cdot m}$

Given the logic circuit, we have the following expressions: $$Y_1 = A \cdot B, \quad Y_2 = \overline{A} + B$$ The output $Y$ is given by: $$Y = \overline{Y_1} \cdot Y_2 = \overline{Y_1} + \overline{Y_2}$$ Substituting the expressions for $Y_1$ and $Y_2$: $$Y = \overline{A \cdot B} + A + \overline{B}$$ This simplifies to: $$Y = \overline{A} + \overline{B} + A \cdot \overline{B}$$ The truth table for $Y$ is: $$ \begin{array}{ccc} A & B & Y \\ 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array} $$

Q14. (A) Mass density = $\frac{M}{V} = M^1 L^{-3}$ ...(iv) (3) (B) Impulse = $M \times u = M^1 L^1 T^{-1}$ ....(ii) (C) Power = $F \cdot V = M^1 L^2 T^{-3}$ ....(i) (D) Moment of inertia = $M r^2 = M^1 L^2$ ...(iii)

Minimum distance between 2 points having same speed is $\frac{\lambda}{2}$. $\lambda = \frac{2\pi}{k} = \frac{1}{10} \, \mathrm{m} = 10 \, \mathrm{cm}$ Distance $= \frac{\lambda}{2} = 5 \, \mathrm{cm}$

Refractive index has no relation with mass density because both have different meaning. Hence reason is incorrect. So (A) is correct but (R) is not correct.

Both statements are correct and reason is also the correct explanation of assertion.

Potential energy is defined for conservative force only. It is not defined for non-conservative force i.e. frictional force.

Q10. (A) Isothermal $\rightarrow \Delta T = 0 \rightarrow \Delta U = 0$ (IV) (B) Adiabatic $\rightarrow \Delta Q = 0$ (II) (C) Isobaric $\rightarrow \Delta P = 0 \rightarrow \Delta U \neq 0$ (III) (D) Isochoric $\rightarrow \Delta V = 0 \rightarrow \Delta W = 0$ (I)

Given $u = 360 \times \frac{5}{18} = 100 \, \mathrm{m/s}$. $x = u \times t = 2 \times 10^3 \, \mathrm{m}$. $t = \sqrt{\frac{2H}{g}} \implies H = \frac{t^2 g}{2}$. $H = \frac{400 \times 10}{2}$. $H = 2000 \, \mathrm{m}$. $D = \sqrt{x^2 + H^2}$. $D = 2\sqrt{2} \, \mathrm{km}$.

Given $F = M \times a$. $v = \frac{4}{\sqrt{x}}$. $v^2 = 16x$. $2v \frac{dv}{dx} = 16$. $v dv = \frac{16}{2} = 8$. $F = 0.5 \times 8 = 4 \mathrm{N}$.

Given $r + \theta_C = 90^\circ$. $\mu_1 \sin \theta = \mu_2 \sin r$. Therefore, $$\sin \theta = \frac{\mu_2}{\mu_1} \sin(90^\circ - \theta_C)$$ $$\sin \theta = \frac{\mu_2}{\mu_1} \cos \theta_C$$ $$\sin \theta_C = \frac{\mu_1}{\mu_2}$$ Now, $$\sin \theta = \frac{\mu_2}{\mu_1} \sqrt{1 - \frac{\mu_1^2}{\mu_2^2}}$$ $$\sin \theta = \sqrt{\frac{\mu_2^2 - \mu_1^2}{\mu_1^2}} = \sqrt{\frac{25}{16} - 1}$$ $$\sin \theta = \frac{3}{4}$$ Therefore, $$\theta = \sin^{-1}\left(\frac{3}{4}\right)$$

Given $Q_{in} = Q_0 \left(1 - \frac{1}{K}\right)$. $4 \times 10^{-6} = 5 \times 10^{-6} \left(1 - \frac{1}{K}\right)$ Solving for $K$: $$1 - \frac{1}{K} = \frac{4}{5}$$ $$K = 5$$

Given $P = V_{rms} I_{rms} \cos \phi$. (1) $P = V_{rms} \times \frac{V_{rms}}{z} \times \frac{R}{z}$ $P = \frac{V_{rms}^2 \times R}{z^2}$ $z = \sqrt{R^2 + (x_L - x_C)^2}$ $Z = 50 \sqrt{2} \, \Omega$ $P = \frac{100 \times 50}{2500 \times 2} = 1 \, W$

Given $R_1 = \frac{\ell_1}{K_1 A_1}$, $R_2 = \frac{\ell_2}{K_2 A_2}$. The rate of heat transfer is given by $\frac{dQ}{dt} = \frac{\Delta T}{R}$. Therefore, $$\left( \frac{dQ}{dt} \right)_1 = \left( \frac{dQ}{dt} \right)_2$$ $$\frac{400 - T}{R_1} = \frac{T - 200}{R_2}$$ $$\frac{400 - T}{T - 200} = \frac{R_1}{R_2} = \left( \frac{\ell_1}{\ell_2} \right) \left( \frac{r_2}{r_1} \right)^2 \times \frac{K_2}{K_1}$$ $$= \frac{1}{2} \times \left( \frac{1}{2} \right)^2 \times 2$$ $$= \left( \frac{1}{4} \right)$$ Solving, $$\frac{400 - T}{T - 200} = \frac{1}{4}$$ $$1600 - 4T = T - 200$$ $$5T = 1800$$ $$T = 360 \, \mathrm{K}$$

Given $\phi = \vec{E} \cdot \vec{A} = (2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3 \cdot A \hat{j}$. $6 = 4 \times 10^3 \, \mathrm{A}$ $A = 1.5 \times 10^{-3} \, \mathrm{m^2}$ $= 15 \, \mathrm{cm^2}$

Without cavity $I_1 = \frac{MR^2}{2}$. Mass of removed disc $= \frac{M}{\pi R^2} \times \left(\frac{R}{3}\right)^2 \pi = \left(\frac{M}{9}\right)$. M.I. of removed disc $I_2 = \frac{M}{9} \left(\frac{R}{3}\right)^2 + \frac{M}{9} \times \left(\frac{2R}{3}\right)^2 = \frac{MR^2}{18}$. $I = I_1 - I_2 = \frac{MR^2}{2} - \frac{MR^2}{18} = \frac{4MR^2}{9}$. $(n = 9)$

(i) Protein does not give $\beta$-amino acid on hydrolysis. (ii) Fibrous proteins are not water soluble. So both statements are wrong.

Chlorobenzene reacts with $\mathrm{NaOH}$ and shows no reaction. Aniline reacts with $\mathrm{NaOH}$ and shows no reaction. Benzoic acid reacts with $\mathrm{NaOH}$ to form sodium benzoate and water. Organic layer in funnel are mixture of chlorobenzene and aniline.

The dissolution of $\mathrm{KCl_{(s)}}$ in water can be represented as: $$\mathrm{KCl_{(s)} + H_2O \xrightarrow{\Delta H \ sol.} K^+_{(aq)} + Cl^-_{(aq)}}$$ The lattice energy (L.E.) and hydration energy (H.E.) are involved in the process. The lattice energy is the energy required to separate $\mathrm{KCl_{(s)}}$ into $\mathrm{K^+_{(g)}}$ and $\mathrm{Cl^-_{(g)}}$. The hydration energy is the energy released when $\mathrm{K^+_{(g)}}$ and $\mathrm{Cl^-_{(g)}}$ are solvated to form $\mathrm{K^+_{(aq)}}$ and $\mathrm{Cl^-_{(aq)}}$. The enthalpy change of solution $\Delta H_{Sol.}$ is given by: $$\Delta H_{Sol.} = L.E. + (H.E.)_{K^+_{(g)}} + (H.E.)_{Cl^-_{(g)}}$$ $$= Z - x - y$$ $$= z - (x + y)$$

For the reaction $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$, the initial conditions and changes in pressure are given as follows: At $t = 0$, the pressure is $P^o$ for $A_{(g)}$, and $0$ for $B_{(g)}$ and $C_{(g)}$. At $t = t$, the pressure is $P^o - x$ for $A_{(g)}$, and $x$ for $B_{(g)}$ and $C_{(g)}$. At $t = \infty$, the pressure is $0$ for $A_{(g)}$, and $P^o$ for $B_{(g)}$ and $C_{(g)}$. The total pressure at time $t$ is given by: $$P_t = P^o + x \Rightarrow x = P_t - P^o = \frac{P_\infty}{2}$$ At equilibrium, $P_\infty = 2P^o \Rightarrow P^o = \frac{P_\infty}{2}$. The rate constant $k$ is given by: $$k = \frac{1}{t} \ln \frac{P^o}{P^o - x}$$ Substituting the expression for $x$: $$k = \frac{1}{t} \ln \frac{P_\infty}{2(P_\infty - P_t)}$$

The compound $P$ is given as $\mathrm{CH_3-C-CH(CH_3)(CH_2CH_3)}$ with a chiral center. It does not give Tollen's test, indicating it is not an aldehyde. The compound reacts with $2,4$-DNP ($\mathrm{NH_2-NH}$) to form a hydrazone derivative, indicating the presence of a carbonyl group. The reaction with $2,4$-DNP confirms the presence of a ketone group.

In a period from left to right atomic size decreases.

Aromatic halide give nucleophilic substitution reaction at high temperature or in presence of $-I/-M$ group rate of reaction high even at low temperature.

For standard state, pressure is 1 bar and temperature is specified only. Therefore, $$ (\Delta H_f^\circ)_{\mathrm{O_2,(g)}} = 0 $$

Given $P_A^o < P_B^o$. Therefore, $\($ $\frac{P_A^o}{P_B^o}$ < 1 $\)$. The equation for $y_A$ is given by: $$ y_A = \frac{P_A^o}{P_B^o} x_A $$ The other variables are $y_B$, $x_A$, and $x_B$.

$CrO_3$ = Acidic $Mn_2O_7$ = Acidic $\therefore\ x=2$ \[ \overset{\mathrm{III}}{[Co(H_2NCH_2CH_2NH_2)_3]_2(SO_4)_3} \;\rightleftharpoons\; 2\overset{\mathrm{III}}{[Co(H_2NCH_2CH_2NH_2)_3]^{3+}} +3SO_4^{2-} \] $\therefore$ Primary valency $=3$ $\therefore\ x+y=5$

The order of basicity is given as $E > D > B > A > C$. The localized electron pair in $[\mathrm{CH_3}]_2\mathrm{NH}$ is more than in $\mathrm{CH_3}-\mathrm{NH}_2$, which makes it more basic. The presence of $+M$ effect in $\mathrm{OMe}$ increases basicity, while $-I$ and $-M$ effects in $\mathrm{NO_2}$ decrease basicity.

Primary valency = Oxidation state. Secondary valency = Co-ordination number. $$ \begin{array}{|c|l|c|c|c|} \hline & \mathrm{Complex} & & \mathrm{Primary} & \mathrm{Secondary} \\ & & & \mathrm{valency} & \mathrm{valency} \\ \hline (A) & [Co(en)_2Cl_2]Cl & (I) & 3 & 6 \\ \hline (B) & [Pt(NH_3)_2Cl(NO_2)] & (II) & 3 & 4 \\ \hline (C) & Hg[Co(SCN)_4] & (III) & 2 & 6 \\ \hline (D) & [Mg(EDTA)]^{2-} & (IV) & 2 & 4 \\ \hline \end{array} $$

$(A)$ Solution of chloroform and acetone shows -ve deviation, so maximum boiling azeotrope. $(B)$ Solution of ethanol & water shows +ve deviation. So minimum boiling azeotrope. $(C)$ Solution of benzene and toluene form ideal solution. $\Delta V_{mix} = 0$. $(D)$ Acetic acid in benzene form dimer.

For $[\mathrm{Fe(CN)}_6]^{3-}$, $\mathrm{Fe}^{3+}$ has $3d^5$ configuration with $t_{2g}^{2,2,1} e_g^{0,0}$ resulting in unpaired $e^- = 1$. For $[\mathrm{FeF}_6]^{3-}$, $\mathrm{Fe}^{3+}$ has $3d^5$ configuration with $t_{2g}^{1,1,1} e_g^{1,1}$ resulting in unpaired $e^- = 5$. For $[\mathrm{CoF}_6]^{3-}$, $\mathrm{Co}^{3+}$ has $3d^6$ configuration with $t_{2g}^{2,1,1} e_g^{1}$ resulting in unpaired $e^- = 4$. For $[\mathrm{Mn(CN)}_6]^{3-}$, $\mathrm{Mn}^{3+}$ has $3d^4$ configuration with $t_{2g}^{2,1,1} e_g^{0,0}$ resulting in unpaired $e^- = 2$.

The given compound undergoes ozonolysis with $\mathrm{O_3}$, $\mathrm{Zn}$, and $\mathrm{H_2O}$. The reaction results in the formation of $2 \ \mathrm{CH_3CHO} + 2 \ \mathrm{HCHO}$. Both products are optically inactive.

Extra stability of half filled is due to: (i) Symmetrical distribution of electrons (ii) Large exchange energy (iii) Smaller coulombic repulsion (iv) Smaller shielding of electrons by one another

(i) True, $\mathrm{Tl}^+$ is more stable than $\mathrm{Tl}^{3+}$, due to inert pair effect. So $\mathrm{Tl}^{3+}$ is a powerful oxidising agent. (ii) True, $E^0_{\mathrm{Al}^{3+}/\mathrm{Al}} = -1.66 \, \mathrm{V}$. So it is difficult to reduce $\mathrm{Al}^{3+}$. So $\mathrm{Al}^{3+}$ is highly stable. (iii) False, as $\mathrm{Tl}^{3+}$ is unstable. (iv) True, $\mathrm{Tl}^+$ is more stable than $\mathrm{Tl}^{3+}$. (v) True, $\mathrm{Al}^{3+}$ and $\mathrm{Tl}^+$ are highly stable.

Statement-II: At cathode, instead of $\mathrm{Mg}$, $\mathrm{H_2O_{(\ell)}}$ will reduce and evolve $\mathrm{H_2}$ gas.

Given $\($ pOH = pK_b + $\log$ $\left$[ $\frac{\mathrm{NH_4^+}}{\mathrm{NH_3}}$ $\right$] $\)$. $\($ pOH = 4.745 $\)$. On adding 0.05 mole HCl: $\($ $\mathrm{NH_3}$ + $\mathrm{H^+}$ $\rightarrow$ $\mathrm{NH_4^+}$ $\)$. $\($ 0.1 0.05 0.1 $\)$. $\($ 0.05 0 0.15 $\)$. $\($ pOH' = 4.745 + $\log$ 3 $\)$. $\($ pOH' - pOH = 0.477 $\)$. $\($ 14 - pH' - 14 + pH = 0.477 $\)$. $\($ $\Delta$ pH = 0.477 $\)$. $\($ = 4.7 $\times$ 10^{-2} $\approx$ 48 $\times$ 10^{-2} $\)$.

$$n_{\mathrm{N_2}} = \frac{P_{\mathrm{N_2}} \cdot V}{RT}$$ Substituting the values: $$n_{\mathrm{N_2}} = \frac{700}{760} \times \frac{50}{1000} \times \frac{1}{0.0821 \times 300}$$ The number of moles of $\mathrm{N}$ is: $$n_{\mathrm{N}} = 2 \times n_{\mathrm{N_2}}$$ The mass of $\mathrm{N}$ is: $$Mass of \mathrm{N} = 2 \times n_{\mathrm{N}} \times 14$$ The percentage of $\mathrm{N}$ is: $$\% \mathrm{N} = \frac{mass of \mathrm{N}}{mass of organic compound} \times 100$$ Substituting the values: $$\% \mathrm{N} = \frac{700}{760} \times \frac{50}{1000} \times \frac{2 \times 14}{0.0821 \times 300} \times \frac{1000}{292} \times 100$$ The percentage of $\mathrm{N}$ is $18\%$.

Given the reaction: $$\mathrm{C_4H_{10} + \frac{13}{2} O_2 \rightarrow 4CO_2 + 5H_2O}$$ Initial amounts: $$3 \times 10^3 10 \times 10^3$$ Moles of $\mathrm{H_2O}$ formed $= n_{H_2O} = 5 \times \frac{2}{13} \times 10 \times 10^3$ Then $w_{H_2O} = \frac{10^5}{13} \times 18$ $$= 1.3846 \times 10^5 \ \mathrm{g}$$ Volume of $\mathrm{H_2O}$ will be $= 138.46$ litre. Ans. 138

The reaction sequence is as follows: Ph-$\mathrm{C}$-$\mathrm{CH_3}$ $\xrightarrow{\mathrm{PCl_5}, \Delta}$ $\mathrm{Ph}$-$\mathrm{C}$-$\mathrm{CH_3}$ + $\mathrm{POCl_3}$ This forms compound [A]. [A] $\xrightarrow{3 eq. \mathrm{NaNH_2} | \mathrm{NH_3}}$ $\mathrm{Ph}$-$\mathrm{C}$ $\equiv$ $\mathrm{CNa}$ This forms compound [B]. [B] $\xrightarrow{Acidify}$ $\mathrm{Ph}$-$\mathrm{C}$ $\equiv$ $\mathrm{CH}$ This forms compound [C]. [C] $\xrightarrow{1. \mathrm{B_2H_6} 2. \mathrm{H_2O_2} | \mathrm{OH^-}}$ $\mathrm{Ph}$-$\mathrm{CH}$ = $\mathrm{CH}$ Tautomerisation leads to: $\mathrm{Ph}$-$\mathrm{CH_2}$-$\mathrm{C}$-$\mathrm{H}$ This forms compound [D]. The number of sp^2 C-atoms in product D = 7 NTA Ans. = 7 ALLEN Ans. = 7