JEE Main 22 January 2025 Shift 2 question paper with solutions

Given $$\left( x + \sqrt{x^3 - 1} \right)^5 + \left( x - \sqrt{x^3 - 1} \right)^5$$ Subtract $$\left[ 5C_0 x^5 + 5C_1 x^4 \left( \sqrt{x^3 - 1} \right) + \ldots + 5C_5 \left( \sqrt{x^3 - 1} \right)^5 \right]$$ Add $$\left[ 5C_0 x^5 - 5C_1 x^4 \left( \sqrt{x^3 - 1} \right) + \ldots - 5C_5 \left( \sqrt{x^3 - 1} \right)^5 \right]$$ This simplifies to $$= 2 \left[ x^5 + 5C_2 x^3 \left( x^3 - 1 \right) + 5C_4 \left( x^3 - 1 \right)^2 \right]$$ $$= 10x^7 + 20x^6 + 2x^5 - 20x^4 + 10x$$ Now $$\alpha = 10, \beta = 2, \gamma = -20, \delta = 10$$ Also, $$\begin{align*} 10u + 2v &= 18 \\ -20u + 10v &= 20 \end{align*}$$ Solving gives $$u = 1, v = 4$$ Thus, $$u + v = 5$$

Total - when $B_1$ and $B_2$ are together $$= 2!(3!4!) - 2!(3!(3!2!)) = 144$$

$(4,4\sqrt3)$ lies on $y^2=4ax$ $\Rightarrow\ 48=4a\cdot4$ $\Rightarrow\ 4a=12$ $\Rightarrow\ y^2=12x$ is equation of parabola Now, parameter of $P$ is $t_1=\dfrac{2}{\sqrt3}$ $\Rightarrow$ Parameter of $Q$ is $t_2=-\dfrac{\sqrt3}{2}$ $\Rightarrow\ Q\left(\dfrac94,-3\sqrt3\right)$ Area of trapezium $PQNM$ $=\dfrac12\,MN\cdot(PM+QN)$ $=\dfrac12\,MN\cdot(PS+QS)$ $=\dfrac12\,MN\cdot PQ$ $=\dfrac12\cdot7\sqrt3\cdot\dfrac{49}{4}$ $=(343)\dfrac{\sqrt3}{8}=3$

Given $\mathrm{tr}(A) = 3$ and $|A| = \frac{1}{2}$. Now, $B = \mathrm{adj}(\mathrm{adj}(2A)) = |2A|^{3-2} \cdot (2A)$. $$= 2^3 |A| \cdot 2A = 8A$$ Therefore, $\mathrm{tr}(B) = 8 \, \mathrm{tr}(A) = 24$ and $|B| = |8A| = 8^3 \cdot \frac{1}{2} = 256$. Thus, $\mathrm{trace}(B) + |B| = 24 + 256 = 280$.

Let the A.P. be $a, a + 2, a + 2d, \ldots, a + (2k - 1)d$ Now, $a + a + 2d + a + 4d + \ldots + a + (2k - 2)d = 40$ $$ka + 2d + 4d + \ldots + (2k - 2)d = 40$$ $$\Rightarrow ka + \frac{k - 1}{2} [2d + 2kd - 2d] = 40$$ $$\Rightarrow ka + k(k - 1)d = 40 \ldots (1)$$ And $a + d + a + 3d + \ldots + a + (2k - 1)d = 55$ $$\Rightarrow ka + \frac{k}{2} (d + 2kd - d) = 55$$ $$\Rightarrow ka + k^2 d = 55 \ldots (2)$$ Also, $a + (2k - 1)d - a = 27$ $$\Rightarrow (2k - 1)d = 27 \Rightarrow d = \frac{27}{2k - 1} \ldots (3)$$ From equation (1) and (2) $$k^2 d - kd - k^2 d = -15$$ $$\Rightarrow d = \frac{15}{k}$$ From equation (3) and (4) $$\frac{27}{2k - 1} = \frac{15}{k}$$ $$27k = 30k - 15$$ $$\Rightarrow 3k = 15$$ $$\Rightarrow k = 5$$

Equation of line $PQ$ is: $$\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2} = r\ (\text{say})$$ Let coordinate of $Q = (3r-2,\ 2r-1,\ 2r+3)$. Since $PR = 5$, then $$(3r-2-1)^2 + (2r-1-3)^2 + (2r+3-3)^2 = 25$$ Therefore, $r = 0$ or $2$. Thus, coordinate of $Q = (4, 3, 7)$. Therefore, square of area of $\triangle PQR = \left|\dfrac{1}{2}\overrightarrow{PQ} \times \overrightarrow{PR}\right|^2$ $$= \left|\frac{1}{2}\left(6\hat{i} + 4\hat{j} + 4\hat{k}\right) \times \left(3\hat{i} + 4\hat{j}\right)\right|^2$$ $$= \left|-8\hat{i} + 6\hat{j} + 6\hat{k}\right|^2 = 136$$

Given $\($ $\alpha$ = $\lim$_{x $\to$ $\infty$} $\left$( $\left$( $\frac{e}{1-e}$ $\right$) $\left$( $\frac{1}{e}$ - $\frac{x}{1+x}$ $\right$) $\right$)^x $\)$ ($\($1^$\infty$$\)$ form). Therefore, $\($ $\alpha$ = e^L $\)$. Where $\($ L = $\lim$_{x $\to$ $\infty$} x $\left$( $\left$( $\frac{e}{1-e}$ $\right$) $\left$( $\frac{1}{e}$ - $\frac{x}{1+x}$ $\right$) - 1 $\right$) $\)$. $\($ $\Rightarrow$ L = $\lim$_{x $\to$ $\infty$} $\left$( $\frac{e}{1-e}$ $\right$) x $\left$( $\frac{1}{e}$ - $\frac{x}{1+x}$ - $\left$( $\frac{1-e}{e}$ $\right$) $\right$) $\)$. $\($ $\Rightarrow$ L = $\frac{e}{1-e}$ $\lim$_{x $\to$ $\infty$} x $\left$( $\frac{1-x}{x(1+x)}$ $\right$) $\)$. $\($ $\Rightarrow$ L = $\frac{e}{1-e}$ $\lim$_{x $\to$ $\infty$} $\frac{x}{x+1}$ $\)$. $\($ $\Rightarrow$ L = $\frac{e}{1-e}$ $\cdot$ 1 $\)$. $\($ $\Rightarrow$ L = $\frac{1-e}{e}$ $\)$. Therefore, $\($ $\alpha$ = e^{$\frac{1-e}{e}$} $\Rightarrow$ $\log$ $\alpha$ = $\frac{e}{1-e}$ $\)$. Therefore, the required value $\($ = $\frac{e}{1-e}$ $\)$ $\($ = e $\)$.

Given $$f(x) = \int_0^{x^2} t^2 - 8t + 15 \, dt, x \in \mathbb{R}$$ $$f'(x) = x^4 - 8x^2 + 15$$ $$e^{x^2} (2x) = 0$$ $$\Rightarrow 2 \times (x^2 - 5)(x^2 - 3) = 0$$ $$\Rightarrow e^{x^2}$$ $$\Rightarrow x(x + \sqrt{5})(x - \sqrt{5})(x + \sqrt{3})(x - \sqrt{3}) = 0$$ By using wavy curve method Number of local maximum = 2 Number of local minimum = 3

Given the point $P(2, -10, 1)$ and the direction vector $\mathbf{n} = 2\hat{i} - \hat{j} + 2\hat{k}$, we find the coordinates of point $A$ on the line. The parametric equations are given by: $$\frac{x - 1}{2} = \frac{y + 2}{-1} = \frac{z + 3}{2} = \lambda (let)$$ This gives the point $A$ as $(2\lambda + 1, -\lambda - 2, 2\lambda - 3)$. The condition for $\overrightarrow{PA} \cdot \overrightarrow{n} = 0$ is: $$\Rightarrow (2\lambda - 1)2 + (-\lambda + 8)(-1) + (2\lambda - 4)2 = 0$$ $$\Rightarrow 4\lambda - 2 + \lambda - 8 + 4\lambda - 8 = 0$$ $$\Rightarrow 9\lambda - 18 = 0 \Rightarrow \lambda = 2$$ Therefore, $A(5, -4, 1)$. The distance $AP$ is: $$AP = \sqrt{3^2 + 6^2 + 0^2} = \sqrt{45} = 3\sqrt{5}$$

$\dfrac{dx}{dy}+\dfrac{x}{1+y^2}=\dfrac{2e^{\tan^{-1}y}}{1+y^2}$ I.F. $=e^{\tan^{-1}y}$ $\therefore\ xe^{\tan^{-1}y} =\int\dfrac{2\left(e^{\tan^{-1}y}\right)^2}{1+y^2}\,dy$ Put $\tan^{-1}y=t,\qquad \dfrac{dy}{1+y^2}=dt$ $\therefore\ xe^{\tan^{-1}y} =\int2e^{2t}\,dt$ $xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+c$ $\therefore\ x=e^{\tan^{-1}y}+ce^{-\tan^{-1}y}$ $\because\ y=0,\;x=1$ $\therefore\ 1=1+c \;\Rightarrow\; c=0$ $\therefore\ x=e^{\tan^{-1}y}$ $\therefore\ y=\dfrac{1}{\sqrt3},\quad x=e^{\pi/6}$

Given $$\frac{d}{dx} \left( \frac{x \cdot \sin^{-1} x}{\sqrt{1-x^2}} \right) - \sin^{-1} x \cdot \left( \frac{1 \cdot \sqrt{1-x^2} + \frac{x \cdot 2x}{2\sqrt{1-x^2}}}{1-x^2} \right)$$ This simplifies to $$= \frac{x}{\sqrt{1-x^2}} \cdot \frac{1}{\sqrt{1-x^2}}$$ $$= \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}$$ Hence, $$I = \int e^x \left( f(x) + f'(x) \right) \, dx$$ $$= e^x \cdot f(x) + C$$ $$I = e^x \cdot \frac{x \cdot \sin^{-1} x}{\sqrt{1-x^2}} + C = g(x) + C$$ Thus, $$g(x) = \frac{x e^x \sin^{-1} x}{\sqrt{1-x^2}}$$ and $$g(1/2) = \frac{\pi}{6} \frac{1}{e} \sqrt{3}$$

To solve the problem, we start by using the properties of the roots of the quadratic equation. The quadratic equation given is $2x^2 + (\cos \theta)x - 1 = 0$. Let $\alpha_\theta$and$\beta_\theta$ be the roots. By Vieta's formulas, we know: $$ \alpha_\theta + \beta_\theta = -\frac{\cos \theta}{2} $$ $$ \alpha_\theta \beta_\theta = -\frac{1}{2} $$ We need to find the minimum and maximum values of $\alpha_\theta^4 + \beta_\theta^4$. We can express $\alpha_\theta^4 + \beta_\theta^4$in terms of$\alpha_\theta + \beta_\theta$and$\alpha_\theta \beta_\theta$. Using the identity for the sum of fourth powers, we have: $$ \alpha_\theta^4 + \beta_\theta^4 = (\alpha_\theta^2 + \beta_\theta^2)^2 - 2(\alpha_\theta \beta_\theta)^2 $$ First, we need to find $\alpha_\theta^2 + \beta_\theta^2$. We can use the identity: $$ \alpha_\theta^2 + \beta_\theta^2 = (\alpha_\theta + \beta_\theta)^2 - 2\alpha_\theta \beta_\theta $$ Substituting the values from Vieta's formulas: $$ \alpha_\theta^2 + \beta_\theta^2 = \left(-\frac{\cos \theta}{2}\right)^2 - 2\left(-\frac{1}{2}\right) = \frac{\cos^2 \theta}{4} + 1 $$ Now, we substitute this back into the expression for $\alpha_\theta^4 + \beta_\theta^4$: $$ \alpha_\theta^4 + \beta_\theta^4 = \left(\frac{\cos^2 \theta}{4} + 1\right)^2 - 2\left(-\frac{1}{2}\right)^2 = \left(\frac{\cos^2 \theta}{4} + 1\right)^2 - 2 \cdot \frac{1}{4} = \left(\frac{\cos^2 \theta}{4} + 1\right)^2 - \frac{1}{2} $$ Next, we expand $\left(\frac{\cos^2 \theta}{4} + 1\right)^2$: $$ \left(\frac{\cos^2 \theta}{4} + 1\right)^2 = \left(\frac{\cos^2 \theta}{4}\right)^2 + 2 \cdot \frac{\cos^2 \theta}{4} \cdot 1 + 1^2 = \frac{\cos^4 \theta}{16} + \frac{\cos^2 \theta}{2} + 1 $$ So, we have: $$ \alpha_\theta^4 + \beta_\theta^4 = \frac{\cos^4 \theta}{16} + \frac{\cos^2 \theta}{2} + 1 - \frac{1}{2} = \frac{\cos^4 \theta}{16} + \frac{\cos^2 \theta}{2} + \frac{1}{2} $$ To find the minimum and maximum values of this expression, we let $u = \cos^2 \theta$. Since $\cos^2 \theta$ranges from 0 to 1,$u$ also ranges from 0 to 1. The expression becomes: $$ f(u) = \frac{u^2}{16} + \frac{u}{2} + \frac{1}{2} $$ We need to find the minimum and maximum values of $f(u)$for$u \in [0, 1]$. First, we find the critical points by taking the derivative of $f(u)$ and setting it to zero: $$ f'(u) = \frac{2u}{16} + \frac{1}{2} = \frac{u}{8} + \frac{1}{2} $$ Setting $f'(u) = 0$: $$ \frac{u}{8} + \frac{1}{2} = 0 \implies \frac{u}{8} = -\frac{1}{2} \implies u = -4 $$ Since $u = -4$is not in the interval$[0, 1]$, we only need to evaluate $f(u)$at the endpoints$u = 0$and$u = 1$: $$ f(0) = \frac{0^2}{16} + \frac{0}{2} + \frac{1}{2} = \frac{1}{2} $$ $$ f(1) = \frac{1^2}{16} + \frac{1}{2} + \frac{1}{2} = \frac{1}{16} + 1 = \frac{17}{16} $$ Thus, the minimum value $m$of$\alpha_\theta^4 + \beta_\theta^4$is$\frac{1}{2}$and the maximum value$M$is$\frac{17}{16}$. We need to find $16(M + m)$: $$ 16\left(\frac{17}{16} + \frac{1}{2}\right) = 16\left(\frac{17}{16} + \frac{8}{16}\right) = 16 \cdot \frac{25}{16} = 25 $$ Therefore, the correct option is $\boxed{b}$.

Given sets $A = \{1, 2, 3, 4\}$ and $B = \{1, 4, 9, 16\}$. Total number of functions is $4^4$. Total number of one-one functions is $4!$. Total number of many-one functions is $4^4 - 4! = 232$. Total number of many-one functions in which $1 \notin f(A)$ is $3 \times 3 \times 3 \times 3 = 81$. Therefore, total number of many-one functions $1 \notin f(A)$ is $$= 232 - 81$$ $$= 151$$

The given equations are $$x + y + 2z = 6$$ $$2x + 3y + az = a + 1$$ $$-x - 3y + bz = 2b,$$ where $a, b, z \in \mathbb{R}$. For infinite many solutions: $$D = D_1 = D_2 = D_3 = 0$$ Therefore, $$D = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{vmatrix} = 2a + b - 6$$ $$D_1 = \begin{vmatrix} 6 & 1 & 2 \\ a + 1 & 3 & a \\ 2b & -3 & b \end{vmatrix} = 12a + 5b + ab - 6$$ $$D_2 = \begin{vmatrix} 1 & 6 & 2 \\ 2 & a + 1 & a \\ -1 & 2b & b \end{vmatrix} = -4a - 3b - ab + 2$$ and $$D_3 = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 3 & a + 1 \\ -1 & -3 & 2b \end{vmatrix} = 2a + 2b - 16$$ From above relations $$a = -2, b = 10$$ Therefore, $7a + 3b = 16$

Given $\hat{a} \cdot \hat{b} = \frac{1}{2}$. Now $(\lambda \hat{a} + 2 \hat{b}) \cdot (3 \hat{a} - \lambda \hat{b}) = 0$. $$3 \lambda \hat{a} \cdot \hat{a} - \lambda^2 \hat{a} \cdot \hat{b} + 6 \hat{a} \cdot \hat{b} - 2 \lambda \hat{b} \cdot \hat{b} = 0$$ $$3 \lambda - \frac{\lambda^2}{2} + 3 - 2 \lambda = 0$$ $$\lambda^2 - 2 \lambda - 6 = 0$$ $$\lambda = 1 \pm \sqrt{7}$$ Therefore, number of values $= 0$.

Given $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ foci are $(ae, 0)$ and $(-ae, 0)$. $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ foci are $(Ae', 0)$ and $(-Ae', 0)$. Therefore, $2ae = 2\sqrt{3} \Rightarrow ae = \sqrt{3}$ and $2Ae' = 2\sqrt{3} \Rightarrow Ae' = \sqrt{3}$. Thus, $ae = Ae' \Rightarrow \frac{e}{e'} = \frac{A}{a}$. This implies $\frac{1}{3} = \frac{A}{a} \Rightarrow a = 3A$. Now $a - A = 2 \Rightarrow a - \frac{a}{3} = 2 \Rightarrow a = 3$ and $A = 1$. $Ae = \sqrt{3} \Rightarrow e = \frac{1}{\sqrt{3}}$ and $e' = \sqrt{3}$. $b^2 = a^2 (1 - e^2)$ $b^2 = 6$ and $B^2 = A^2 ((e')^2 - 1) = (2) \Rightarrow B^2 = 2$. Sum of LR = $\frac{2b^2}{a} + \frac{2B^2}{A} = 8$.

Given $P(A \cap B) = 0.1$, $P(A \mid B)$ and $P(B \mid A)$ are the roots of the equation $12x^2 - 7x + 1 = 0$. $$P(A \mid B)P(B \mid A) = \frac{1}{12}$$ Therefore, $$\frac{P(A \cap B)}{P(B)} \times \frac{P(A \cap B)}{P(A)} = \frac{1}{12}$$ This implies $$P(A)P(B) = 12(0.1)^2 = 0.12$$ Also, $$P(A \mid B) + P(B \mid A) = \frac{7}{12}$$ Thus, $$P(A \cap B) \left( \frac{1}{P(B)} + \frac{1}{P(A)} \right) = \frac{7}{12}$$ Therefore, $$P(A) + P(B) = \frac{7}{12} \times \frac{0.12}{0.1}$$ This gives $$P(A) + P(B) = 0.7$$ Now, $$\frac{P(\bar{A} \cup \bar{B})}{P(\bar{A} \cap \bar{B})} = \frac{P(A \cap B)}{P(A \cup B)}$$ This simplifies to $$= \frac{1 - P(A \cap B)}{1 - P(A \cup B)}$$ Which is $$= \frac{1 - 0.1}{1 - (0.7 - 0.1)} = \frac{0.9}{0.4} = \frac{9}{4}$$

Given $2 \sin^2 \theta = \cos 2\theta$. $2 \sin^2 \theta = 1 - 2 \sin^2 \theta$ $$4 \sin^2 \theta = 1$$ $$\sin^2 \theta = \frac{1}{4}$$ $$\sin \theta = \pm \frac{1}{2}$$ $2 \cos^2 \theta = 3 \sin \theta$ $$2 - 2 \sin^2 \theta + 3 \sin \theta - 2 = 0$$ $$(2 \sin \theta - 1)(2 \sin \theta - 2) = 0$$ $$\sin \theta = \frac{1}{2}$$ So common equation which satisfy both equations is $\sin \theta = \frac{1}{2}$. $$\theta = \frac{\pi}{6}, \frac{5\pi}{6} (\theta \in [0, 2\pi])$$ Sum = $\pi$

Let $z = x + iy$. $$(x + iy)(1 + i) + (x - iy)(1 - i) = 4$$ $$x + ix + iy - y + x - ix - iy - y = 4$$ $$2x - 2y = 4$$ $$x - y = 2$$ $$|z - 3| \leq 1$$ $$(x - 3)^2 + y^2 \leq 1$$ Area of shaded region $$= \frac{\pi \cdot 1^2}{4} - \frac{1}{2} \cdot 1 \cdot 1 = \frac{\pi}{4} - \frac{1}{2}$$ Area of unshaded region inside the circle $$= \frac{3}{4} \pi \cdot 1^2 + \frac{1}{2} \cdot 1 \cdot 1 = \frac{3\pi}{4} + \frac{1}{2}$$ Therefore, difference of area $$= \left( \frac{3\pi}{4} + \frac{1}{2} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \right)$$ $$= \frac{\pi}{2} + 1$$

Area = $\int$_0^2 $\left$( $\sqrt{16 - 8x}$ - (x^2 - 4x + 4) $\right$) $\,$ dx = $\left$[ $\frac{-(16 - 8x)^{3/2}}{12}$ - $\frac{x^3}{3}$ + 2x^2 + 4x $\right$]_0^2 = $\frac{8}{3}$

I.F. $e^{\frac{1}{2} \int \frac{2x}{1-x^2} \, dx} = e^{-\frac{1}{2} \ln(1-x^2)} = \sqrt{1-x^2}$. $y \times \sqrt{1-x^2} = \int (x^6 + 4x) \, dx = \frac{x^7}{7} + 2x^2 + c$. Given $y(0) = 0 \Rightarrow c = 0$. $$y = \frac{\frac{x^7}{7} + 2x^2}{\sqrt{1-x^2}}$$ Now, $6 \int_0^{\frac{\pi}{2}} \frac{x^7}{7} + 2x^2}{\sqrt{1-x^2}} \, dx = 6 \int_0^{\frac{1}{2}} \frac{2x^2}{\sqrt{1-x^2}} \, dx$. $$= 24 \int_0^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} \, dx$$ Put $x = \sin \theta$, $dx = \cos \theta \, d\theta$. $$= 24 \int_0^{\frac{\pi}{6}} \frac{\sin^2 \theta \cos \theta \, d\theta}{\cos \theta}$$ $$= 24 \int_0^{\frac{\pi}{6}} \left( \frac{1 - \cos 2\theta}{2} \right) \, d\theta = 12 \left[ \theta - \frac{\sin 2\theta}{2} \right]_0^{\frac{\pi}{6}}$$ $$= 12 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)$$ $$= 2\pi - 3\sqrt{3}$$ $$\alpha^2 = (3\sqrt{3})^2 = 27$$

Given the points, we have: $$\frac{a + 0}{3} = h \Rightarrow a = 3h$$ $$\frac{9 + 0}{3} = k \Rightarrow k = 3$$ Therefore, $$(h, k) = \left( \frac{6 + 10 \cos \alpha - 10 \sin \alpha}{3}, \frac{8 - 10 \sin \alpha + 10 \cos \alpha}{3} \right)$$ From the equations: $$6 + 10 \cos \alpha - 10 \sin \alpha = 3h$$ $$10 \cos \alpha - 10 \sin \alpha = 3h - 6$$ $$10(\cos \alpha - \sin \alpha) = 1$$ And, $$\frac{8 - 10 \sin \alpha + 10 \cos \alpha}{3} = k$$ Solving further: $$\Rightarrow 100 \sin 2\alpha = 99$$ $$h = \frac{7}{3}$$ $$\Rightarrow a = 7$$ Now, $$5a - 3h + 6k + 100 \sin 2\alpha = 35 - 7 + 18 + 99 = 145$$

Given $PR = \csc \theta$, $PQ = 4 \sec(30^\circ + \theta)$. For equilateral $d = PR = PQ$. Therefore, $$\cos(\theta + 30^\circ) = 4 \sin \theta$$ which implies $$\frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta = 4 \sin \theta$$ leading to $$\tan \theta = \frac{1}{3\sqrt{3}}.$$ Thus, $$QR^2 = d^2 = \csc^2 \theta = 28.$$

Given the expression $$\sum_{r=1}^{30} r^2 \left( \binom{30}{r} \right)^2$$ we start by rewriting it as $$\sum_{r=1}^{30} r^2 \left( \frac{31-r}{r} \right) \cdot \frac{30!}{r!(30-r)!}$$ which simplifies to $$\sum_{r=1}^{30} \frac{(31-r)!}{(r-1)!(30-r)!}$$ This can be rewritten as $$30 \sum_{r=1}^{30} \frac{(31-r)29!}{(r-1)!(30-r)!}$$ which simplifies further to $$30 \sum_{r=1}^{30} (30-r+1) \binom{30}{30-r}$$ This is equal to $$30 \left( \sum_{r=1}^{30} (31-r) \binom{30}{30-r} + \sum_{r=1}^{30} \binom{30}{30-r} \right)$$ which simplifies to $$30 (29 \times 2^{28} + 2^{29}) = 30(29+2)2^{28}$$ Simplifying further, we have $$= 15 \times 31 \times 2^{29}$$ which equals $$= 465 \times 2^{29}$$ Therefore, $$\alpha = 465$$

Given $R$ is reflexive $\Rightarrow R$ have $(1,1), (2,2), (3,3)$. $R$ is transitive. Since $(1,2), (2,3) \in R$ and $(1,3) \in R$, therefore, $R_1 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$. Clearly $R_1$ is reflexive and transitive but not symmetric. Similarly, $R_2 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (3,2)\}$ and $R_3 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3), (2,1)\}$. Therefore, 3 relations are possible.

For the first lens, $$\frac{1}{f_1} = (\mu - 1) \frac{2}{R} = P = 4D$$. For the second lens, $$\frac{1}{f_2} = (\mu - 1) \frac{1}{R} = \frac{P}{2} = 2D$$.

Given the forces acting on the object, we have the equation: $$mg - F_B - f = 0$$ Substituting for $F_B$, we get: $$mg - \frac{mg}{2} - f = 0$$ Therefore, $$f = \frac{mg}{2}$$

Given $d = \frac{m}{vol.} = \frac{m}{\pi R^2 \ell}$, we have $\frac{d\rho}{\rho} = \frac{dm}{m} + \frac{2 \, dR}{R} + \frac{d\ell}{\ell}$. Therefore, $$\frac{d\rho}{\rho} = \left( \frac{0.003}{0.6} + \frac{2 \times 0.01}{0.5} + \frac{0.05}{10} \right) 100 = 5\%$$

Initially, $I_0 = \frac{\varepsilon_m}{R}$. Finally, $I_0^1 = \frac{\varepsilon_m}{2R} = \frac{I_0}{2}$.

Given: - The electric potential of A is $V_0$ - The electric field of A is $E_0$. We need to determine the electric potential and electric field at point 8. Step E1: Electric Potential of a Dipole The electric potential $V$ due to a short dipole at a point at distance $r$ is: $$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{pr^2}{r^2}$$ On the axial line (Point A at distance $r$): On the equatorial line (Point B at distance $2r$): Since the dipole potential is given by: $$V_A = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r^2} = V_0$$ $$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p \cos \theta}{r^2}$$ and on the equatorial line $\theta = 90^\circ \Rightarrow \cos 90^\circ = 0$, $$V_g = 0$$ Thus, the electric potential at 8 is zero. Step 2: Electric Field of a Dipole The magnitude of the electric field at a distance $r$ from a dipole: On the axial line: $$B_{axial} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^3}$$ Given that of A (on the axial line of $r$), the field is $E_0$: $$E_A = E_n = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r_1^3}$$ On the equatorial line: $$E_{equatorial} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r^3}$$ At B (distance $2r$): $$E_8 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{(2r)^3} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{6r^3}$$ Since $E_0 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^3}$, we can express $E_n$ in terms of $E_0$: $$E_n = \frac{E_0}{16}$$ Final Answer: - Electric potential at E = 0 - Electric field to B = $\frac{E_0}{10}$

Energy is given by $$\frac{Q^2}{2C}$$. Therefore, the dimensional formula for $$[F]$$ is $$[C^2]$$ which is $$[ML^2T^{-2}]$$. This simplifies to $$[C^2M^{-1}L^{-2}T^{+2}]$$.

Given $$mvr = \frac{nh}{2\pi} ...(i)$$ $$r = \frac{vm}{Bq} ...(ii)$$ Let $\($ n = 2 $\)$. Then $$mr \left( \frac{rBq}{m} \right) = \frac{2h}{2\pi}$$ Solving for $\($ r $\)$, we get $$r = \sqrt{\frac{h}{\pi Bq}}$$ Given $\($ q = e $\)$, we have $$r = \sqrt{\frac{h}{\pi Be}}$$

For rigid diatomic molecules $$f = 5$$ $$\therefore \gamma_1 = \frac{7}{5} = 1.4$$ For non-rigid diatomic molecules $$f = 5 + 2 = 7$$ $$\gamma_2 = \frac{9}{7} = 1.28$$ $$\therefore \gamma_1 > \gamma_2$$

The electromotive force is given by the negative rate of change of magnetic flux. $$E = -\frac{d\phi}{dt}$$ The magnetic flux is given by $$\phi = Bbx$$ The magnitude of the electromotive force is $$|E| = Bbv$$

Einstein's photoelectric equation $$KE = \frac{hc}{\lambda} = \phi_0$$ $$2 \, eV = \frac{hc}{\lambda} - 1 \, eV$$ $$\frac{hc}{\lambda} = 3 \, eV$$ $$KE' = \frac{hc}{(\lambda/2)} - \phi_0 = 6 \, eV - 1 \, eV$$ $$= 5 \, eV$$

Given $g = \frac{GM}{R^2}$. $$g' = \frac{G(4M)}{(2R)^2} = g$$ A is correct, $R$ is correct; but since $T = 2\pi \sqrt{\frac{\ell}{g}}$ doesn't depend on mass; $R$ doesn't explain A.

Given $\tau = \mathbf{r} \times \mathbf{F}$. $$\tau = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$$ $$\tau = \hat{i}(2 - 1) - \hat{j}[0] + \hat{k}(1 - 2)$$ $$= \hat{i} - \hat{k}$$

$\therefore$ Truth table \begin{tabular}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 0 \\ \hline 1 & 1 & 1 \\ \hline \end{tabular} For NOR gate: $\overline{AB} = \overrightarrow{A + B}$.

Given $w = 0$. Therefore, $\vec{F} \cdot S = 0$. $$(2\hat{i} + b\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} - \hat{k}) = 0$$ $$2 - 2b - 1 = 0$$ $$b = \frac{1}{2}$$

Given $\beta = \frac{\lambda D}{d}$. We have $\lambda_{red} > \lambda_{blue}$. Assertion is false. Reason is true.

The initial kinetic energy $k_i$ is given by $$k_i = \frac{1}{2} mv^2.$$ The final kinetic energy $k_f$ is given by $$k_f = \frac{1}{2} m (v \cos 60^\circ)^2 = \frac{1}{8} mv^2.$$ The change in kinetic energy $\Delta k$ is $$\Delta k = k_i - k_f = \frac{3}{8} mv^2 = \frac{3}{8} \times 0.1 \times 400 = 15 \, \mathrm{J}.$$

For transmitted green light to be maxima, reflected green should be minima. $$\Delta P = 2 \mu_0 t = n \lambda$$ $$\Rightarrow t = \frac{n \lambda}{2 \mu_0} \therefore t_{\min} = \frac{\lambda}{2 \mu_0} = \frac{550}{2 \times 2} = 137.5$$

Given $A_1 V_1 = A_2 V_2$. Therefore, $2\pi (2R)^2 = V_2 \pi R^2$. Thus, $V_2 = 8 \, \mathrm{m/s}$.

Extensive variables depend on size and amount of system. Extensive: Volume, mass, internal energy. Intensive: Pressure, temperature, density.

Given the equation: $\($ $\frac{1}{2}$ m $\times$ 100 + 0 = $\frac{1}{2}$ m V_B^2 + mg $\left$( R - $\frac{R \sqrt{3}}{2}$ $\right$) $\)$. Solving for $\($ V_B^2 $\)$: $\[$ 100 = V_B^2 + 2gR $\left$( 1 - $\frac{\sqrt{3}}{2}$ $\right$) $\]$ $\[$ V_B^2 = 100 - 20(2 - $\sqrt{3}$) $\]$ $\[$ V_B^2 = 60 + 20 $\sqrt{3}$ $\]$ The kinetic energy at point B is: $\[$ K.E_B = $\frac{1}{2}$ m V_B^2 = $\frac{m}{2}$ (60 + 20 $\sqrt{3}$) $\]$ For point C: $\[$ $\frac{1}{2}$ m (100) = $\frac{1}{2}$ m V $\]$ $\[$ 100 = mg $\left$( $\frac{3R}{2}$ $\right$) $\]$ $\[$ 100 V_C^2 = 60 $\]$ $\[$ V_C^2 = 40 $\]$ The kinetic energy at point C is: $\[$ K.E_C = $\frac{1}{2}$ m V V_C^2 = $\frac{1}{2}$ m (40) $\]$ Comparing kinetic energies: $\[$ $\frac{60 + 20 \sqrt{3}}{40}$ = $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ = $\frac{3 + \sqrt{3}}{2}$ $\]$

Given $Bvq = Eq$ and $E = Bv$. We have $vm = Bq$ and $r = \frac{Bq}{mv}$. Therefore, $B = \frac{mv}{rq}$. Now, $E = \left(\frac{mv}{rq}\right)v = \frac{mv^2}{rq}$. Calculating, $$1.6 \times 10^{-27} \times 4 \times 10^{10}$$ $$= \frac{2 \times 10^{-2} \times 1.6 \times 10^{-19}}{2 \times 10^4 \, \mathrm{N/C}}$$ Finally, $x = 2$.

The magnetic field at point P due to the current in wire X is given by: $$B_X = \frac{\mu_0 (5)}{2\pi \times 0.01}$$ The magnetic field at point P due to the current in wire Y is given by: $$B_Y = \frac{\mu_0 (4)}{2\pi \times 0.04}$$ The net magnetic field at point P is: $$B = B_X - B_Y$$ Substituting the values, we get: $$B = \frac{\mu_0 (5)}{2\pi \times 0.01} - \frac{\mu_0 (4)}{2\pi \times 0.04}$$ Simplifying, we have: $$B = \frac{100 \mu_0}{4\pi}$$ $$B = -100 \times 10^{-7}$$ $$B = -1 \times 10^{-5} \, \mathrm{T}$$

Given $i_d = i_c$. Total displacement current $= 6 \, \mathrm{A}$. Through $A_0$, $$i = \left( \frac{A_0}{A} \right) 6$$ $$= \frac{3.2}{16} \times 6 = 1.2 \, \mathrm{A} = 1200 \, \mathrm{mA}$$

Step 1: Consider the Force at a Distance $x$. When the tube rotates with angular velocity $\omega$, each element of the liquid experiences a centrifugal force. The force at a distance $x$ from the pivot can be found by considering the differential force due to an element of liquid. $$dF = \rho A dx \omega^2 x$$ where $\rho$ is the volumetric mass density of the liquid. Step 2: Find the Volumetric Mass Density. Since the mass of the liquid is $2M$, the length of the tube is $1 \, \mathrm{m}$, and let the area of cross-section is $A$, the volumetric mass density is: $$\rho = \frac{2M}{AL} = \frac{2M}{A}$$ Step 3: Calculate the Total Force at the Other End. The force at $x = L$ is obtained by integrating: $$F = \int_0^L \rho x \omega^2 dx$$ $$F = 2M \omega^2 \int_0^1 x dx$$ $$F = \frac{2M \omega^2}{2} = M \omega^2$$ Step 4: Solve for $\omega$. $$\omega^2 = \frac{F}{M}$$ $$\omega = \sqrt{\frac{F}{M}}$$

The equivalent resistance $R_{eq} = 2 \, \Omega$. The current $I = \frac{2}{2} = 1 \, A$.

Given $\mu_{\mathrm{HBr}} = 0.78 \, \mathrm{D}$, $\mu_{\mathrm{H_2 S}} = 0.95 \, \mathrm{D}$, $\mu_{\mathrm{NF_3}} = 0.24 \, \mathrm{D}$, $\mu_{\mathrm{CHCl_3}} = 1.01 \, \mathrm{D}$. Hence dipole moment of $\mathrm{NF_3} < \mathrm{HBr} < \mathrm{H_2 S} < \mathrm{CHCl_3}$.

In A, B, D - OH group in right hand side then D-configuration is assigned.

N - N < P - P: single ($\sigma$) bond strength. Due to L.P.-L.P. repulsion and maximum possible covalency of nitrogen is 4.

Activation energy of forward reaction is given by $E_1 + E_2$. The energy of the product is greater than the energy of the reactant. Stability is such that reactant is greater than product.

Formation of more stable free radical intermediate.

Given $\mathrm{ClO_4^-} \rightarrow x + \{(-2) \times 4\} = -1 \Rightarrow x = +7$. Chlorine is in its maximum oxidation state, so disproportionation not possible in $\mathrm{ClO_4^-}$.

Option (A) involves the reduction of an amide to an amine using $\mathrm{LiAlH_4}$ and $\mathrm{H_2O}$. Option (B) involves the reduction of a nitro group to an amine using $\mathrm{Sn}$ and $\mathrm{HCl}$. Option (C) involves the reduction of a nitrile to an amine using $\mathrm{H_2/Ni}$. Option (D) involves the hydrolysis of a cyclic imide to a dicarboxylate salt using $\mathrm{aq.\ NaOH}$.

$RBr \xrightarrow[\text{dry Ether}]{Mg} RMgBr \xrightarrow{H_2O} R-H$ Hence, RBr can be Total 4 structural isomers

Since $dH = dq$ (at $P = constant$), $$dH = C_p dT$$ $$\left( \frac{dH}{dT} \right)_P = C_P$$ $dU = dq$ (at $V = constant$), $$dU = C_v dT$$ $$\left( \frac{dU}{dT} \right)_V = C_V$$ Since $dG = VdP - SdT$, at $P = constant$, $dP = 0$, $$\left( \frac{dG}{dT} \right)_P = -S$$ at $T = constant$, $$\left( \frac{dG}{dP} \right)_T = V$$

Crystal field splitting energy ($\Delta$) is proportional to the charge or oxidation state of the central metal atom. Crystal field splitting energy ($\Delta$) is also proportional to the field strength of the ligand (and chelation). Crystal field stabilisation energy (CFSE) is given by: $$\left[-0.4 t_{2g} + 0.6 eg \right] \Delta_\circ (for octahedral)$$ For, $$[Co(en)_3]^{3+} : Co^{3+} : t_{2g}^6 e_g^0; CFSE = -2.4(\Delta_0)_1$$ $$[Co(NH_3)_6]^{3+} : Co^{3+} : t_{2g}^6 e_g^0; CFSE = -2.4(\Delta_0)_2$$ $$[Co(NH_3)_6]^{2+} : Co^{2+} : t_{2g}^5 e_g^2; CFSE = -0.8(\Delta_0)_3$$ $$[Co(NH_3)_4]^{2+} : Co^{2+} : e_g^4 t_{2g}^2; CFSE = -1.2\Delta_t$$ As: $$\left(\Delta_t = \frac{4}{9}(\Delta_0)_3\right)$$ Therefore, $$\Delta_t < (\Delta_0)_3 < (\Delta_0)_2 < (\Delta_0)_1$$

Given $3M \mathrm{NaCl}$, $d_{sol} = 1.25 \, \mathrm{gm/mol}$. Molality $= \frac{M \times 1000}{1000 \, d - M \times M_W}$ $$= \frac{3000}{1250 - 175.5} = 2.79$$

The dissolution reaction is given by: $$\mathrm{Zr_3(PO_4)_4 (s) \rightleftharpoons 3Zr^{+4} (aq) + 4PO_4^{-3} (aq)}$$ Let the solubility be $s$. The solubility product $K_{sp}$ is given by: $$K_{sp} = (3s)^3 (4s)^4 = 6912 \, s^7$$ Solving for $s$: $$s = \left( \frac{K_{sp}}{6912} \right)^{1/7}$$

Due to the +M effect of -OMe at the para position, it will be most stable among these.

Analyzing Statement (I) The extreme left of the periodic table corresponds to the alkali metals (Group 1) and alkaline earth metals (Group 2). These metals typically form basic (or occasionally amphoteric, in the case of some Group 2 elements) oxides, not acidic oxides. For example, $\mathrm{Na_2O}$, $\mathrm{K_2O}$, $\mathrm{MgO}$, $\mathrm{CaO}$, etc., all form basic solutions (e.g., $\mathrm{Na_2O} + \mathrm{H_2O} \rightarrow 2\mathrm{NaOH}$). Hence, Statement (I) - that an element in the extreme left forms acidic oxides - is false. Analyzing Statement (II) The extreme right of the periodic table corresponds to the nonmetals in Groups 15, 16, 17 (and noble gases in Group 18). Nonmetal oxides (such as those of sulfur, phosphorus, chlorine) are generally acidic. When these oxides dissolve in water, they typically form acids. Example: $\mathrm{SO_3} + \mathrm{H_2O} \rightarrow \mathrm{H_2SO_4}$ (sulfuric acid) Example: $\mathrm{P_2O_5} + 3\mathrm{H_2O} \rightarrow 2\mathrm{H_3PO_4}$ (phosphoric acid) Example: $\mathrm{Cl_2O_7} + \mathrm{H_2O} \rightarrow 2\mathrm{HClO_4}$ (perchloric acid) Thus, Statement (II) - that acid is formed when water reacts with an oxide of a reactive element in the extreme right - is true. Conclusion Statement (I) is false. Statement (II) is true.

2p_x and 2p_y are degenerated orbitals hence having equal energy and therefore no spectral line will be observed for 2p_x $\rightarrow$ 2p_y transition.

The elements present in the compound are converted from covalent form into ionic form by fusing the compound with sodium in Lassaigne's test.

(A) $[\mathrm{Fe(CN)_5NO}]^{-2} \rightarrow$ Heteroleptic, $\mathrm{Fe^{+2}}$, $3\, d^6$, $t_{2g}^6 e_g^0$, $d^2 sp^3$, Low spin (3 d series + SFL) (B) $[\mathrm{CoF_6}]^{-3} \rightarrow$ Homoleptic, $sp^3 d^2$, High spin, $\mathrm{Co^{+3}}$, $3\, d^6$ (3 d series + WFL) (C) $[\mathrm{Fe(CN)_6}]^{-4} \rightarrow$ Homoleptic $\mathrm{Fe^{+2}}$, $3\, d^6$, $d^2 sp^3$, $t_{2g}^6 e_g^0$ Low spin (3 d series + SFL) (D) $[\mathrm{Co(NH_3)_6}]^{+3} \rightarrow$ Homoleptic, $\mathrm{Co^{+3}} 3\, d^6$, $d^2 sp^3$, $t_{2g}^6 e_g^0$, Low spin (3d series + SFL) (E) $[\mathrm{Cr(H_2O)_6}]^{+2} \rightarrow$ Homoleptic $\mathrm{Cr^{+2} 3\, d^4}$, $d^2 sp^3$, High spin $t_{2g}^3 e_g^1$ (3d series + WFL)

The reaction starts with the oxidation of toluene to benzaldehyde using $\mathrm{CrO_2Cl_2}$ in $\mathrm{CS_2}$ and $\mathrm{H_3O^+}$. The benzaldehyde is then reacted with $\mathrm{NaHSO_3}$ to form a bisulfite adduct. This adduct, labeled as (A), is treated with dilute $\mathrm{HCl}$ to regenerate benzaldehyde, labeled as (B).

Statement I: Corrosion is an example of electrochemical phenomenon in which pure metal acts as anode and impure metal (rusted metal) acts as cathode. Statement II: Corrosion is more favourable in acid medium than alkaline so rate of corrosion is high in acid medium than alkaline.

Since 4 moles of $\mathrm{H_2}$ is being added for complete hydrogenation, the degree of unsaturation $= 4$. No. of $\pi$ electrons in $\mathrm{C_6H_6} = 4 \times 2 = 8$.

For Z = 41, Nb (Niobium) has the electron configuration $[\mathrm{Kr}] 4d^4 5s^1$. The number of electrons in 4d is 4, so $x = 4$. For Z = 44, Ru (Ruthenium) has the electron configuration $[\mathrm{Kr}] 4d^7 5s^1$. The number of electrons in 4d is 7, so $y = 7$. Given $x + y = 11$.

Ni^{2+} with (dmg) forms [$\mathrm{Ni(dmg)_2}$]^{2+} having 2 H-Bonds as shown. The number of H atoms is 14.

Given the reaction $2\mathrm{C_{(graphite)}} + 3\mathrm{H_2(g)} \rightarrow \mathrm{C_2H_6(g)}$, we need to find $\Delta H_f$. For the reaction $\mathrm{C_2H_6(g)} + \frac{7}{2}\mathrm{O_2(g)} \rightarrow 2\mathrm{CO_2(g)} + 3\mathrm{H_2O(l)}$, $\Delta H_1 = -1550$. For the reaction $\mathrm{C_{(graphite)}} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}$, $\Delta H_2 = -393.5$. For the reaction $\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}$, $\Delta H_3 = -286$. The formation enthalpy $\Delta H_f = 2\Delta H_2 + 3\Delta H_3 - \Delta H_1 = 95 \, \mathrm{kJ/mole^2}$.

Given $$M_F = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$$ Substitute the values: $$= \frac{2 \times 20 + 0.5 \times 400}{420} = 0.571M$$ $$= 57.1 \times 10^{-2} M$$ $$= 57$$