JEE Main 11 April 2023 Shift 2 question paper with solutions

Given that, $$\begin{vmatrix} x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2 \end{vmatrix} = \frac{9}{8}(103x + 81)$$ Put $x = 0$ as $x \in \mathbb{R}$ $$\begin{vmatrix} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^2 \end{vmatrix} = \frac{9}{8}(103(0) + 81)$$ On expanding the determinant we get, $$\Rightarrow \lambda^3 = \frac{9}{8} \times 81$$ $$\Rightarrow \lambda^3 = \frac{9^3}{2^3}$$ $$\Rightarrow \lambda = \frac{9}{2} and \frac{\lambda}{3} = \frac{9}{6}$$ Now the required quadratic equation is $$x^2 - \left(\frac{9}{2}\right)x + \frac{9}{2} \times \frac{9}{6} = 0$$ $$\Rightarrow x^2 - (6)x + \frac{27}{4} = 0$$ $$\Rightarrow 4x^2 - 24x + 27 = 0$$ Hence, the required quadratic equation is $4x^2 - 24x + 27 = 0$

Given, the line passing through the points $P(2, -1, 2)$ and $Q(5, 3, 4)$. So, the equation of line $PQ$ will be, $$\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{2} = \lambda$$ Now let point $R$ be $(3\lambda + 2, 4\lambda - 1, 2\lambda + 2)$. Given, $R$ lies on plane $x - y + z = 4$. Therefore, $$3\lambda + 2 - 4\lambda + 1 + 2\lambda + 2 = 4$$ $$\Rightarrow \lambda = -1$$ Therefore, $R(-1, -5, 0)$. Now let line $SR$ be: $$\frac{x+1}{2} = \frac{y+5}{2} = \frac{z}{1} = k$$ as it is parallel to $$\frac{x-7}{2} = \frac{y+3}{2} = \frac{z-2}{1}$$ and passing through $(-1, -5, 0)$. So, the point $S: (2k - 1, 2k - 5, k)$. Now $S$ lies on plane: $x + 2y + 3z + 2 = 0$. $$\Rightarrow (2k - 1) + (4k - 10) + 3k + 2 = 0$$ $$\Rightarrow 9k - 9 = 0 \Rightarrow k = 1$$ Hence, $S(1, -3, 1)$. So, distance $SR = \sqrt{4 + 4 + 1} = 3$.

Given, the 1011th term from the end in the binomial expansion of $\left( \frac{4x}{5} - \frac{5}{2x} \right)^{2022}$ is 1024 times the 1011th term from the beginning. So, according to the question we have, $$\frac{T_{1013}}{T_{1011}} = 1024$$ $$\Rightarrow \frac{{^{2022}C_{1012} \left( \frac{4x}{5} \right)^{1010} \cdot \left( -\frac{5}{2x} \right)^{1012}}}{{^{2022}C_{1010} \left( \frac{4x}{5} \right)^{1012} \cdot \left( -\frac{5}{2x} \right)^{1010}}} = 1024$$ $$\Rightarrow \left( -\frac{5}{2x} \right)^2 = 1024 \left( \frac{4x}{5} \right)^2$$ $$\Rightarrow \frac{5^4}{2^6 x^4} = 2^{10}$$ $$\Rightarrow \left( \frac{5}{x} \right)^4 = 2^{16}$$ $$\Rightarrow |x| = \frac{5}{16}$$ $$\Rightarrow 32|x| = 10$$ Note this question was bonus in Jee Main 2023 April session as answer key was not matching, so we have modified the question.

Given, $$f(x) = \begin{cases} \min \{ x^2, x - \lfloor x \rfloor \}, & x \in [0, 1) \\ e^{x - \log_e x}, & x \in [1, 2] \end{cases}$$ $$\Rightarrow f(x) = \begin{cases} e^{x^2} & x \in [0, 1) \\ e, & x \in [1, 2] \end{cases}$$ As $x - \ln x \in [1, 2)$ for $x \in [1, 2]$, so $[x - \ln x] = \{x\}$, so $x \in (0, 1)$, $x^2 < \{x\}$. Now solving the integral we get, $$\int_0^2 x f(x) = \int_0^1 x \cdot e^{x^2} \, dx + \int_1^2 x \cdot e \, dx$$ Now in first integral, let $x^2 = t \Rightarrow 2x \, dx = dt$ we get, $$\Rightarrow \int_0^2 x f(x) = \left[ \frac{1}{2} \int_0^1 e^t \, dt \right] + \left[ e \cdot \frac{x^2}{2} \right]_1^2$$ $$\Rightarrow \int_0^2 x f(x) = \left( e - \frac{1}{2} \right) + \frac{3e}{2}$$ $$\Rightarrow \int_0^2 x f(x) = 2e - \frac{1}{2}$$

Given: $$ \frac{dy}{dx} + \frac{5}{x(1+x^5)}y = \frac{(1+x^5)^2}{x^7} $$ This is a linear differential equation. I.F. = $$ e^{\int \frac{5}{1+x^5} dx} $$ $\($ $\Rightarrow$ $\)$ I.F. = $$ e^{\int \frac{5x^5}{x(1+x^5)} dx} $$ $\($ $\Rightarrow$ $\)$ I.F. = $$ e^{\int \frac{5}{x(1+x^5)} dx} $$ $\($ $\Rightarrow$ $\)$ I.F. = $$ e^{-\int \frac{-5}{x^5+1} dx} $$ $\($ $\Rightarrow$ $\)$ I.F. = $$ e^{-\int \frac{d(x^{-5}+1)}{x^{-5}+1}} $$ $\($ $\Rightarrow$ $\)$ I.F. = $$ e^{-\log_e (x^{-5}+1)} $$ $\($ $\Rightarrow$ $\)$ I.F. = $$ (x^{-5} + 1)^{-1} = \frac{x^5}{x^5+1} $$ So, solution is given by $$ y \frac{x^5}{x^5+1} = \int \frac{x^5}{x^5+1} \times \frac{(x^5+1)^2}{x^7} dx $$ $\($ $\Rightarrow$ $\)$ $$ \frac{y x^5}{x^5+1} = \int \frac{x^5+1}{x^2} dx $$ $\($ $\Rightarrow$ $\)$ $$ \frac{y x^5}{x^5+1} = \int (x^3 + \frac{1}{x^2}) dx $$ $\($ $\Rightarrow$ $\)$ $$ \frac{y x^5}{x^5+1} = \frac{x^4}{4} - \frac{1}{x} + C $$ Since, $\($ y(1) = 2 $\)$, so $$ \frac{2}{2} = \frac{1}{4} - 1 + C \Rightarrow C = \frac{7}{4} $$ So, $$ \frac{y x^5}{x^5+1} = \frac{x^4}{4} - \frac{1}{x} + \frac{7}{4} $$ Put $\($ x = 2 $\)$, then we get $$ y(2) \times 32 = \frac{16}{4} - \frac{1}{2} + \frac{7}{4} $$ $$ \Rightarrow y(2) = \frac{33}{128} $$

Given that $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar. Therefore, $\vec{b} - \vec{a}, \vec{c} - \vec{a}, \vec{d} - \vec{a}$ are also coplanar. Hence, $$\left[ \vec{b} - \vec{a} \vec{c} - \vec{a} \vec{d} - \vec{a} \right] = 0$$ This implies $$\left( \vec{b} - \vec{a} \right) \cdot \left( \left( \vec{c} - \vec{a} \right) \times \left( \vec{d} - \vec{a} \right) \right) = 0$$ Therefore, $$\left( \vec{b} - \vec{a} \right) \cdot \left( \vec{c} \times \vec{d} - \vec{c} \times \vec{a} - \vec{a} \times \vec{d} \right) = 0 \left( Since \vec{a} \times \vec{a} = 0 \right)$$ This leads to $$\vec{b} \cdot \left( \vec{c} \times \vec{d} \right) - \vec{b} \cdot \left( \vec{c} \times \vec{a} \right) - \vec{b} \cdot \left( \vec{a} \times \vec{d} \right) + \vec{a} \cdot \left( \vec{c} \times \vec{a} \right) + \vec{a} \cdot \left( \vec{a} \times \vec{d} \right) = 0$$ Therefore, $$\left[ \vec{b} \vec{c} \vec{d} \right] - \left[ \vec{b} \vec{c} \vec{a} \right] - \left[ \vec{b} \vec{a} \vec{d} \right] = 0$$ We know that $\left[ \vec{x} \vec{y} \vec{z} \right] = \left[ \vec{y} \vec{z} \vec{x} \right] = \left[ \vec{z} \vec{x} \vec{y} \right]$ and $\left[ \vec{x} \vec{y} \vec{z} \right] = -\left[ \vec{x} \vec{z} \vec{y} \right]$ Therefore, $$\left[ \vec{b} \vec{c} \vec{d} \right] + \left[ \vec{b} \vec{a} \vec{c} \right] = \left[ \vec{a} \vec{b} \vec{c} \right]$$ and $$\left[ \vec{c} \vec{d} \vec{b} \right] + \left[ \vec{d} \vec{b} \vec{a} \right] = \left[ \vec{a} \vec{b} \vec{c} \right]$$ Hence, the required answer is $$\left[ \vec{d} \vec{c} \vec{a} \right] + \left[ \vec{c} \vec{d} \vec{b} \right]$$

Given, $$\int_0^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$$ Now, let $$I = \int_0^{\frac{\pi}{2}} f(\sin 2x) \cdot \sin x \, dx$$ Now using property $$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx$$ we get, $$\Rightarrow I = \int_0^{\frac{\pi}{4}} f(\sin 2x) \sin x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx$$ Now using property $$\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx$$ we get, $$\Rightarrow I = \int_0^{\frac{\pi}{4}} f(\cos 2x) \sin \left( \frac{\pi}{4} - x \right) \, dx + \int_0^{\frac{\pi}{4}} f(\cos 2x) \sin \left( \frac{\pi}{4} + x \right) \, dx$$ $$\Rightarrow I = \int_0^{\frac{\pi}{4}} f(\cos 2x) \left( \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x \right) \, dx + \int_0^{\frac{\pi}{4}} f(\cos 2x) \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \right) \, dx$$ $$\Rightarrow I = \int_0^{\frac{\pi}{4}} f(\cos 2x) \left( \sqrt{2} \cos x \right) \, dx$$ So, putting the value of $I$ in $$\int_0^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$$ we get, $$\alpha = -\sqrt{2}$$

Given, system of linear equations, $$7x + 11y + \alpha z = 13$$ $$5x + 4y + 7z = \beta$$ $$175x + 194y + 57z = 361$$ Now we know that, for infinite solution $D = D_1 = D_2 = D_3 = 0$. Now solving $D = 0$ we get, $$\begin{vmatrix} 7 & 11 & \alpha \\ 5 & 4 & 7 \\ 175 & 194 & 57 \end{vmatrix} = 0$$ Now applying operation $R_3 \rightarrow R_3 - 25R_1$ we get, $$\begin{vmatrix} 7 & 11 & \alpha \\ 5 & 4 & 7 \\ 0 & -81 & 57 - 25\alpha \end{vmatrix} = 0$$ $$\Rightarrow 81(49 - 5\alpha) + (57 - 25\alpha)(-27) = 0$$ $$\Rightarrow 270\alpha = -2430 \Rightarrow \alpha = -9$$ And now solving $D_1 = 0$ we get, $$\begin{vmatrix} 13 & 11 & -9 \\ \beta & 4 & 7 \\ 361 & 194 & 57 \end{vmatrix} = 0$$ $$\Rightarrow 13(4 \times 57 - 7 \times 194) - \beta(11 \times 57 + 9 \times 194) + 361(11 \times 7 + 9 \times 4) = 0$$ $$\Rightarrow 13(-1130) - \beta(2373) + 361(113) = 0$$ $$\Rightarrow 13(-10) - \beta(21) + 361 = 0$$ $$\Rightarrow 21\beta = 231$$ $$\Rightarrow \beta = 11$$ Therefore, $\alpha + \beta + 2 = 4$

Given: $$f(x) = \frac{1}{\sqrt{[x]^2 - 3[x] - 10}}$$ For $f(x)$ to be defined, $$[x]^2 - 3[x] - 10 > 0$$ $$\Rightarrow ([x] - 5)([x] + 2) > 0$$ $$\Rightarrow [x] 5$$ $$\Rightarrow x < -2 and x \geq 6$$ So, domain is $(-\infty, -2) \cup [6, \infty)$.

Given, $P$ be the plane passing through the points $(5, 3, 0), (13, 3, -2)$ and $(1, 6, 2)$. Now, $\overrightarrow{AB} = 8\hat{i} - 2\hat{k}$, $\overrightarrow{AC} = -4\hat{i} + 3\hat{j} + 2\hat{k}$. So, normal vector to the plane will be, $$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & 0 & -2 \\ -4 & 3 & 2 \end{vmatrix} = 6\hat{i} - 8\hat{j} + 24\hat{k}$$ Now equation of plane will be: $6x - 8y + 24z = d$. And given plane passes through $(5, 3, 0)$, so, $6 \times 5 - 8 \times 3 + 24 \times 0 = d \Rightarrow d = 6$. Hence, equation of plane will be, $$6x - 8y + 24z = 6 \Rightarrow 3x - 4y + 12z = 3$$ Now using the formula distance of point $(3, 4, \alpha)$ from the plane $3x - 4y + 12z = 3$, we get, $$\frac{9 - 16 + 12\alpha - 3}{\sqrt{9 + 16 + 144}} = 2 \Rightarrow \alpha = 3$$ Now again using the distance of point $(2, \alpha, a) = (2, 3, a)$ from the plane formula we get, $$\frac{3 \times 2 - 4 \times 3 + 12 \times a - 3}{13} = 3$$ $$\Rightarrow 12a - 9 = 39$$ $$\Rightarrow 12a = 48$$ $$\Rightarrow a = 4$$

We know that, Converse of $A \Rightarrow B$ is $B \Rightarrow A$. So, using the above formula we get, Converse of $\sim p \land q \Rightarrow r$ $$\equiv r \Rightarrow (\sim p \land q)$$ $$\equiv (\sim r) \lor (\sim p \land q)$$ $$\equiv (\sim p \land q) \lor (\sim r)$$ $$\equiv \sim (p \lor \sim q) \lor (\sim r) \{ as \sim (A \lor B) \equiv \sim A \land \sim B \}$$ $$\equiv (p \lor (\sim q)) \Rightarrow (\sim r)$$

Given that the elevation to the top of the tower from the feet of one person due south is $45^\circ$. Similarly the angle of elevation to the top of tower from the feet of another person to the west of tower is $30^\circ$. The given information can be represented in the form of a diagram as below: We need to find the distance $AB$. Let the distance of $A, B$ from the tower be $x_1, x_2$. $$\Rightarrow \tan 45^\circ = \frac{5}{x_1}, \tan 30^\circ = \frac{5}{x_2}$$ $$\Rightarrow x_1 = 5, \; x_2 = 5\sqrt{3}.$$ $$\Rightarrow AB = \sqrt{x_1^2 + x_2^2} = \sqrt{(5)^2 + (5\sqrt{3})^2} = 10 \, \mathrm{m}$$ Hence this is the correct option.

Given: $a + b + c + d = 11$ We know that, AM $\geq$ GM $$\frac{a}{5} + \frac{a}{5} + \frac{a}{5} + \frac{a}{5} + \frac{a}{5} + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + \frac{c}{2} + \frac{c}{2} + d \geq \sqrt[11]{\frac{a^5b^3c^2d}{5^5 3^3 2^2}}$$ $$\Rightarrow \frac{a+b+c+d}{11} \geq \sqrt[11]{\frac{a^5b^3c^2d}{5^5 3^3 2^2}}$$ $$\Rightarrow \frac{11}{11} \geq \sqrt[11]{\frac{a^5b^3c^2d}{5^5 3^3 2^2}}$$ $$\Rightarrow \sqrt[11]{\frac{a^5b^3c^2d}{5^5 3^3 2^2}} \leq 1$$ $$\Rightarrow a^5b^3c^2d \leq 5^5 3^3 2^2$$ $$\Rightarrow a^5b^3c^2d \leq 90 \cdot 3750$$ So, $\beta = 90$

Given equation of ellipse is $\frac{x^2}{36} + \frac{y^2}{9} = 1$. It is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a = 6$, $b = 3$. The centre of the circle is $C(2, 0)$. Let $P(a \cos \theta, b \sin \theta)$ be the common point for ellipse and circle. $P \equiv (6 \cos \theta, 3 \sin \theta)$. Equation of normal to the ellipse will be $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2 - b^2$. $$\Rightarrow 6x \sec \theta - 3y \csc \theta = 36 - 9$$ But the normal through $P$ passes through the centre of the circle $C(2, 0)$. $$\Rightarrow 6(2) \sec \theta - 3(0) \csc \theta = 27$$ $$\Rightarrow \sec \theta = \frac{27}{12}$$ $$\Rightarrow \cos \theta = \frac{4}{9} and \sin \theta = \frac{\sqrt{65}}{9}$$ $$P \equiv \left( 6 \times \frac{4}{9}, 3 \times \frac{\sqrt{65}}{9} \right)$$ $$P \equiv \left( \frac{8}{3}, \frac{\sqrt{65}}{3} \right)$$ Now $CP = r = \sqrt{\left( 2 - \frac{8}{3} \right)^2 + \left( 0 - \frac{\sqrt{65}}{3} \right)^2}$. $$\Rightarrow r^2 = \frac{69}{9}$$ $$\Rightarrow 12r^2 = 12 \times \frac{69}{9} = 92$$ Therefore, the required answer is 92.

Consider the data 1, 2, 4, 5, x, y. So, Mean $$= \frac{1+2+4+5+x+y}{6}$$ $$\Rightarrow 5 = \frac{x+y+12}{6}$$ $$\Rightarrow x + y = 18$$ And, Variance $$= \left( \frac{\sum x_i^2}{n} \right) - \left( \frac{\sum x_i}{n} \right)^2$$ $$\Rightarrow 10 = \left( \frac{1^2+2^2+4^2+5^2+x^2+y^2}{6} \right) - (5)^2$$ $$\Rightarrow x^2 + y^2 = 164$$ We have $x + y = 18$ and $x^2 + y^2 = 164$. Now solving above equations we get, $x = 10$, $y = 8$. Now mean deviation about mean will be $$= \frac{|1-5| + |2-5| + |4-5| + |5-5| + |10-5| + |8-5|}{6}$$ $$= \frac{16}{6}$$ $$= \frac{8}{3}$$ Therefore, the required answer is $\frac{8}{3}$.

Ratio of three consecutive terms in the binomial expansion of $(1+x)^{n+2}$ is ${}^{n+2}C_{r-1} : {}^{n+2}C_{r} : {}^{n+2}C_{r+1} :: 1 : 3 : 5$. So, $$\frac{{}^{n+2}C_{r}}{{}^{n+2}C_{r-1}} = \frac{3}{1}$$ $$\Rightarrow \frac{n-r+3}{r} = 3$$ $$\Rightarrow n + 3 = 4r \ldots(1)$$ And, $$\frac{{}^{n+2}C_{r+1}}{{}^{n+2}C_{r}} = \frac{5}{3}$$ $$\Rightarrow \frac{n+2-r}{r+1} = \frac{5}{3}$$ $$\Rightarrow 3n + 6 - 3r = 5r + 5$$ $$\Rightarrow 3n + 1 = 8r \ldots(2)$$ So, $3(4r - 3) + 1 = 8r$ $$\Rightarrow 12r - 9 + 1 = 8r$$ $$\Rightarrow 4r = 8$$ $$\Rightarrow r = 2$$ So, $n = 5$. Hence, we have $$(1+x)^{n+2} = (1+x)^7$$ Sum of consecutive terms is $${}^{7}C_{1} + {}^{7}C_{2} + {}^{7}C_{3}$$ $$= 7 + 21 + 35 = 63$$

The given word is MATHS. Arranging the letters alphabetically, we get AHMST. When the word starts with any of the letters A/H/M/S, the number of possibilities is $4! \times 4 = 96$. Now when the word starts with TA, then the number of possibilities is $3! = 6$. Now when the word starts with THAMS, then the number of possibilities is $1$. Rank $= 96 + 6 + 1 = 103$. Hence, rank of the word THAMS is 103.

Let $a = x_1 + iy_1$, $\bar{a} = x_1 - iy_1$ and $z = x_2 + iy_2$, $\bar{z} = x_2 - iy_2$. Now, $$a + \bar{z} = (x_1 + x_2) + i(y_1 - y_2)$$ and $$\bar{a} + z = (x_1 + x_2) + i(y_2 - y_1)$$ Now according to the question, $$\mathrm{Re}(a + \bar{z}) = x_1 + x_2$$ and $$\mathrm{Im}(\bar{a} + z) = y_2 - y_1$$ $$A = \{ z : x_1 + x_2 > y_2 - y_1 \} = \{ z : x_1 + y_1 + x_2 > y_2 \}$$ $$B = \{ z : x_1 + x_2 0$ then $$A = \{ z : x_2 > -(x_1 + y_1) \}$$ $A$ covers a part of negative real axis and therefore, does not contain whole real axis. Similarly if $y_2 = 0$, and $x_1, y_1 < 0$, then $$B = \{ z : x_2 < -(x_1 + y_1) \}$$ $B$ covers part of positive real axis and therefore does not cover whole real axis. Hence, both are false.

Given, $A = \{1, 3, 4, 6, 9\}$ and $B = \{2, 4, 5, 8, 10\}$. And relation is given by, $$R = \{(a_1, b_1), (a_2, b_2): a_1 \leq b_2 and b_1 \leq a_2\}$$ Now taking cases for $a_1 \leq b_2$ we get, $a_1 = 1$, $b_2 \in \{2, 4, 5, 8, 10\} \rightarrow 5$ cases $a_1 = 3$, $b_2 \in \{4, 5, 8, 10\} \rightarrow 4$ cases $a_1 = 4$, $b_2 \in \{4, 5, 8, 10\} \rightarrow 4$ cases $a_1 = 6$, $b_2 \in \{8, 10\} \rightarrow 2$ cases $a_1 = 9$, $b_2 \in \{10\} \rightarrow 1$ cases So, total 16 cases will be there. Now finding cases of $b_1 \leq a_2$ we get, $b_1 = 2$, $a_2 \in \{3, 4, 6, 9\} \rightarrow 4$ cases $b_1 = 4$, $a_2 \in \{4, 6, 9\} \rightarrow 3$ cases $b_1 = 5$, $a_2 \in \{6, 9\} \rightarrow 2$ cases $b_1 = 8$, $a_2 \in \{9\} \rightarrow 1$ cases So, here total 10 cases. Hence, total elements in relation $= 16 \times 10 = 160$

Given, $f(x) = x + 1,\quad x < 0$ $f(x) = |x - 1|,\quad x \geq 0$ and $g(x) = x + 1,\quad x < 0$ $g(x) = 1,\quad x \geq 0$ $( \Rightarrow f(x) = )$ $f(x) = x + 1,\quad x < 0$ $f(x) = 1 - x,\quad 0 \leq x < 1$ $f(x) = x - 1,\quad x \geq 1$ and $g(x) = 1 + x,\quad x < 0$ $g(x) = 1,\quad x \geq 0$ Now, for $( gof(x) )$ $gof(x) = 1 + f(x),\quad f(x) < 0,\; x < 0$ $gof(x) = 1,\quad f(x) \geq 0,\; x \geq 0$ $( \Rightarrow gof(x) = )$ $gof(x) = x + 2,\quad x < -1$ $gof(x) = 1,\quad x \geq -1$ Now plotting the diagram we get, Now from above diagram we can say that function is continuous but not differentiable at $( x = -1 )$ due to sharp corner.

Given, $$e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1 = 0$$ $$\Rightarrow e^{4x} - e^{2x} - 3 - \frac{1}{e^{2x}} + \frac{1}{e^{4x}} = 0$$ $$\Rightarrow e^{4x} + \frac{1}{e^{4x}} + 2 - \left(e^{2x} + \frac{1}{e^{2x}}\right) = 5$$ $$\Rightarrow \left(e^{2x} + \frac{1}{e^{2x}}\right)^2 - \left(e^{2x} + \frac{1}{e^{2x}}\right) = 3$$ Now let $\left(e^{2x} + \frac{1}{e^{2x}}\right) = t$ So, the equation becomes $$t^2 - t - 5 = 0$$ $$\Rightarrow t = \frac{1 + \sqrt{21}}{2},$$ ignoring negative sign as exponential function are positive, Now $e^{2x} + \frac{1}{e^{2x}} = \frac{1 + \sqrt{21}}{2}$ $$\Rightarrow e^{4x} - \frac{1 + \sqrt{21}}{2} e^{2x} + 1 = 0$$ which is quadratic equation in $e^{2x}$ with upward parabola, Now by $A.M \geq G.M$ we get, $e^{2x} + \frac{1}{e^{2x}} \geq 2$ So, $y = \frac{1 + \sqrt{21}}{2} > 2$, hence it will cut at two distinct point, Hence, there will be two solution.

Given a quadratic equation $64x^2 + 5nx + 1 = 0$ has no real roots, so $D < 0$. $$\Rightarrow 25N^2 - 4 \times 64 < 0$$ $$\Rightarrow N < \frac{16}{5} \Rightarrow N = 1, 2, 3$$ And, for a biased coin, the probability of getting head is $P = \frac{1}{4}$, and probability of getting tail will be $Q = \frac{3}{4}$. So, required probability is $$\frac{1}{4} + \frac{3}{4} \cdot \frac{1}{4} + \left( \frac{3}{4} \right)^2 \cdot \frac{1}{4} + \left( \frac{3}{4} \right)^3 \cdot \frac{1}{4} = \frac{p}{q}$$ $$\Rightarrow \frac{1}{4} \left( 1 - \left( \frac{3}{4} \right)^3 \right) = \frac{p}{q}$$ $$\Rightarrow \frac{1}{1 - \left( \frac{3}{4} \right)} = \frac{p}{q}$$ $$\Rightarrow 1 - \left( \frac{3}{4} \right)^3 = \frac{p}{q}$$ $$\Rightarrow 1 - \frac{27}{64} = \frac{p}{q}$$ $$\Rightarrow \frac{p}{q} = \frac{37}{64}$$ $$q - p = 27$$

Given, $\($ $\vec{a}$ = $\hat{i}$ + 2 $\hat{j}$ + 3 $\hat{k}$, $\vec{b}$ = $\hat{i}$ + $\hat{j}$ - $\hat{k}$, $\vec{a}$ $\cdot$ $\vec{c}$ = 11, $\vec{b}$ $\cdot$ ($\vec{a}$ $\times$ $\vec{c}$) = 27 $\)$ and $\($ $\vec{b}$ $\cdot$ $\vec{c}$ = -$\sqrt{3}$ |$\vec{b}$| $\)$. Now finding, $\($ $\left$( $\vec{b}$ $\times$ $\vec{a}$ $\right$) $\)$ we get, $$ \left( \vec{b} \times \vec{a} \right) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & 2 & 3 \end{vmatrix} = 5 \hat{i} - 4 \hat{j} + \hat{k} $$ Let $\($ $\vec{c}$ = c_1 $\hat{i}$ + c_2 $\hat{j}$ + c_3 $\hat{k}$ $\)$ Now solving, $\($ $\vec{a}$ $\cdot$ $\vec{c}$ = 11 $\)$ we get, $c_1$ + $2c_2$ + $3c_3$ = 11 $\ldots$ (1) Now solving, $\vec{b}$ $\cdot$ $\vec{c}$ = -$\sqrt{3}$ |$\vec{b}$| we get, $c_1$ + $c_2$ - $c_3$ = -$\sqrt{3}$ $\sqrt{3}$ $c_1$ + $c_2$ - $c_3$ = -3 $\ldots$ (2) Now solving $\left$( $\vec{b}$ $\times$ $\vec{a}$ $\right$) $\cdot$ $\vec{c}$ = 27 we get, 5$c_1$ - 4$c_2$ + $c_3$ = 27 $\ldots$ (3) Now on solving equation (1), (2) $\&$ (3) we get, \[ \vec{c}=3\hat{i}-2\hat{j}+4\hat{k} \] Hence, \[ \left|\vec{a}\times\vec{c}\right|^2 \] = \[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -2 & 4 \end{vmatrix}^2 \] = \[ \left|14\hat{i}+5\hat{j}-8\hat{k}\right|^2 \]$\($ $\Rightarrow$ \[ \left|\vec{a}\times\vec{c}\right|^2 \] = $14^2$ + $5^2$ + $8^2$ = 285 $\)$

Put $z = x + iy$ in the given expression and equate the imaginary part to zero. $$\Rightarrow \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} = \frac{(x + iy)^2 + 8i(x + iy) - 15}{(x + iy)^2 - 3i(x + iy) - 2}$$ $$= \frac{x^2 - y^2 - 8y - 15 + i(2xy + 8x)}{x^2 - y^2 + 3y - 2 + i(2xy - 3x)}$$ $$= \frac{x^2 - y^2 - 8y - 15 + i(2xy + 8x)}{(x^2 - y^2 + 3y - 2) + i(2xy - 3x)} \times \frac{(x^2 - y^2 + 3y - 2) - i(2xy - 3x)}{(x^2 - y^2 + 3y - 2) - i(2xy - 3x)}$$ $$= \frac{(x^2 - y^2 - 8y - 15)(x^2 - y^2 + 3y - 2) + (2xy + 8x)(2xy - 3x) - i(2xy - 3x)(x^2 - y^2 - 8y - 15)}{(x^2 - y^2 + 3y - 2)^2 - (2xy - 3x)^2}$$ But $\mathrm{Im}\left(\frac{z^2 + 8iz - 15}{z^2 - 3iz - 2}\right) = 0$, $$\Rightarrow -(x^2 - y^2 - 8y - 15)(2xy - 3x) + (2xy + 8x)(x^2 - y^2 + 3y - 2) = 0$$ $$\Rightarrow (x^2 - y^2)(2xy + 8x - 2xy + 3x) + (8y + 15)(2xy - 3x) = 0$$ $$\Rightarrow 11x^3 - 11xy^2 + 16xy^2 - 24xy + 30xy - 45x = 0$$ $$\Rightarrow 11x^3 + 11xy^2 + 26xy - 61x = 0$$ $$\Rightarrow 11x^2 + 11y^2 + 26y - 61 = 0$$ Therefore, $\alpha \neq 0$ So, put $y = -\frac{13}{11}, x = \alpha$ $$11\alpha^2 + 11 \cdot \frac{132}{121} \cdot 26 \cdot \frac{13}{11} - 61 = 0$$ $$\Rightarrow 121\alpha^2 = 840$$ $$\Rightarrow 242\alpha^2 = 1680$$ Hence this is the required answer.

Given, $$10 = 1 + \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \cdots \ldots (1)$$ $$\frac{10}{k} = \frac{1}{k} + \frac{4}{k^2} + \frac{8}{k^3} + \frac{13}{k^4} + \cdots \ldots (2)$$ On subtracting above two equations we get, $$10 - \frac{10}{k} = 1 + \frac{3}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \cdots \ldots \infty$$ $$\Rightarrow 10 \left(1 - \frac{1}{k}\right) = 1 + \frac{3}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \cdots \ldots \infty \ldots (3)$$ $$\Rightarrow 10 \left(1 - \frac{1}{k}\right) \left(\frac{1}{k}\right) = \frac{1}{k} + \frac{3}{k^2} + \frac{4}{k^3} + \frac{5}{k^4} + \frac{6}{k^5} + \cdots \ldots \infty \ldots (4)$$ Again subtracting both equations we get, $$10 \left(1 - \frac{1}{k}\right)^2 = 1 + \frac{2}{k} + \frac{1}{k^2} + \frac{1}{k^3} + \frac{1}{k^4} + \cdots \ldots \infty$$ $$\Rightarrow 10 \left(1 - \frac{1}{k}\right)^2 = \left(1 + \frac{1}{k} + \frac{1}{k^2} + \frac{1}{k^3} + \frac{1}{k^4} + \cdots \ldots \infty\right) + \frac{1}{k}$$ $$\Rightarrow 10 \left(1 - \frac{1}{k}\right)^2 = \left(\frac{1}{1 - \frac{1}{k}}\right) + \frac{1}{k}$$ $$\Rightarrow 10(k - 1)^3 = k(k^2 + k - 1)$$ $$\Rightarrow 9k^3 - 31k^2 + 31k - 10 = 0$$ $$\Rightarrow k = 2$$ Hence, $k = 2$

Given, $$f(1) + f(2) + 1 = f(4)$$ $$\Rightarrow f(1) + f(2) \leq 5 \{ as f(4) \leq 6 \}$$ Now taking cases we get, Case (1), $f(1) = 1 \Rightarrow f(2) = \{1, 2, 3, 4\} \rightarrow 4$ mappings Case (2), $f(1) = 2 \Rightarrow f(2) = \{1, 2, 3\} \rightarrow 3$ mappings Case (3), $f(1) = 3 \Rightarrow f(2) = \{1, 2\} \rightarrow 2$ mappings Case (4), $f(1) = 4 \Rightarrow f(2) = \{1\} \rightarrow 1$ mappings And $f(5)$ and $f(6) \rightarrow 6$ mapping each So, total number of function will be $=(4 + 3 + 2 + 1) \times 6 \times 6 = 360$

Given, the tangent to the parabola $y^2 = 12x$ at the point $(3, \alpha)$ be perpendicular to the line $2x + 2y = 3$. So, Slope of tangent $= 1 = \frac{6}{\alpha} \Rightarrow \alpha = 6$. So, equation of hyperbola will be, $$36x^2 - 9y^2 = 324$$ $$\Rightarrow \frac{x^2}{9} - \frac{y^2}{36} = 1$$ Now equation of tangent at $(5, 8)$ will be, $$\frac{5x}{9} - \frac{8y}{36} = 1$$ $$\Rightarrow 5x - 2y = 9$$ So, slope of normal $= -\frac{2}{5}$. Now finding, equation of normal $y - 8 = -\frac{2}{5}(x - 5)$. $$\Rightarrow 5y - 40 = -2x + 10$$ $$\Rightarrow 5y + 2x = 50$$ Now finding, distance from $(6, -4)$, we get, $$= \frac{|12 - 20 - 50|}{\sqrt{29}} = \frac{58}{\sqrt{29}}$$ $$= 2\sqrt{29}$$ Hence, $$\left(2\sqrt{29}\right)^2 = 116$$

The given equations are, $$L: \frac{x-0}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda} \& P: x + 2y + 3z = 4$$ So, Vector parallel to line: $$ = \vec{b}$$ Normal vector to plane $$P: = \vec{n}$$ Angle between plane and line is $\theta$. Then angle between the line and normal to the plane will be $90 - \theta$. Then, $$\Rightarrow \cos(90 - \theta) = \frac{(\hat{i} + 2\hat{j} + \lambda \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k})}{\sqrt{1^2 + 2^2 + \lambda^2} \cdot \sqrt{1^2 + 2^2 + 3^2}}$$ $$\Rightarrow \frac{3}{\sqrt{14}} = \frac{1 + 4 + 3\lambda}{\sqrt{\lambda^2 + 5} \sqrt{14}} \Rightarrow \lambda = \frac{2}{3}$$ $$L_1 = \frac{x-0}{3} = \frac{y-1}{6} = \frac{z-3}{2} = \mu$$ $$(x, y, z) \equiv (3\mu, 6\mu + 1, 2\mu + 3)$$ But this point lies on the plane: $$(3\mu + 12\mu + 2 + 6\mu + 9) = 4$$ $$\Rightarrow \mu = \frac{-1}{3}$$ Hence, $$\alpha = 3\mu = -1, \beta = 6\mu + 1 = -1, \gamma = 2\mu + 3 = \frac{7}{3}$$ Now, $$\alpha + 2\beta + 6\gamma = 11$$ Hence this is the required answer.

Given, $L_1 : 3y - 2x = 3$ is angular bisector of $L_2 : x - y + 1 = 0$ and $L_3 : \alpha x + \beta y + 17 = 0$ Now finding point of intersection of $L_1$ and $L_2$ we get, $(0, 1)$ And point will lie on $L_3$, so $\alpha \times 0 + \beta \times 1 + 17 = 0$ $$\Rightarrow \beta = -17$$ Any point, say $\left(-\frac{3}{2}, 0\right)$ on $L_1$ should be equidistant from lines $L_2$ and $L_3$ Now using the formula of distance of a point from the line we get, $$\Rightarrow \frac{\left|-\frac{3}{2} - 0 + 1\right|}{\sqrt{1^2 + 1^2}} = \frac{\left|-\frac{3\alpha}{2} - 0 + 17\right|}{\sqrt{\alpha^2 + (-17)^2}}$$ $$\Rightarrow (\alpha - 7)(\alpha - 17) = 0$$ Now, for $\alpha = 17$, $L_2$ and $L_3$ coincides So, $\alpha = 7$ Now putting the value of $\alpha$ and $\beta$ in given expression we get, $$\alpha^2 + \beta^2 - \alpha - \beta = (7^2) + (-17)^2 - 7 + 17 = 348$$

Given, $y = 2x^2 + 1$ Now tangent to above equation at $(1, 3)$ is given by, $$y - 3 = 4(x - 1)$$ $$\Rightarrow y = 4x - 1$$ Now plotting the diagram we get, Required area of bounded region will be, $$A = \int_0^1 (2x^2 + 1) \, dx - area of \triangle QOT - area of \triangle PQR + area of \triangle QRS$$ $$\Rightarrow A = \left( \frac{2}{3} + 1 \right) - \frac{1}{2} - \frac{9}{8} + \frac{9}{40} = \frac{16}{40}$$ $$\Rightarrow 60A = 16$$

The terminal velocity can be given by $$v_T = \frac{2(\rho - \sigma) r^2 g}{9 \eta}.$$ Clearly, $v_T \propto r^2$. Let the radius of the small drops be $r_1$ and the radius of the big drop be $r_2$. So, $$8 \left( \frac{4\pi}{3} \right) r_1^3 = \left( \frac{4\pi}{3} \right) r_2^3$$ $$\Rightarrow r_1 = \frac{r_2}{2}$$ Hence, taking the ratio $$\left( \frac{v_T}{v_T} \right)_1 = \left( \frac{r_1}{r_2} \right)^2$$ $$\Rightarrow \frac{10}{(v_T)_2} = \frac{1}{4}$$ $$\Rightarrow (v_T)_2 = 40 \, \mathrm{cm} \, \mathrm{s}^{-1}$$

Let velocity of sound in air be $v \, \mathrm{m \, s^{-1}}$. Using Doppler formula, $$f = f_0 \left( \frac{v + v_0}{v - v_s} \right)$$ $$= f_0 \left( \frac{v + 40}{v - 20} \right)$$ $$= 400 \left( \frac{360 + 40}{340} \right)$$ $$\approx 471 \, \mathrm{Hz}$$

It is given that $\vec{E} = 6.6 \hat{j} \left( \mathrm{V \, m^{-1}} \right)$. The relation between the electric, magnetic field and speed of light is $\vec{E} \times \vec{B} = \vec{c}$. Clearly, $\vec{B} = B_0 \left( \hat{k} \right)$. Using the relation, $cB_0 = E$. Therefore, $$B_0 = \left( \frac{E}{c} \right) = \frac{6.6}{3 \times 10^8} = 2.2 \times 10^{-8} \, \mathrm{T}$$

The charge enclosed is given by $q_{inc} = CV$. Gauss' law states that the flux through any closed surface is equal to the charge enclosed by the surface. Using Gauss' law the flux is, $$\phi = \frac{q_{inc}}{\varepsilon_0} = \left( \frac{CV}{\varepsilon_0} \right)$$

The dimensional formula of force is $F = MLT^{-2}$. The dimensional formula of velocity is $V = LT^{-1}$. The formula for density is $$Density = \frac{M}{L^3} = ML^{-3}$$ Comparing the density dimensions with the assumed fundamental quantities in the question, $$ML^{-3} = [F]^x [V]^y [T]^z$$ $$\Rightarrow ML^{-3} = [MLT^{-2}]^x [LT^{-1}]^y [T]^z$$ $$\Rightarrow ML^{-3} = M^x L^{(x+y)} T^{-2x-y+z}$$ Comparing the exponents, $$x = 1 \cdots (i)$$ $$x + y = -3 \cdots (ii)$$ $$-2x - y + z = 0 \cdots (iii)$$ Solving equations (i), (ii) $\&$ (iii), we get $y = -4$, $z = -2$. The dimension is derived as $F V^{-4} T^{-2}$.

In satellite telecommunication, a downlink is the link from a satellite down to one or more ground stations or receivers, and an uplink is the link from a ground station up to a satellite. Uplink frequency band for satellite communication is $5.925 - 6.425 \, \mathrm{GHz}$.

The magnitude of gravitational potential of a solid sphere at a distance $r$ from the centre is $$V = \frac{-GM}{2R^3} \left( 3R^2 - r^2 \right)$$ At $r = R$, $V = \frac{-GM}{R}$. At $r = 0$, $V' = \frac{-GM}{2R^3} \left( 3R^2 \right) = \frac{-3GM}{2R}$. Clearly, $V' = \frac{3V}{2}$.

It is given that $v = 10 \sqrt{x} \, \mathrm{m \, s^{-1}}$ Force is defined by $F = m \frac{dv}{dt} \cdots (i)$ Multiplying and dividing equation (i) by $dx$ $$F = m \frac{dv}{dt} \frac{dx}{dx}$$ $$\Rightarrow F = m \frac{dx}{dt} \frac{dv}{dx} = mv \frac{dv}{dx} \cdots (ii)$$ Finding the values of the terms in equation (ii) $$\frac{dv}{dx} = \frac{10}{2\sqrt{x}} = \frac{5}{\sqrt{x}}$$ $$v \frac{dv}{dx} = a = 10 \sqrt{x} \times \frac{5}{\sqrt{x}} = 50 \, \mathrm{m \, s^{-2}}$$ Hence, the force is $$F = \frac{1}{2} \times 50 = 25 \, \mathrm{N}.$$

It is given that $v = 40 \, \mathrm{m \, s^{-1}}$. The components of the velocity will be $$v_y = v \sin 30^\circ = \frac{40 \, \mathrm{m \, s^{-1}}}{2} = 20 \, \mathrm{m \, s^{-1}}$$ $$v_x = v \cos 30^\circ = \frac{40 \sqrt{3}}{2} \, \mathrm{m \, s^{-1}} = 20 \sqrt{3} \, \mathrm{m \, s^{-1}}$$ The time taken to reach maximum height is $T = \frac{u \sin 30^\circ}{g} = \frac{20 \, \mathrm{m \, s^{-1}}}{10 \, \mathrm{m \, s^{-2}}} = 2 \, \mathrm{s}$. Hence, at $T = 2 \, \mathrm{s}$, as $v_y = 0 \, \mathrm{m \, s^{-1}}$ therefore $v_{net} = v_x = 20 \sqrt{3} \, \mathrm{m \, s^{-1}}$.

It is given that $i = 30^\circ$. The formula for angle of deviation is $$\delta = 180^\circ - 2i$$ Substituting the values, $$\delta = 180^\circ - 60^\circ$$ $$= 120^\circ$$

The orbital velocity of satellite is $$v_{orb} = \sqrt{\frac{GM}{R}}$$ The escape velocity is $$v_{esc} = \sqrt{\frac{2GM}{R}}$$ Hence, the additional velocity required is $$\Delta v = v_{esc} - v_{orb} = \sqrt{\frac{2GM}{R}} - \sqrt{\frac{GM}{R}} = \left(\sqrt{2} - 1\right) \sqrt{\frac{GM}{R}}$$ $$= \left(\sqrt{2} - 1\right) \sqrt{gR}$$ $$= \left(\sqrt{2} - 1\right) \sqrt{10 \times 6400 \times 10^3}$$ $$= 8 \left(\sqrt{2} - 1\right) \, km \, s^{-1}$$

The de-Broglie wavelength is given by $$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \left( As, p^2 = 2mK \right).$$ It is given that $$(K)_e = (K)_p$$ Thus, the ratio of the wavelengths of the proton and electron is $$\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2K m_e}}{\sqrt{2K m_p}} = \sqrt{\frac{m_e}{m_p}} = \sqrt{\frac{1}{1849}} = \frac{1}{43}.$$

An isothermal process is a thermodynamic process in which the temperature of a system remains constant. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained. $$\Delta T = 0, \Delta U = nC_V \Delta T = 0.$$ According to the first law of thermodynamics, $$\Delta Q = \Delta U + \Delta W$$ $$\Rightarrow \Delta Q = \Delta W$$

Energy of Hydrogen like atoms is, $E = \frac{-13.6Z^2}{n^2} eV$. Since Helium is in the first state, $Z = 2$, $n = 2$. The energy of $\mathrm{He}^+$ ion is, $$E = \frac{-13.6 \times 2^2}{2^2} = -13.6 eV$$

As shown in the figure the various logic gates have been mentioned. Let the output before the NOT gate be $Z$. Using the expression for the logic gates, it can be written $$Z = \overline{(A + B)} \cdot (A \cdot B)$$ Hence, the output $Y$ is $$Y = \overline{Z} = \overline{(A + B) \cdot (A \cdot B)}$$ $$\Rightarrow Y = A \cdot B$$ Hence the digital circuit is equivalent to an AND gate.

The RMS value of a gas is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ $$= \sqrt{\frac{3k_B T}{m}}$$ Substituting the given values, $$v_{rms} = \sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{4.6 \times 10^{-26}}}$$ $$\approx 523 \, \mathrm{m \, s^{-1}}$$

As can be seen from the first circuit above, $CB$ and $AB$ are in series which is in parallel with $CA$. So, the net resistance is $R' = \frac{6}{2} \, \Omega = 3 \, \Omega$ as shown in the second diagram. From the second diagram, $DC$ and $CA$ is in series which is parallel to $DA$. Hence the resistance is $$R'' = \left( \frac{6 \times 3}{6 + 3} \right) \, \Omega = 2 \, \Omega$$ as shown in the third diagram. From the third diagram, both $ED$ and $DA$ are in series which is parallel to $EA$. The equivalent resistance as can be seen from the above simplified circuit is $R_{eq} = 2 \, \Omega$. The current $I$ through battery, $I = \frac{8}{2} = 4 \, \mathrm{A}$. From the second diagram, the part $EDACD$ has a resistance of $4 \, \Omega$, the total current will divide equally between $EDACD$ and $EA$. So consider $I_1 = 2 \, \mathrm{A}$ flowing through $ED$. Note that $I_1$ will be the same current flowing in the first diagram through $ED$. From the first diagram, when the resistances are in parallel the current is divided in the inverse ratio. Thus, $I'$ through $CB = \frac{2}{3} \times 2 = \frac{4}{3} \, \mathrm{A}$ $I''$ through $R_2 = \frac{1}{2} \left( \frac{2}{3} \right) = \frac{1}{3} \, \mathrm{A}$

It is given that $\vec{A} = 2\hat{i} + 3\hat{j} + 2\hat{k}$. By the given condition, $$\vec{B} - \left(2\hat{i} + 3\hat{j} + 2\hat{k}\right) = 2\hat{j}$$ $$\Rightarrow \vec{B} = 2\hat{i} + 5\hat{j} + 2\hat{k}$$ The magnitude is given by the formula, $$\left| \vec{B} \right| = \sqrt{B_x^2 + B_y^2 + B_z^2}$$ $$\Rightarrow \left| \vec{B} \right| = \sqrt{4 + 25 + 4} = \sqrt{33}$$

Due to change in flux, eddy current is produced which opposes the motion in the metallic pipe for the bar magnet. Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday's law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field. Hence, R is the correct explanation and both statements are true.

The Lorentz force is given by $\vec{F} = q \left( \vec{v} \times \vec{B} \right) = qvB \sin \theta \hat{n}$. The angle $\theta = 0^\circ$. Hence the force experienced is zero. Thus, the electron will not experience any magnetic force and the electron will continue to move along the axis.

From the diagram of the circuit, $C_3$, $C_4$, $C_5$ are in series, so the total capacitance is $$\frac{1}{C'} = \frac{1}{2} + \frac{1}{4} + \frac{1}{2}$$ $$\Rightarrow C' = \frac{4}{5} \, \mu F$$ $C'$ and $C_2$ are parallel, the total capacitance is $$C'' = \left( \frac{4}{5} + 0.2 \right) \, \mu F = 1 \, \mu F$$ $C_1$, $C''$, $C_6$ are in series. So the equivalent capacitance is, $$\frac{1}{C_{eq}} = \frac{1}{2} + 1 + \frac{1}{2}$$ $$\Rightarrow C_{eq} = 0.5 \, \mu F$$ Using $Q = C_{eq} V$. The charge through the battery, $Q = 10 \times \frac{1}{2} = 5 \, \mu C$ From the diagram above, $$Q' = \frac{5 \, \mu C \times 0.8}{0.8 + 0.2} = 4 \, \mu C$$

The kinetic energy is given by $$E = \frac{1}{2} I \omega^2 = \left( \frac{L^2}{2I} \right)$$ It is given that the new moment of inertia $$I' = 3I$$ Clearly, the final kinetic energy is $$E_f = \frac{L^2}{2I'} = \frac{L^2}{6I}$$ $$\Rightarrow E_f = \left( \frac{E}{3} \right)$$

Using the law of Momentum conservation $$m_1 v_1 + m_2 (-v_2) = 0$$ Thus, $$m_1 v_1 = m_2 v_2$$ Therefore, $$\left( \frac{m_1}{m_2} \right) = \left( \frac{v_2}{v_1} \right)$$ Substituting, $$\frac{4 \pi r_1^3}{3} = \frac{4 \pi r_2^3}{3}$$ Thus, $$\left( \frac{r_1}{r_2} \right)^3 = \left( \frac{v_2}{v_1} \right)$$ It is given $$\frac{r_1}{r_2} = \frac{1}{2^{\frac{1}{3}}}$$ and $$\left( \frac{r_1}{r_2} \right)^3 = \left( \frac{v_2}{v_1} \right)$$ Therefore, $$\left( \frac{1}{2} \right) = \left( \frac{v_2}{v_1} \right)$$ Thus, $$\frac{v_1}{v_2} = \frac{2}{1}$$

Let $R_1 = 10 \, \Omega$. Using $V = iR$ $$i = \left( \frac{3}{10 + 2r} \right)$$ The voltmeter reading is $V' = 1.5 \, \mathrm{V}$. Thus, $$iR_1 = V'$$ $$\Rightarrow 10 \times \frac{3}{10 + 2r} = 1.5$$ $$\Rightarrow 20 = 10 + 2r$$ $$\Rightarrow 2r = 10$$ $$\Rightarrow r = 5 \, \Omega$$

Let's consider the following free-body diagram: From the above diagram, it can be written that $$F - mg \sin 30^\circ = ma \ldots (1)$$ Substitute the values of the known parameters into equation (1) and solve to calculate the force applied on the block. $$F - 50 \sin 30^\circ = 5 \times 1$$ $$\Rightarrow F = (5 + 25) \, \mathrm{N}$$ $$= 30 \, \mathrm{N}$$ The velocity of the block $(v)$ at $t = 10 \, \mathrm{s}$ can be written as $$v = 0 + 1 \, \mathrm{m/s^2} \times 10 \, \mathrm{s}$$ $$= 10 \, \mathrm{m/s}$$ Hence, the required power is given by $$P = Fv$$ $$= 30 \, \mathrm{N} \times 10 \, \mathrm{m/s}$$ $$= 300 \, \mathrm{W}$$

The data given is $L = 2 \, \mathrm{H}$ $R = 4 \, \Omega$ $V = 10 \, \mathrm{V}$ Using Ohm's law, $V = IR$, $$I = \frac{10 \, \mathrm{V}}{4 \, \Omega} = \frac{5}{2} \, \mathrm{A}$$ The energy of an inductor is given by $$E = \frac{1}{2} LI^2 = \frac{1}{2} \times 2 \times \left( \frac{25}{4} \right) = 625 \times 10^{-2} \, \mathrm{J}$$

Using the equation $$qE = qvB$$ $$\Rightarrow E = vB$$ Using the relation $\Delta V = Ed$ $$\Rightarrow \Delta V = vBd$$ $$= 2 \times \frac{1}{2} \times 0.15 \, \mathrm{V}$$ $$= 0.15 \, \mathrm{V} = 150 \, \mathrm{mV}$$

Using the relation of Young's modulus, $$\frac{T}{A} = Y \left( \frac{\Delta L}{L} \right)$$ which implies $$T = \left( \frac{Y \Delta L}{L} \times A \right).$$ The linear mass density is $$\mu = \left( \frac{m}{L} \right).$$ So, $$\frac{T}{\mu} = \frac{Y \Delta L A}{L \left( \frac{m}{L} \right)} = \frac{Y (\Delta L) \times L A}{L (m)} = \left( \frac{Y \Delta L}{L} \right) \times \left( \frac{1}{\rho} \right).$$ Substituting the values, $$\frac{T}{\mu} = \frac{8 \times 10^{10} \times 3.2 \times 10^{-4}}{0.5} \times \left( \frac{1}{8 \times 10^3} \right) = 6.4 \times 10^3$$ which implies $$\frac{T}{\mu} = 64 \times 10^2.$$ The fundamental frequency is given by $$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}.$$ Therefore, $$\sqrt{\frac{T}{\mu}} = 8 \times 10 = 80 \, \mathrm{m \, s^{-1}}.$$ Therefore, $$f = \left( \frac{80}{1} \right) = 80 \, \mathrm{Hz}.$$

The amount of work done is equal to the change in the surface energy to increase the radius of the bubble. The formula to calculate the change in surface energy is given by $$\Delta U = 2(T \times \Delta A) \ldots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the required work done. $$\Delta U = 3.5 \times 10^{-2} \times 4 \pi \left[ (0.2)^2 - (0.1)^2 \right] \times 2 \, \mathrm{J}$$ $$= 3.5 \times 10^{-2} \times 4 \times \frac{22}{7} \times 0.03 \times 2$$ $$= 2 \times 1.32 \times 10^{-2}$$ $$= 264 \times 10^{-4} \, \mathrm{J}$$

The data given is $d = 8 \, \mathrm{cm}$ $\mu = \frac{4}{3}$ Apparent depth is $$h' = \frac{d}{\mu} = \frac{8}{\frac{4}{3}} = 6 \, \mathrm{cm}$$ Distance of object from mirror $= 42 + 6 = 48 \, \mathrm{m}$ Distance of image from bottom of tank $= 48 + 50 = 98 \, \mathrm{m}$

BeH$_2$ is indeed considered to be less stable compared to other hydrides among the given due to its significant covalent character and electron deficiency. Beryllium, being a small and highly charged cation, attracts the electrons in the Be–H bonds more strongly, resulting in a greater degree of electron sharing between beryllium and hydrogen. This covalent character is in contrast to other alkaline earth metal hydrides, which tend to have more ionic character.

The reagent hydrazine in alkaline medium is used in the Wolff-Kishner reduction reaction. In this reaction aldehydes and ketone converts to respective alkanes. This reaction is not good for base sensitive groups in the carbonyl compounds. The following chloro carbonyl compound gives alkene as elimination reaction expected in the presence of base.

The carbonyl compounds undergo Clemmensen's reduction in the presence of zinc amalgam and concentrated hydrochloric acid to form alkanes. The reagent in the Clemmensen reduction is not compatible with acid-sensitive functional groups. In the given reactant, the hydroxy group undergoes dehydration to give alkene as shown below.

$\mathrm{C_2Cl_2F_2}$ is an example of freon. It can act as an aerosol propellant, refrigerant, or organic solvent because it is a gas at room temperature and a liquid when cooled or compressed. Freon is a term that was commonly used to refer to a group of chlorofluorocarbon (CFC) and hydrochlorofluorocarbon (HCFC) compounds.

To determine the values of $x$ and $y$, we can use the information provided and the rate expression for the given reaction, which is first order with respect to both A and B: Rate $r = K[A]^1[B]^1$ $$0.1 = K(20)^1(0.5)^1 \cdots (i)$$ $$0.40 = K(x)^1(0.5)^1 \cdots (ii)$$ $$0.80 = K(40)^1(y)^1 \cdots (iii)$$ From (i) and (ii) $$x = 80$$ From (i) and (iii) $$y = 2$$

The threshold frequency is the minimum frequency required for emission of electrons from the outermost shell. The light with frequencies lower than threshold would not eject photoelectrons from the shell of an atom. As for the emission kinetic energy of an electron should be greater than or equal to zero. Hence, the photoelectric emission will take place only when the energy of a photon is greater than the energy possessed by the electron. Atoms are ionised resulting in transfer of energy when photons of sufficient energy strike the metal surface.

When glycine is treated with chlorine in presence of red phosphorus, HVZ reaction takes place i.e., substitution of chlorine takes place at alpha carbon. $$\mathrm{H_2N - CH_2 - COOH \xrightarrow{Red P \atop Cl_2} H_2N - \overset{*}{CH} - COOH \atop (HVZ)}$$ Optically active A molecule with two chiral carbons is not always optically active. Suppose the following compound is having two chiral carbons but optically inactive due to internal compensation of optical rotation.

The alcohol is reacted with a mixture of sodium iodide and concentrated phosphoric acid, $\mathrm{H_3PO_4}$, and the iodoalkane is distilled off. The 2-iodobutane formed in the previous reaction gives Grignard reagent with magnesium metal in dry ether. The Grignard reagent formed reacts with $\mathrm{D_2O}$ to give 2-deuterobutane.

Low Density Polyethylene is obtained by the polymerisation of ethene under high pressure of $1000 - 2000 \, \mathrm{atm}$ at $350 - 570 \, \mathrm{K}$ in the presence of an initiator. High Density Polyethylene is obtained when polymerisation is done in the presence of Ziegler-Natta catalyst at $333 - 343 \, \mathrm{K}$ under $6 - 7 \, \mathrm{atm}$ pressure. Nylon-6 is manufactured from the monomer called caprolactam. The monomer caprolactam is heated at $533 - 543 \, \mathrm{K}$ in an inert atmosphere, it polymerises to give nylon-6.

Nitrous acid gives nitrosonium ion as electrophile ($\mathrm{NO^+}$) which attacks phenol at less hindered p-position of form p-nitrosophenol. The nitroso group undergoes reduction and forms amine in the presence of $\mathrm{NH_4SH}$.

SO$_2$ is a bent molecule with a central sulfur atom bonded to two oxygen atoms. The oxygen atoms are more electronegative than sulfur, causing a separation of charge. This results in a bent molecular geometry and a net dipole moment, making SO$_2$ a polar molecule. NH$_3$ is a trigonal pyramidal molecule with a central nitrogen atom bonded to three hydrogen atoms. Again, the electronegativity difference between nitrogen and hydrogen leads to a separation of charge. The arrangement of the atoms in NH$_3$ gives it a net dipole moment, making it a polar molecule.

Thionyl chloride reacts with white phosphorus to produce phosphorus trichloride. The other side products in this reaction are sulphur dioxide and sulphur monochloride. $$\mathrm{P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2}$$ Hence $A = \mathrm{PCl_3}$, $x = 4$, $B = \mathrm{S_2Cl_2}$.

Compounds which result in the formation of stable carbocation upon treating with $\mathrm{AgNO_3}$ will produce precipitate. $$Br \xrightarrow{\mathrm{AgNO_3}} Aromatic$$ $$Br \xrightarrow{\mathrm{AgNO_3}} Unstable$$ $$Br \xrightarrow{\mathrm{AgNO_3}} Benzylic$$ $$CH = CH - Br \xrightarrow{\mathrm{AgNO_3}} Allylic$$ In the above carbocations, the second carbocation is unstable, hence, it cannot give precipitate with silver nitrate.

To calculate the mass percent of "X" in the solution, we need to determine the mass of "X" and the total mass of the solution. Given: Solute (X) = 2 $\mathrm{g}$ Solvent ($\mathrm{H_2O}$) = 1 mole = 18 $\mathrm{g}$ Total mass = 2 + 18 = 20 $\mathrm{g}$ Mass percent of X = $\frac{2}{2+18}$ $\times$ 100 = 10

Both Statement I and Statement II are correct. In many metallurgical processes, sulphide ores are first converted to oxide ores through a process known as roasting or calcination. The conversion of sulphide ore to oxide ore is done to remove the sulphur and simplify the subsequent reduction process. Oxide ores are generally easier to reduce compared to sulphide ores. This is because oxygen is already bonded to the metal in oxide ores, making it easier to remove the oxygen and extract the metal through reduction reactions.

On moving down the group in alkali metals, the melting point generally decreases. As you move down the alkali metal group (Group 1 of the periodic table), the atomic size or atomic radius increases. This increase in atomic size leads to a decrease in the strength of metallic bonds between the alkali metal atoms. Weaker metallic bonding results in lower melting points because less energy is required to overcome these weaker bonds and convert the solid metal into a liquid state. On moving down the group in alkali metals, the melting point generally decreases. As you move down the alkali metal group (Group 1 of the periodic table), the atomic size or atomic radius increases. This increase in atomic size leads to a decrease in the strength of metallic bonds between the alkali metal atoms. Weaker metallic bonding results in lower melting points because less energy is required to overcome these weaker bonds and convert the solid metal into a liquid state. \[ \begin{array}{c|c} \text{Element} & \text{M.P. (K)} \\ \hline \mathrm{Na} & 371 \\ \mathrm{K} & 336 \\ \mathrm{Rb} & 312 \\ \mathrm{Cs} & 302 \end{array} \]

Sol. Using the formula $$\frac{P^o - P_s}{P^o} = \frac{Moles of solute}{Moles of solvent}$$ (for dilute solution) where $P^o$ is the vapour pressure of pure water and $P_s$ is the vapour pressure of the solution. $$\frac{54.2 - 54}{54.2} = \frac{w/180}{100/18}$$ $$w = \frac{0.2 \times 1000}{54.2}$$ $$= \frac{200}{54.2} = 3.69$$

The spin only magnetic moment can be calculated as follows $$\mu = \sqrt{n(n+2)} BM$$ $n =$ Number of unpaired electrons $$[Fe (H_2O)_6]^{3+} = d^5 (High spin complex) = 5 unpaired electrons$$ $$\mu = \sqrt{5 \times 7} = \sqrt{35}$$ $$= 5.92 B.M.$$ $$[Fe (CN)_6]^{3+} = d^5 (Low spin complex) = 1$$ $$\mu = \sqrt{1 \times 3} = \sqrt{3}$$ $$= 1.732 B.M.$$

The formula $\mathrm{Mg(NH_4)PO_4}$ represents magnesium ammonium phosphate; it is white in colour. $\mathrm{K_3[Co(NO_2)_6]}$ is known as potassium hexanitritocobaltate(III). It is yellow in colour. The yellow colour is due to the presence of the cobalt ion ($\mathrm{Co^{3+}}$) in the complex, which absorbs certain wavelengths of light and reflects yellow light. $\mathrm{MnO(OH)_2}$ is brown in colour. $\mathrm{Fe_4[Fe(CN)_6]_3}$ is blue in colour.

When $\mathrm{Ni^{2+}}$ is replaced by $\mathrm{Pt^{2+}}$ in the complex $[\mathrm{NiCl_2 \, Br_2}]^{2-}$, the following properties are expected to be changed: A. Geometry: Since $\mathrm{Pt^{2+}}$ and $\mathrm{Ni^{2+}}$ have different coordination preferences and atomic sizes, the replacement is likely to result in a change in the geometry of the complex. That is it changes to square planar. B. Geometrical isomerism: If the replacement of $\mathrm{Ni^{2+}}$ with $\mathrm{Pt^{2+}}$ leads to a change in the ligand arrangement, it can result in the generation of cis and trans isomers. C. Magnetic properties: $[\mathrm{PtCl_2 \, Br_2}]^{2-}$ is diamagnetic. Both the complex species are optically inactive.

Except sucrose and amylose all others, give orange red precipitate with Benedict solution. Sucrose is a non-reducing disaccharide composed of glucose and fructose. It does not have a free aldehyde or ketone group and, therefore, cannot undergo oxidation and reduce the Benedict's reagent. As a result, sucrose will not produce an orange-red precipitate with the Benedict's solution. Amylose is a polysaccharide composed of multiple glucose units linked together. While glucose, a monosaccharide, is a reducing sugar, the glycosidic bonds in amylose prevent the free aldehyde or ketone group from being available for oxidation. As a result, amylose, as a part of starch, does not give a positive Benedict's test and will not produce an orange-red precipitate.

The equilibrium moles of each reactant and product can be calculated as follows, $$\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$$ Initial: $$\begin{array}{ccc} 4.5 & 4.5 & 0 \\ \end{array}$$ At equilibrium: $$\begin{array}{ccc} 4.5 - 1.5 & 4.5 - 1.5 & 3 \\ \end{array}$$ Now, using equilibrium moles, equilibrium constant calculated as shown below. $$K_{eq} = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}] [\mathrm{I_2}]} = \frac{(3)^2}{3 \times 3} = 1$$

Adsorption theory of Heterogeneous catalysis involves five steps- 1. Diffusion of reactants to the surface of the catalyst. 2. Adsorption of reactant molecules on the surface of the catalyst. (Occurrence of chemical reaction on the catalyst's surface through formation of an intermediate.) 3. Desorption of reaction products from the catalyst surface. 4. Making the surface available again for more reaction to occur. 5. Diffusion of reaction products away from the catalyst's surface. So, we can deduce from Adsorption theory of catalysis that as the concentration of reactants at active centres increases due to adsorption and thus the rate of the reaction is also increased.

Probability density curves for 1s, 2s, and 2p orbitals are given below. For 1s orbital, the probability density is maximum at the nucleus. For 2s orbital, the probability density first decreases and then increases. At any distance from the nucleus, the probability density of finding an electron is never zero, and it always has some finite value. The number of angular nodes is equal to the azimuthal quantum number. For p-orbital, it is equal to 1, and for d-orbital, it is equal to 2. Probability density of p-orbital is zero at the nucleus.

3-chloro-1-butene undergo protonation and gives secondary carbocation according Markonikov's addition in the first step. In the second step the lone pair on the chlorine attacks the positively charged carbon to form cyclic intermediate. Now the nucleophile in the reaction i.e., chloride ion attacks cyclic intermediate and forms two structural products as shown below. 2,2-dichlorobutane exit in single isomer only, but 2,3-dichlorobutane exhibits three isomers.

Magnesium nitrate crystallises with six molecules of water, whereas barium nitrate is anhydrous salts. The hydrated form of magnesium nitrate is $\mathrm{Mg(NO_3)_2} \cdot 6\mathrm{H_2O}$ and anhydrous salt of barium nitrate is $\mathrm{Ba(NO_3)_2}$ respectively. Therefore, $x = 6$, $y = 0$.

An intensive property is a physical quantity whose value does not depend on the amount of substance which was measured. Molar heat capacity, molarity, $E^0_{cell}$ and molar mass are intensive properties. An extensive property of a system depends on the system size or the amount of matter in the system. Volume, Gibbs free energy change, Mole are extensive properties.

$\mathrm{ClO_3^-}$: The central atom is Cl. There is one lone pair of electrons on the central atom. $\mathrm{XeF_4}$: The central atom is Xe. It has two lone pairs on the central atom. $\mathrm{SF_4}$: The central atom is S. It contains one lone pair on the central atom. $\mathrm{I_3^-}$: The central atom is I. It contains three lone pairs on the central atom.

The stoichiometric equation of the reaction between magnesium and hydrogen chloride can be written as follows. One mole magnesium can liberate one mole of hydrogen gas according to the following equation. $$\mathrm{Mg} + 2\mathrm{HCl} \rightarrow \mathrm{MgCl_2} + \mathrm{H_2}$$ $$2.4 \, \mathrm{g} = 0.1 \, mole 0.1 \, mole$$ Volume of 1 mole $\mathrm{H_2}$ at STP = 22.4 L Therefore, 0.1 mole $\mathrm{H_2}$ at STP will occupy = 2.24 L

An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system. $$\Delta G = -nF E_{cell}$$ The $$\Delta G$$ in the above equation is an extensive property, $$E_{cell}$$ is an intensive property. A negative $$E^0$$ means that the redox couple is a stronger reducing agent than the $$\mathrm{H^+/H_2}$$ couple. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.