JEE Main 15 April 2023 Shift 1 question paper with solutions

Given, three-digit number to be formed which are divisible by 3 using the digits 1, 3, 5, 8 and repetition is allowed. Now taking case (1), where all digits are same, we get $(1, 1, 1)$, $(3, 3, 3)$, $(5, 5, 5)$, $(8, 8, 8) \rightarrow 4$ ways. Now taking case (2), where 2-digit are same and one is distinct, we get $(5, 5, 8) \rightarrow \frac{3!}{2!} = 3$ ways, $(8, 8, 5) \rightarrow 3$ ways. Now taking case (3), where all are distinct, we get $(1, 3, 5) \rightarrow 6$ ways, $(1, 8, 3) \rightarrow 6$ ways. So, total ways will be $4 + 3 + 3 + 6 + 6 = 22$.

The given lines are $\frac{x - \lambda}{0} = \frac{y - 3}{4} = \frac{z + 6}{1}$ and $\frac{x + \lambda}{3} = \frac{y}{-4} = \frac{z - 6}{0}$. We know that the shortest distance between two lines is $d = \frac{\left| (\vec{a_2} - \vec{a_1}) \cdot (\vec{n_1} \times \vec{n_2}) \right|}{|\vec{n_1} \times \vec{n_2}|}$ and $d = 13$ (given). $$\Rightarrow \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & 1 \\ 3 & -4 & 0 \end{vmatrix} = 4\hat{i} + 3\hat{j} - 12\hat{k}$$ Now using the above formula of distance we get, $$\Rightarrow \frac{\left| (2\lambda \hat{i} + 3\hat{j} - 12\hat{k}) \cdot (4\hat{i} + 3\hat{j} - 12\hat{k}) \right|}{\sqrt{16 + 9 + 144}} = 13$$ $$\Rightarrow \frac{\left| (2\lambda \hat{i} + 3\hat{j} - 12\hat{k}) \cdot (4\hat{i} + 3\hat{j} - 12\hat{k}) \right|}{13} = 13$$ $$\Rightarrow |8\lambda + 9 + 144| = 169$$ $$\Rightarrow |8\lambda + 153| = 169$$ $$\Rightarrow 8\lambda = \pm 169 - 153$$ $$\Rightarrow \lambda = \frac{16}{8}, \frac{-322}{8}$$ $$\Rightarrow 8\left| \sum_{\lambda \in S} \lambda \right| = 8 \left| \frac{16}{8} - \frac{322}{8} \right|=306$$ Hence this is the correct option.

Incorrect mean is $$\frac{\sum_{i=1}^{10} x_i}{10} = 20$$ which implies $$\sum_{i=1}^{10} x_i = 200 \cdots (i)$$ So, corrected mean is $$\frac{\sum_{i=1}^{10} x_i - 50 + 40}{10} = \frac{200 - 10}{10} = 19$$ Incorrect variance is $$\frac{\sum_{i=1}^{10} x_i^2}{10} - 400 = 64$$ which implies $$\sum_{i=1}^{10} x_i^2 = 4640$$ Now, corrected variance is $$= \frac{\sum_{i=1}^{9} x_i^2 + 40^2 - 50^2}{10} - (19)^2$$ $$= \frac{4640 - 900}{10} - (19)^2$$ $$= 374 - 361$$ $$= 13$$

Given, $\($ABCD$\)$ be a quadrilateral. If $\($E$\)$ and $\($F$\)$ are the midpoints of the diagonals $\($AC$\)$ and $\($BD$\)$ respectively and $\($$\left$( $\overrightarrow{AB}$ - $\overrightarrow{BC}$ $\right$) + $\left$( $\overrightarrow{AD}$ - $\overrightarrow{DC}$ $\right$) = k $\overrightarrow{FE}$$\)$, Now plotting the diagram we get, Now let position vector of $\($A, B, C$\)$ and $\($D$\)$ are $\($$\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$$\)$ and $\($$\overrightarrow{d}$$\)$ respectively. Now by midpoint formula, Position vector of $\($E = $\frac{\overrightarrow{OC} + \overrightarrow{OA}}{2}$ = $\frac{\overrightarrow{c} + \overrightarrow{a}}{2}$$\)$ And Position vector of $\($F = $\frac{\overrightarrow{b} + \overrightarrow{d}}{2}$$\)$ Now, $\($$\left$( $\overrightarrow{AB}$ - $\overrightarrow{BC}$ $\right$) + $\overrightarrow{AD}$ - $\overrightarrow{DC}$$\)$ $\($$\Rightarrow$ $\overrightarrow{b}$ - $\overrightarrow{a}$ - $\left$( $\overrightarrow{c}$ - $\overrightarrow{b}$ $\right$) + $\overrightarrow{d}$ - $\overrightarrow{a}$ - $\left$( $\overrightarrow{c}$ - $\overrightarrow{d}$ $\right$)$\)$ $\($$\Rightarrow$ 2 $\overrightarrow{b}$ - 2 $\overrightarrow{a}$ - 2 $\overrightarrow{c}$ + 2 $\overrightarrow{d}$$\)$ $\($$\Rightarrow$ 2 $\left$( $\overrightarrow{b}$ + $\overrightarrow{d}$ $\right$) - 2 $\left$( $\overrightarrow{a}$ + $\overrightarrow{c}$ $\right$)$\)$ $\($$\Rightarrow$ 4 $\left$[ $\frac{\overrightarrow{b} + \overrightarrow{d}}{2}$ - $\frac{\overrightarrow{a} + \overrightarrow{c}}{2}$ $\right$] = 4 $\left$[ $\overrightarrow{OF}$ - $\overrightarrow{OE}$ $\right$]$\)$ $\($$\Rightarrow$ 4 $\overrightarrow{EF}$ = -4 $\overrightarrow{FE}$$\)$ $\($$\therefore$ k = -4$\)$

Given, $$2(y+2)\log_e(y+2)dx+(x+4-2\log_e(y+2))dy=0$$ Let $x+4=u$, $y+2=v$ $$dx=du, \; dy=dv$$ So, the equation becomes, $$(2v \ln v)du = -(u-2 \ln v)dv$$ $$\Rightarrow 2v \ln v \frac{du}{dv} + u = 2 \ln v$$ $$\Rightarrow \frac{du}{dv} + \frac{1}{2v \ln v} \cdot u = \frac{1}{v}$$ Which is a linear differential equation, So, $IF = e^{\int \frac{1}{v \ln v} = e^{\frac{1}{2} \ln (\ln v)} = (\ln v)^{\frac{1}{2}}}$ Now solution of differential equation is given by, $$u \cdot (\ln v)^{\frac{1}{2}} = \int \frac{1}{v} \cdot (\ln v)^{\frac{1}{2}} dv$$ $$\Rightarrow u \cdot (\ln v)^{\frac{1}{2}} = \frac{2}{3} (\ln v)^{\frac{3}{2}} + c \ldots \ldots \ldots (i)$$ Now using given value, $y = e^4 - 2 \Rightarrow x = 1$ $$\therefore \; v = e^4 \Rightarrow u = 5$$ $$5 \cdot \left(\frac{1}{4}\right)^{\frac{1}{2}} = \frac{2}{3} \cdot (4)^{\frac{3}{2}} + c$$ $$\Rightarrow 10 = \frac{16}{3} + c$$ $$\Rightarrow c = \frac{14}{3}$$ Now finding, $y = e^9 - 2 \Rightarrow v = y + 2 = e^9$ Now putting the value in equation (i) we get, $$\Rightarrow u \cdot 3 = 2 \times 27 + \frac{14}{3} = 18 + \frac{14}{3}$$ $$\Rightarrow x + 4 = u = 6 + \frac{14}{9}$$ $$\Rightarrow x = 2 + \frac{14}{9} = \frac{32}{9}$$

Given, $$f(x) = \max \{ 1 + x + \lfloor x \rfloor,\; 2 + x,\; x + 2 \lfloor x \rfloor \},\; 0 \leq x \leq 2$$ $$\Rightarrow f(x) =$$ $\max \{ x + 1,\; x + 2,\; x \}$ , $0 \leq x < 1$ $\max \{ x + 2,\; x + 2,\; x + 2 \}$ , $1 \leq x < 2$ $\max \{ 5,\; 4,\; 6 \}$ , $x = 2$ $$\Rightarrow f(x) =$$ $x + 2$, $0 \leq x < 1$ $x + 2$, $1 \leq x < 2$ $6$, $x = 2$ $$\Rightarrow f(x) =$$ $x + 2$, $0 \leq x < 2$ $6$, $x = 2$ So, $f$ is not continuous at $x = 2$ as $f(2^-) = 4$ and $f(2) = 6$. And $f$ is differentiable in $(0, 2)$ as $x + 2$ is a linear function, $$\therefore\ m = 1,\; n = 0$$ $$\Rightarrow (m + n)^2 + 2 = 1 + 2 = 3$$ Hence this is the required option.

Given, $$x|x|-5|x+2|+6=0$$ Now taking case (1) when $x < -2$ we get, $$-x^2 + 5x + 10 + 6 = 0$$ $$\Rightarrow x^2 - 5x - 16 = 0$$ $$\Rightarrow x = \frac{5-\sqrt{89}}{2}, ignoring x = \frac{5+\sqrt{89}}{2} as x < -2, so one solution is possible here,$$ Now taking case (2) when $-2 \leq x < 0$ we get, $$-x^2 - 5x - 10 + 6 = 0$$ $$\Rightarrow x^2 + 5x + 4 = 0$$ $$\Rightarrow x = -1, ignoring x = -4 as -2 \leq x < 0$$ Now taking case (3), when $x \geq 0$ we get, $$x^2 - 5x - 10 + 6 = 0$$ $$\Rightarrow x^2 - 5x - 4 = 0$$ $$\Rightarrow x = \frac{5+\sqrt{41}}{2}, ignoring x = \frac{5-\sqrt{41}}{2} as x \geq 0$$ So, from all cases we get total 3 solutions.

Given that $(a + bx + cx^2)^{10} = \sum_{i=10}^{20} p_i x^i$, $a, b, c \in \mathbb{N}$. General term: $$\frac{10!}{r_1! \, r_2! \, r_3!} (a)^{r_1} (bx)^{r_2} (cx^2)^{r_3}$$ For coefficient of $x$: $r_2 + 2r_3 = 1$. Let $$\begin{array}{ccc} r_1 & r_2 & r_3 \\ 9 & 1 & 0 \end{array}$$ Therefore, coefficient of $x = \frac{10!}{9!} a^9 b^1 = 20$. $$\Rightarrow a^9 \cdot b = 2 \ldots \ldots (i)$$ Coefficient of $x^2$: $$\frac{10!}{8! \, 2! \, 0!} a^8 \cdot b^2 + \frac{10!}{9! \, 0! \, 1!} a^9 \cdot c = 210$$ $$\Rightarrow 45a^8 \cdot b^2 + 10 \cdot a^9 \cdot c = 210$$ $$\Rightarrow 9a^8 b^2 + 2a^9 \cdot c = 42 \ldots \ldots (ii)$$ As $a, b, c \in \mathbb{N}$, so by hit and trial in equation (i) $\&$ (ii) we get, $a = 1, b = 2, c = 3$. Hence, $2(a + b + c) = 2(3 + 2 + 1) = 12$.

We have been given that $|A| = m - n$, where $4m + n = 22 \ldots (i)$ and $17m + 4n = 93 \ldots (ii)$. Solving (i) and (ii) $m = 5$, $n = 2$. Order $= 5$. Therefore, $|A| = 3$. We know that $|adj A| = |A|^{w-1} |adj (adj A)| = |A|^{(w-1)^2}$ where $w \times w$ is the order of the matrix. Therefore, $\det (n adj (adj (mA))) = |2 adj (adj (5A))| = 2^5 |5A|^{16} = 2^5 5^{80} |A|^{16} = 2^5 \cdot 3^{16} \cdot 5^{80} = 3^{11} 5^{80} 6^5$. So, $a + b + c = 96$. Hence this is the required option.

Let the two numbers are $a$, $b$. $a$, $A_1$, $A_2$, $b$ are in AP. $b = a + (4 - 1)d$ $d = \frac{b-a}{3}$ $A_1 = a + \frac{b-a}{3} = \frac{2a+b}{3}$ $A_2 = a + \frac{b-a}{3} \cdot 2 = \frac{a+2b}{3}$ Similarly $a$, $G_1$, $G_2$, $G_3$, $b$ are in GP. $b = a(r)^{5-1}$ $r = \left(\frac{b}{a}\right)^{\frac{1}{4}}$ $G_1 = a\left(\frac{b}{a}\right)^{\frac{1}{4}}$ $G_2 = a\left(\frac{b}{a}\right)^{\frac{2}{4}}$ $G_3 = a\left(\frac{b}{a}\right)^{\frac{3}{4}}$ $(G_1)^4 + (G_2)^4 + (G_3)^4 + (G_1)^2 \cdot (G_3)^2$ $= a^4 \cdot \frac{b}{a} + a^4 \cdot \frac{b^2}{a^2} + a^4 \cdot \frac{b^3}{a^3} + a^4 \cdot \frac{b^2}{a^2}$ $= ba^3 + b^2a^2 + b^3a + a^2b^2$ $= ab(a^2 + b^2 + 2ab) = ab(a+b)^2$ $(A_1 + A_2)^2 \cdot G_1 G_3 = (a+b)^2 \cdot ab$ Hence this is the correct option.

Given that the set $\left\{ \Re \left( \frac{z - \bar{z} + z^2}{2 - 3z + 5\bar{z}} \right) : z \in \mathbb{C}, \ \Re z = 3 \right\}$ is equal to the interval $(\alpha, \beta)$. Let $z = x + iy$. $$= \Re \left( \frac{x + iy - (x - iy) + x^2 + y^2}{2 - 3(x + iy) + 5(x - iy)} \right)$$ $$= \Re \left( \frac{x^2 + y^2 + i(2y)}{2 + 2x - 8iy} \right)$$ $$= \Re \left( \frac{(x^2 + y^2 + 2yi)(2(1 + x) + 8iy)}{(2(1 + x))^2 + (8y)^2} \right)$$ $$= \frac{2(x^2 + y^2)(1 + x) - 16y^2}{4(1 + x)^2 + (8y)^2}$$ Now using, $\Re(z) = 3 \Rightarrow x = 3$ we get, $$\frac{8(9 + y^2) - 16y^2}{64 + 64y^2}$$ $$\Rightarrow f(y) = \frac{1}{8} \left( \frac{9 - y^2}{1 + y^2} \right)$$ Let $t = \frac{1}{8} \cdot \frac{1}{1 + y^2}$ $$\Rightarrow 8t + 8ty^2 = 9 - y^2$$ $$\Rightarrow y^2(8t + 1) + 8t - 9 = 0$$ Now $D \geq 0$ $$\Rightarrow 0^2 - 4(8t + 1)(8t - 9) \geq 0$$ $$\Rightarrow \left( t - \left( -\frac{1}{8} \right) \right) \left( t - \frac{9}{8} \right) \leq 0$$ $$\Rightarrow t \in \left( -\frac{1}{8}, \frac{9}{8} \right) as t \neq -\frac{1}{8}$$ Range of $f(y) = (-0.125, 1.125)$ $$\Rightarrow \alpha = -0.125$$ $$\Rightarrow \beta = 1.125$$ $$\Rightarrow \beta - \alpha = 1.25$$ $$\Rightarrow 24(\beta - \alpha) = 30$$ Hence this is the required solution.

If $S = x^2 + y^2 + 2gx + 2fy + c = 0$ $C = (-g,\,-f)$ and $r = \sqrt{g^2 + f^2 - c}$ The given circles are $x^2 + y^2 - 18x - 15y + 131 = 0$ $\Rightarrow C_1(9,\; \frac{15}{2}),\quad r_1 = \sqrt{81 + \frac{225}{4} - 131} = \frac{3}{2}$ and $x^2 + y^2 - 6x - 6y - 7 = 0$ $\Rightarrow C_2(3,\; 3),\quad r_2 = \sqrt{9 + 9 + 7} = 5$ $$d = C_1C_2 = \sqrt{(9-3)^2 + \left(\frac{15}{2}-3\right)^2} = \sqrt{36 + \frac{81}{4}} = \frac{15}{2}$$ $$r_1 + r_2 = \frac{3}{2} + 5 = \frac{13}{2}$$ $$\therefore C_1C_2 = r_1 + r_2$$ $\therefore$ Circles touch externally, 3 common tangents. Hence this is the correct option.

Given, Expression $p \land (q \land \sim (p \land q))$ So, negation of the above expression will be, $$\sim [p \land (q \land \sim (p \land q))]$$ Now using demorgan's law $\sim (A \land B) \equiv \sim A \lor \sim B$ we get, $$\sim [p \land (q \land \sim (p \land q))]$$ $$\equiv \sim [p \land (q \land (\sim p \lor \sim q))]$$ $$\equiv \sim [p \land ((q \land \sim p) \lor (q \land \sim q))]$$ $$\equiv \sim [p \land ((q \land \sim p) \lor F)]$$ $$\equiv \sim [p \land (q \land \sim p)]$$ $$\equiv \sim p \lor \sim (q \land \sim p)$$ $$\equiv \sim p \lor (\sim q \lor p)$$ $$\equiv (\sim p \lor \sim q) \lor p$$ $$\equiv p \lor \sim q \lor p$$

Given, System of linear equation, $$-x + 2y - 9z = 7 \ldots \ldots (i)$$ $$-x + 3y + 7z = 9 \ldots \ldots (ii)$$ $$-2x + y + 5z = 8 \ldots \ldots (iii)$$ $$-3x + y + 13z = \lambda \ldots \ldots (iv)$$ Now solving, equations (i), (ii) & (iii) we get, $$x = -3, \ y = 2, \ z = 0$$ Now substituting in equation (iv) we get, $$3 \times 3 + 2 = \lambda$$ $$\Rightarrow \lambda = 11$$ Now finding the distance of Point $(-3, 2, 0)$ from the plane $2x - 2y + z = 11$ we get, $$d = \left| \frac{-6 - 4 - 11}{\sqrt{2^2 + 2^2 + 1}} \right| = \left| \frac{21}{3} \right| = 7$$

The required diagram will be: Equation of $AD$: $y + 7 = \frac{-5}{3}(x - 3)$ $$\Rightarrow 3y + 21 = -5x + 15$$ $$\Rightarrow 5x + 3y + 6 = 0 \ldots \ldots (i)$$ Equation of $BE$: $y - 2 = \frac{-1}{12}(x + 1)$ $$\Rightarrow 12y - 24 = -x - 1$$ $$\Rightarrow x = 23 - 12y \ldots (ii)$$ By (ii) $115 - 60y + 3y + 6 = 0$ $$\Rightarrow 57y = 121$$ $$y = \frac{121}{57}, x = 23 - 12 \times \frac{121}{57}$$ Therefore, $9\alpha - 6\beta + 60 = 9 \times 23 - 108 \times \frac{121}{57} - 6 \times \frac{121}{57} + 60$ $$= 207 - 242 + 60 = 25$$ Hence this is the required option.

Given that three points lie on a plane, $$\overrightarrow{AC} = 10\hat{i} + 4\hat{k}$$ $$\overrightarrow{AB} = 10\hat{i} + 5\hat{j} + 7\hat{k}$$ $$\overrightarrow{AC} \times \overrightarrow{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 10 & 0 & 4 \\ 10 & 5 & 7 \end{vmatrix}$$ $$\overrightarrow{AC} \times \overrightarrow{AB} = -20\hat{i} - 30\hat{j} + 50\hat{k}$$ $$\Rightarrow \overrightarrow{AC} \times \overrightarrow{AB} = 2\hat{i} + 3\hat{j} - 5\hat{k}$$ Equation of plane will be, $$\Rightarrow 2x + 3y - 5z = d$$ And the plane passing through, $(-1, -2, -3)$ $$\Rightarrow -2 - 6 + 15 = d$$ $$\Rightarrow d = 7$$ $$\therefore 2x + 3y - 5z = 7$$ Now finding the foot of perpendicular we get, $$\frac{x-3}{2} = \frac{y+2}{3} = \frac{z+9}{-5} = -\left(\frac{38}{38}\right)$$ $$x = 1, \ y = -5, \ z = -4$$ $Q(1, -5, -4)$ Distance from origin = $\sqrt{1^2 + 25 + 16} = \sqrt{42}$ Hence this is the required option.

Given, Bag have 6 white and 4 black balls. Now also given die is rolled and the number denotes the number of ball drawn from the bag. So, probability all drawn balls are white will be, $$= \frac{1}{6} \left[ \frac{{^6C_1}}{{^{10}C_1}} + \frac{{^6C_2}}{{^{10}C_2}} + \frac{{^6C_3}}{{^{10}C_3}} + \frac{{^6C_4}}{{^{10}C_4}} + \frac{{^6C_5}}{{^{10}C_5}} + \frac{{^6C_6}}{{^{10}C_6}} \right]$$ $$= \frac{1}{6} \left[ \frac{6}{10} + \frac{15}{45} + \frac{20}{120} + \frac{15}{210} + \frac{6}{252} + \frac{1}{210} \right]$$ $$= \frac{504}{2520} = \frac{1}{5}$$

Given, $$\int_0^1 \frac{1}{(5+2x-2x^2)(1+e^{1-4x})} \, dx = \frac{\alpha}{\beta} \log_e \left( \frac{\alpha+1}{\beta} \right)$$ Now let, $$I = \frac{1}{2} \int_0^1 \frac{1}{\left( 5 - (x^2 - x) \right) \left( 1 + e^{1-4x} \right)} \, dx$$ $$\Rightarrow I = \frac{1}{2} \int_0^1 \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2} \frac{1}{\left( 1 + e^{1-4x} \right)} \, dx$$ Now let $x = \frac{1}{2} + t \Rightarrow dx = dt$ So, integral becomes, $$I = \frac{1}{2} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \frac{1}{\left( 1 + e^{1-4t} \right)} \, dt$$ $$\Rightarrow I = \frac{1}{2} \int_0^{\frac{1}{2}} \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \frac{1}{\left( 1 + e^{1-4t} \right)} \, dt + \frac{1}{2} \int_0^{\frac{1}{2}} \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \frac{1}{\left( 1 + e^{4t-1} \right)} \, dt$$ Now solving $I_2 = \frac{1}{2} \int_0^{\frac{1}{2}} \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \frac{1}{\left( 1 + e^{4t-1} \right)} \, dt,$ Let $t = -z$, we get $dt = -dz$ So, $$I_2 = -\frac{1}{2} \int_{\frac{1}{2}}^0 \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - (-z)^2} \frac{1}{\left( 1 + e^{4(-z)-1} \right)} \, dz$$ $$\Rightarrow I_2 = \frac{1}{2} \int_0^{\frac{1}{2}} \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \frac{1}{\left( 1 + e^{1-4t} \right)} \, dt$$ Now putting the value of $I_2$ in $I$ we get, $$I = \frac{1}{2} \int_0^{\frac{1}{2}} \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \frac{1}{\left( 1 + e^{1-4t} \right)} \, dt + \left( \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \right) \frac{1}{\left( 1 + e^{1-4t} \right)} \, dt$$ $$\Rightarrow I = \int_0^{\frac{1}{2}} \frac{e^{4t}}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \frac{1}{(e^{4t}+1)} \, dt + \left( \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \right) \frac{1}{\left( 1 + e^{1-4t} \right)} \, dt$$ $$\Rightarrow I = \frac{1}{2} \int_0^{\frac{1}{2}} \frac{1}{\left( \frac{\sqrt{11}}{2} \right)^2 - t^2} \frac{1}{\left( 1 + e^{1-4t} \right)} \, dt$$ $$\Rightarrow I = \frac{1}{2} \times \frac{1}{\sqrt{11}} \left[ \ln \left| \frac{\sqrt{11}+1}{\sqrt{11}-1} \right| \right]_0^{\frac{1}{2}}$$ $$\Rightarrow I = \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{11}-1} \right) = \frac{1}{\sqrt{11}} \ln \left( \frac{(\sqrt{11})^2}{10} \right)$$ $$\Rightarrow I = \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{11}-1} \right)$$ Now on comparing with $\frac{\alpha}{\beta} \log_e \left( \frac{\alpha+1}{\beta} \right)$ we get, $$\Rightarrow \alpha = \sqrt{11}, \beta = \sqrt{10} \Rightarrow \alpha^4 - \beta^4 = 21$$

The given equations are coplanar. $$\begin{vmatrix} \lambda & -1 & 1 \\ 1 & 2 & \mu \\ 3 & -4 & 5 \end{vmatrix} = 0$$ $$\Rightarrow \lambda [10 + 4\mu] + 1[5 - 3\mu] + 1[-10] = 0$$ $$\Rightarrow 4\lambda \mu + 10\lambda - 3\mu = 5 \ldots \ldots (i)$$ And $\lambda - \mu = 5$ So, by (1) $$\Rightarrow 4(5 + \mu)\mu + 10(5 + \mu) - 3\mu = 5$$ $$\Rightarrow 4\mu^2 + 27\mu + 45 = 0$$ $$\Rightarrow 4\mu^2 + 15\mu + 12\mu + 45 = 0$$ $$\therefore \mu = -3, \frac{-15}{4}$$ $$\Rightarrow \lambda = 2, \frac{5}{4}$$ $$\therefore \sum_{\lambda, \mu \in S} 80(\lambda^2 + \mu^2) = 80 \left[ (9 + 4) + \frac{225}{16} + \frac{25}{16} \right]$$ $$= 80 \left[ 13 + \frac{125}{8} \right] = 10 \times 229$$ $$= 2290$$ Hence this is the correct option.

Given that $f(x) = \log(4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1}\left(\frac{10x + 6}{3}\right)$. We know that the domain of $A(x) + B(x) + C(x)$ is $D(A) \cap D(B) \cap D(C)$. Let us find the domain of $\log(4x^2 + 11x + 6)$. We know that domain of $\log(f(x))$ is $f(x) > 0$. $$4x^2 + 11x + 6 > 0$$ $$4x^2 + 8x + 3x + 6 > 0$$ $$(4x + 3)(x + 2) > 0$$ $$x \in (-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right).$$ Domain of $\log(4x^2 + 11x + 9)$ is $x \in (-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right)$. Let us find the domain of $\sin^{-1}(4x + 3)$ $$-1 \leq 4x + 3 \leq 1$$ $$-1 \leq x \leq -\frac{1}{2}$$ $$x \in \left[-1, -\frac{1}{2}\right]$$ Let us find the domain of $\cos^{-1}\left(\frac{10x+6}{3}\right)$. $$-1 \leq \frac{10x+6}{3} \leq 1$$ $$-3 \leq 10x + 6 \leq 3$$ $$-\frac{9}{10} \leq x \leq -\frac{3}{10}$$ Now the common domain is $$-\frac{3}{4} \leq x \leq -\frac{1}{2}.$$ $$x \in \left[-\frac{3}{4}, -\frac{1}{2}\right].$$ But given that the domain of $f(x)$ is $x \in [\alpha, \beta]$. $$\alpha = -\frac{3}{4}, \ \beta = -\frac{1}{2}.$$ $$\Rightarrow |36(\alpha + \beta)| = \left|36\left(-\frac{3}{4} + \left(-\frac{1}{2}\right)\right)\right|$$ $$\Rightarrow |36(\alpha + \beta)| = \left|36 \times \frac{-5}{4}\right|$$ $$\Rightarrow |36(\alpha + \beta)| = 45$$ Therefore, the required answer is 45.

Given, Series $\left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{2^2} - \frac{1}{2 \cdot 3} + \frac{1}{3^2} \right) + \left( \frac{1}{2^3} - \frac{1}{2^2 \cdot 3} + \frac{1}{2 \cdot 3^2} - \frac{1}{3^3} \right) + \left( \frac{1}{2^4} - \frac{1}{2^3 \cdot 3} + \frac{1}{2^2 \cdot 3^2} - \frac{1}{2 \cdot 3^3} + \frac{1}{3^4} \right) + \ldots$ Now let $a = \frac{1}{2}$, $b = \frac{1}{3}$ So, series will be: $(a - b) + (a^2 - ab + b^2) + (a^3 - a^2 b + ab^2 - b^3) + \ldots$ $$= \frac{1}{a+b} \left( (a^2 - b^2) + (a^3 + b^3) + (a^4 - b^4) + \ldots \right)$$ $$= \left( \frac{1}{a+b} \right) \left( a^2 + a^3 + a^4 \ldots - (b^2 - b^3 + b^4 \ldots) \right)$$ Now using infinite G.P. formula we get, $$= \left( \frac{1}{a+b} \right) \left( a^2 + a^3 + a^4 \ldots - (b^2 - b^3 + b^4 \ldots) \right)$$ $$= \left( \frac{1}{a+b} \right) \left( \frac{a^2}{1-a} - \frac{b^2}{1+b} \right)$$ Now putting the value of $a$ and $b$ we get, $$= \frac{1}{\frac{1}{2} + \frac{1}{3}} \left( \frac{1}{4} - \frac{1}{9} \right) \left( \frac{1}{1 - \frac{1}{2}} - \frac{1}{1 + \frac{1}{3}} \right)$$ $$= \frac{6}{5} \left( \frac{1}{2} - \frac{1}{12} \right)$$ $$= \frac{6}{5} \left( \frac{5}{12} \right) = \frac{1}{2} = \frac{\alpha}{\beta}$$ Hence, $\alpha + 3 \beta = 1 + 6 = 7$

\[ \lambda=7,\quad (a,b)\to(0,7)\ \text{or}\ (7,0), \] then \[ (c,d)\to(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \] and vice-versa, so \[ 2\times 12=24 \] numbers. \[ \lambda=8,\quad (a,b)\to(1,7)\ \text{or}\ (7,1), \] then \[ (c,d)\to(2,6),(3,5),(6,2),(5,3) \] and vice-versa, so total \[ 2\times 8=16 \] numbers. \[ \lambda=9,\quad (a,b)\to(2,7)\ \text{or}\ (7,2), \] then \[ (c,d)\to(3,6),(4,5),(5,4),(6,3) \] and vice-versa, so total \[ 2\times 8=16 \] numbers. \[ \lambda=10,\quad (a,b)\to(3,7),(7,3), \] so \[ (c,d)\to(6,4),(4,6) \] \[ \Rightarrow 8\ \text{numbers} \] \[ \lambda=11,\quad (a,b)\to(4,7),(7,4), \] so \[ (c,d)\to(6,5),(5,6) \] \[ \Rightarrow 8\ \text{numbers} \] \[ \lambda=12,13\ \&\ 14 \] are not possible. $\therefore$ Total numbers = 24+16+16+8+8 = 72.

Let the Plane containing the line $2x + y - 3 - 3 = 0 = 5x - 3y + 4z + 9$ is given by, $$2x + y - z - 3 + \lambda(5x - 3y + 4z + 9) = 0$$ $$\Rightarrow x(2 + 5\lambda) + y(1 - 3\lambda) + z(4\lambda - 1) + 9\lambda - 3 = 0$$ Also, this plane is parallel to the line $$\frac{x+2}{2} = \frac{3y-4}{-3} = \frac{z-5}{-7}$$ $$\Rightarrow \frac{1}{2} = \frac{-3}{-3} = \frac{-7}{5}$$ Now normal of the plane will be perpendicular to the line, so using perpendicular condition we get, $$\therefore (2 + 5\lambda)(2) + (1 - 3\lambda)(4) + (4\lambda - 1)5 = 0$$ $$\Rightarrow 4 + 10\lambda + 4 - 12\lambda + 20\lambda - 5 = 0$$ $$\Rightarrow 18\lambda = -3 \Rightarrow \lambda = -\frac{1}{6}$$ So, equation of the Plane will be, $$P : \frac{7}{6}x + \frac{3}{2}y - \frac{5}{2}z - \frac{9}{2} = 0$$ $$\Rightarrow 7x + 9y - 10z - 27 = 0$$ Now to find $AB$, Let $\frac{x-8}{-3} = \frac{y+1}{4} = \frac{z+19}{12} = k$ So, the point $B = (-3k + 8, 4k - 1, 12k - 19)$ And $B$ lies on plane $P$. $$7(-3k + 8) + 9(4k - 1) - 10(12k - 19) = 27$$ $$\Rightarrow -21k + 56 + 36k - 9 - 120k + 190 = 27$$ $$\Rightarrow -105k = -210$$ $$\Rightarrow k = 2$$ $$\therefore B = (2, 7, 5)$$ Hence, Distance is $AB = \sqrt{36 + 64 + 576} = \sqrt{676} = 26$ Hence this is the required answer.

Given, An ellipse with centre $(1, 0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along x-axis And its minor axis subtends an angle $60°$ at the foci. So, on plotting the diagram we get, Now from diagram we get, $\frac{b^2}{a} = \frac{1}{2} \Rightarrow b^2 = \frac{a}{2}$ ...(1) And $\tan 30° = \frac{b}{c}$ $\Rightarrow \left(\frac{1}{\sqrt{3}}\right) = \left(\frac{b}{c}\right)^2$ $\Rightarrow \frac{1}{3} = \frac{b^2}{c^2} = \frac{b^2}{a^2 - b^2}$ $\Rightarrow a^2 - b^2 = 3b^2$ $\Rightarrow b^2 = \frac{a^2}{4}$ .....(2) Now from equation (1) & (2) we get, $\Rightarrow a = 1, b^2 = \frac{1}{2}, b = \frac{1}{\sqrt{2}}$ Hence, $(2a + 2b)^2 = 9$

Given, Set $A = \{1, 2, 3, 4\}$ and relation on $A \times A$ as $2a + 3b = 4c + 5d$. Now, maximum value of $2a + 3b = 20$ at $(4, 4)$ and minimum value of $4c + 5d = 9$ at $(1, 1)$. So, $4c + 5d$ can be equal to $9, 13, 14, 17, 18, 19$. Now $2a + 3b$ can be $9$ if $(a, b) = (3, 1) \Rightarrow (c, d) = (1, 1)$. Similarly, $2a + 3b$ can be $13$, if $(a, b) = (2, 3) \Rightarrow (c, d) = (2, 1)$. And $2a + 3b$ can be $14$, if $(a, b) = (4, 2)$ or $(1, 4)$ $\Rightarrow$ (c, d) = (1, 2)$. Similarly, $2a + 3b$ can be $17$, if $(a, b) = (4, 3) $\Rightarrow$ (c, d) = (3, 1)$. And $2a + 3b$ can be $18$ if $(a, b) = (3, 4) $\Rightarrow$ (c, d) = (2, 2)$. Hence, there are total $6$ elements which satisfy the given relation.

Given that $n \in [10, 100]$ and $n \in \mathbb{N}$. Let us take the values of $n = 1, 2, 3, \ldots$ and check whether $3^n - 3$ is divisible by 7 or not. $$3^1 - 3 = 0$$ is divisible by 7 $$3^2 - 3 = 3, \; 3 is not divisible by 7$$ $$3^6 = 7k + 1 is not divisible by 7$$ $$3^7 = 7\alpha + 3,$$ $$3^7 - 3 = 7\alpha is divisible by 7$$ Similarly $$3^{13} = (7k + 1)(7\alpha + 3)$$ $$= 7\beta + 3$$ $$3^{13} - 3 = 7\beta is divisible by 7$$ If we observe the pattern for $n = 1, 7, 13, 3^n - 3$ is divisible by 7. That means the series forms an AP with $d = 6$. Also $n \in [10, 100]$ The required progression is $13, 19, \ldots, a_n$ $$a_n = 13 + (n - 1)6$$ But $13 + (n - 1)6 < 100$ $$n < 15.5$$ $$n = 15 \in \mathbb{N}$$ Therefore, the required answer is 15.

Given, the line $x = y = z$ intersects the line $x \sin A + y \sin B + z \sin C - 18 = 0 = x \sin 2A + y \sin 2B + z \sin 2C - 9$. So, let $x = y = z = k$. Now putting the value in $x \sin A + y \sin B + z \sin C = 18$ we get, $$k(\sin A + \sin B + \sin C) = 18$$ Now we know that, if $A, B \& C$ are angles of triangle then $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}$. $$\Rightarrow k \left( 4 \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2} \right) = 18 \ldots (i)$$ Also $k(\sin 2A + \sin 2A + \sin 2A) = 9$. And similarly $\sin 2A + \sin 2A + \sin 2A = 4 \sin A \cdot \sin B \cdot \sin C$. $$\Rightarrow k(4 \sin A \cdot \sin B \cdot \sin C) = 9 \ldots \ldots (ii)$$ Now dividing equation (i) $\&$ (ii) we get, $$8 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = \frac{9}{18}$$ $$\Rightarrow 80 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = 5$$

Given: $2y^2 = 3x$, $x + y = 3$ (slope is $-1$) and $(x - 3)^2 + y^2 = 2$. Now plotting the diagram of the above function we have, So, required area is $$A = \int_0^{\frac{3}{2}} \left(3 - y - \frac{2y^2}{3}\right) \, dy - \frac{\pi (\sqrt{2})^2 \times 45^\circ}{360^\circ}$$ $$\Rightarrow A = \left[3y - \frac{y^2}{2}\right]_0^{\frac{3}{2}} - \frac{2}{3} \left[y^3\right]_0^{\frac{3}{2}} - \frac{\pi}{4}$$ $$\Rightarrow A = \frac{21}{8} - \frac{\pi}{4}$$ So, $$A = \frac{1}{4} \left(\frac{21}{2} - \pi\right)$$ $$\Rightarrow \pi + 4A = \frac{21}{2}$$ $$\Rightarrow 4(\pi + 4A) = 42$$

Given, the triangles with vertices $A(2, 1)$, $B(0, 0)$ and $C(t, 4)$, $t = [0, 4]$. And the maximum and the minimum perimeters of such triangles are obtained at $t = \alpha$ and $t = \beta$ respectively. Now to minimise $CA + CB$, in the below diagram take the image of $B$ in $y = 4$. We get, $B' = (0, 8)$. Now finding the equation of $AB'$ we get, $$y - 8 = -\frac{7}{2}(x - 0)$$ Now, putting $y = 4$ in the above equation we get, $$-4 = -\frac{7}{2}(x)$$ $$\Rightarrow x = \frac{8}{7} \Rightarrow \beta = \frac{8}{7}$$ Now, the maximum perimeter will be possible if $\alpha = 0$ or $4$. Now taking $\alpha = 0$ we get, $$AB = \sqrt{5}, \ BC = 4 \ & \ AC = \sqrt{13}$$ Now when $\alpha = 4$ we get, $$AB = \sqrt{5}, \ BC = 4\sqrt{2} \ & \ AC = \sqrt{13}$$ Now on comparing the perimeter we get, the maximum perimeter at $\alpha = 4$. Hence, $6\alpha + 21\beta = 48$.

Given, $$f(x) = \int \frac{dx}{(3 + 4x^2) \sqrt{4 - 3x^2}}$$ Put $x = \frac{1}{t}$, $dx = -\frac{1}{t^2} dt$ So, $$f(x) = -\int \frac{-dt}{t^2 \left(3 + \frac{4}{t^2}\right) \sqrt{4 - \frac{3}{t^2}}}$$ $$\Rightarrow f(x) = \int \frac{1}{(3t^2 + 4) \sqrt{4t^2 - 3}} \, dt$$ Now let, $4t^2 - 3 = \lambda^2 \Rightarrow 8t \, dt = 2\lambda \, d\lambda$ $$\Rightarrow f(x) = -\int \frac{\lambda \, d\lambda}{4 \left(3 \left(\frac{\lambda^2 + 3}{4}\right) + 4\right) \lambda}$$ $$\Rightarrow f(x) = -\int \frac{\lambda \, d\lambda}{3\lambda^2 + 9 + 16}$$ $$\Rightarrow f(x) = -\int \frac{\lambda \, d\lambda}{\lambda^2 + 25}$$ $$\Rightarrow f(x) = -\frac{1}{3} \int \frac{\lambda \, d\lambda}{\lambda^2 + \left(\frac{25}{3}\right)}$$ $$\Rightarrow f(x) = -\frac{1}{3} \times \frac{\sqrt{3}}{5} \tan^{-1} \left(\frac{\sqrt{3}}{5}\right) + C$$ $$\Rightarrow f(x) = -\frac{\sqrt{3}}{15} \tan^{-1} \left(\frac{\sqrt{3}(4 - 3x^2)}{5x}\right) + C$$ Now using, $f(0) = 0 \Rightarrow c = + \frac{\sqrt{3}\pi}{30}$ Hence, $$f(1) = -\frac{\sqrt{3}}{15} \tan^{-1} \left(\frac{\sqrt{3}}{5}\right) + \frac{\sqrt{3}}{15} \times \frac{\pi}{2}$$ $$\Rightarrow f(1) = \frac{\sqrt{3}}{15} \left(\tan^{-1} \left(\frac{\sqrt{3}}{5}\right) - \frac{\pi}{2}\right)$$ $$\Rightarrow f(1) = \frac{\sqrt{3}}{15} \left(\frac{\pi}{2} - \tan^{-1} \left(\frac{\sqrt{3}}{5}\right)\right)$$ $$\Rightarrow f(1) = \frac{\sqrt{3}}{15} \tan^{-1} \left(\frac{\sqrt{3}}{5}\right)$$ $$\Rightarrow f(1) = \frac{1}{\sqrt{3}} \tan^{-1} \left(\frac{5}{\sqrt{3}}\right)$$ $$\Rightarrow f(1) = \frac{1}{\sqrt{3}} \tan^{-1} \left(\frac{5}{\sqrt{3}}\right)$$ Now comparing with given value of $f(1)$ we get, $\alpha = 5$ and $\beta = \sqrt{3}$ $$\Rightarrow \alpha^2 + \beta^2 = 28$$

An electric dipole is a system of two charges of equal and opposite magnitude placed at a short distance. The dipole moment is, $p = 2ql$. The electric field due to the dipole is, $$E = \frac{2kp}{r^3}$$ Clearly, the electric field varies as $\frac{1}{r^3}$.

The equations of SHM is given by $$F = -kx$$ $$\Rightarrow a = -\omega^2 x$$ Velocity is given by, $v = \omega \sqrt{A^2 - x^2}$. Clearly, velocity is maximum at mean position and acceleration is maximum at extreme positions. Hence, this is the correct option.

The gravitational force ($F_G$) between the particles is given by $$F_G = \frac{Gm^2}{(2a)^2}$$ $$= \frac{Gm^2}{4a^2} \ldots (1)$$ The centripetal force ($F_C$) of each particle can be written as $$F_C = m\omega^2 a \ldots (2)$$ Under balanced condition, equate equation (1) and equation (2) and simplify to obtain the angular speed of each particle. $$m\omega^2 a = \frac{Gm^2}{4a^2}$$ $$\Rightarrow \omega^2 = \frac{Gm}{4a^3}$$ $$\Rightarrow \omega = \sqrt{\frac{Gm}{4a^3}}$$

The maximum distance for line of sight communication is $$d = \sqrt{2hR} + \sqrt{2h'R}$$ The given data is $$R = 6400 \, \mathrm{km}$$ $$h = 180 \, \mathrm{m}$$ $$h' = 245 \, \mathrm{m}$$ Thus, the value is $$d = \sqrt{2 \times 180 \times 6400 \times 10^3} + \sqrt{2 \times 245 \times 6400 \times 10^3}$$ $$= 6 \times 80 \times 10^2 + 7 \times 80 \times 10^2$$ $$= 104 \, \mathrm{km}$$

The formula to calculate the number of half lives covered by the radioactive material is given by $$n = \frac{t}{T_{\frac{1}{2}}} \ldots (1)$$ Substitute the values of the known parameters into equation (1) to calculate the number of half lives covered. $$n = \frac{15 \, years}{5 \, years}$$ $$= 3$$ The relation between the amount of radioactive substance after certain time can be written as $$N = N_0 2^{-n} \ldots (2)$$ Substitute the value of $n$ into equation (2) to obtain the fraction of the remaining substance. $$N = N_0 2^{-3}$$ $$\Rightarrow \frac{N}{N_0} = \frac{1}{8}$$ Hence, the amount of the decayed substance can be written as $$1 - \frac{N}{N_0} = 1 - \frac{1}{8}$$ $$= \frac{7}{8}$$

The de-Broglie wavelength in the first case is $$\lambda = \frac{h}{\sqrt{2mE}}$$ In the second case the kinetic energy is $K = \frac{E}{4}$. Let the de-Broglie wavelength be $\lambda'$. The value of wavelength is $$\lambda' = \frac{h}{\sqrt{2m \frac{E}{4}}} = \frac{2h}{\sqrt{2mE}} = 2\lambda$$

The resistance is given by $$R = \frac{V}{I_g} - G$$ $$\Rightarrow \frac{50}{10^{-3}} - 54 \approx 50 \, \mathrm{k\Omega}$$ For ammeter, $$S = \frac{I_g G}{I - I_g}$$ $$\Rightarrow S = \frac{10^{-3} \times 54}{(10 - 1) \times 10^{-3}} = 6 \, \Omega$$

The formula to calculate the average kinetic energy per molecule of a gas at any temperature $T$ is given by $$K_{avg} = \frac{3}{2} k_B T$$ where, $k_B$ is Boltzmann constant. From above equation, it can be concluded that the average kinetic energy per molecule is independent of the molecular weight of the gas. Hence, the required ratio is 1 : 1.

The formula for series equivalent resistance is given by $R_s = R_1 + R_2 + \ldots + R_n$ Hence, the equivalent resistance will be the highest value compared to other resistances. The resistivity of a material is given by, $\rho = \rho_0 [1 + \alpha \Delta T]$ Clearly, the resistivity is dependent on temperature. Hence, option (1) is correct.

Using conservation of mechanical energy, $$U_i + K_i = U_f + K_f$$ $$\Rightarrow -\frac{GMm}{2R} + 0 = -\frac{GMm}{R} + \frac{1}{2}mv^2$$ $$\Rightarrow \frac{1}{2}mv^2 = \frac{GMm}{2R}$$ $$\Rightarrow v = \sqrt{\frac{GM}{R}} = \sqrt{gR}$$

The current $(i)$ through the circuit is given by $$i = \frac{V}{R}$$ $$= \frac{12 \, \mathrm{V}}{6 \, \Omega}$$ $$= 2 \, \mathrm{A}$$ The formula to calculate the induced emf in the coil can be written as $$\varepsilon = L \frac{di}{dt} \ldots (1)$$ Substitute the values of the known parameters into equation (1) and solve to calculate the inductance of the coil. $$20 \, \mathrm{V} = L \times \frac{2 \, \mathrm{A}}{1 \times 10^{-3} \, \mathrm{s}}$$ $$\Rightarrow L = \frac{20 \, \mathrm{V} \times 1 \times 10^{-3} \, \mathrm{s}}{2 \, \mathrm{A}}$$ $$= 10 \, \mathrm{mH}$$

Young's modulus is same for both the wires since the material is same. For the first wire, $$Y = \frac{fL}{\pi r^2 l} \cdots (i)$$ For the second wire, $$Y = \frac{2f(2L)}{4\pi r^2 l'} \cdots (ii)$$ Dividing (i) by (ii) $$\frac{fL}{\pi r^2 l} \times \frac{4\pi r^2 l'}{4fL} = 1$$ $$\Rightarrow l' = l$$

It is given that $x = 5t^2 - 4t + 5$ The velocity is given by $$v = \frac{dx}{dt} = \frac{d(5t^2 - 4t + 5)}{dt} = 10t - 4$$ At $t = 2 \, \mathrm{s}$, $$v = 10(2) - 4 = 16 \, \mathrm{m \, s^{-1}}$$

Given, the position vector of the particle is $\vec{r} = 10t \hat{i} + 15t^2 \hat{j} + 7 \hat{k}$. The velocity of the particle can be calculated as follows: $$\vec{v} = \frac{d\vec{r}}{dt}$$ $$= \frac{d}{dt} \left( 10t \hat{i} + 15t^2 \hat{j} + 7 \hat{k} \right)$$ $$= 10 \hat{i} + 30t \hat{j}$$ And, the acceleration of the particle is given by $$\vec{a} = \frac{d\vec{v}}{dt}$$ $$= \frac{d}{dt} \left( 10 \hat{i} + 30t \hat{j} \right)$$ $$= 30 \hat{j}$$ Hence, the force on the particle can be written as $$\vec{F} = m \vec{a}$$ $$= 30m \hat{j}$$

The electromagnetic spectrum is the range of all types of EM radiation. Radiation is energy that travels and spreads out with different energies. In daily life, two commonly known EM waves are the light wave and the radio wave. The other types of EM radiation that make up the electromagnetic spectrum are microwaves, infrared light, ultraviolet light, X-rays and gamma-rays. The wavelengths of different parts of the electromagnetic spectrum are related by the following relation: $$\lambda_X < \lambda_{\mathrm{UV}} < \lambda_{\mathrm{IR}} < \lambda_{\mathrm{MW}}$$

The angle made by the vector on the $x$ axis is $\theta = 90^\circ - 30^\circ = 60^\circ$. Let the vector be $\vec{a}$. It is given that $$a_y = 2\sqrt{3}$$ $$\Rightarrow a \sin 60^\circ = 2\sqrt{3}$$ $$\Rightarrow \frac{a \sqrt{3}}{2} = 2\sqrt{3}$$ $$\Rightarrow a = 4$$ The $x$ component is $$a_x = a \cos 60^\circ$$ $$\Rightarrow a_x = 4 \times \frac{1}{2} = 2$$

The dimensional form of velocity can be written as $[v] = [\lambda^a g^b \rho^c]$. Therefore, $$[M^0 L T^{-1}] = [L]^a [L T^{-2}]^b [M L^{-3}]^c$$ Comparing the exponents for $M, L, T$, For $M$, $$M^c = M^0$$ For $L$, $$L^{a+b-3c} = L^1$$ For $T$, $$T^{-2b} = T^{-1}$$ So, $$c = 0 \ldots (i)$$ $$a + b - 3c = 1 \ldots (ii)$$ $$-2b = -1 \ldots (iii)$$ From equations (i), (ii), (iii), $$a = \frac{1}{2}, b = \frac{1}{2}, c = 0$$

The work done is given by the area of the $PV$ graph of the loop. $$W = \frac{1}{2} (V_E - V_C)(P_D - P_C) = \frac{2 \times 300}{2} \, \mathrm{J}$$ $$= 300 \, \mathrm{J}$$

For the first minima $a \sin \theta = \lambda$ The given data is $\theta = 30^\circ$ $\lambda = 600 \, \mathrm{nm}$ The value of $a$ is $$a = \frac{\lambda}{\sin 30^\circ} = \frac{600 \times 10^{-9}}{\frac{1}{2}} = 1.2 \, \mu \mathrm{m}$$

In the first circuit diagram, the diodes $D_1$ and $D_3$ are forward biased and $D_2$ is reverse biased (hence, no current through it). So the circuit is simplified to the second diagram. Hence, the equivalent resistance is $$R_{eq} = \frac{20 \times 10}{20 + 10}$$ $$\Rightarrow R_{eq} = \frac{20}{3} \, \Omega$$ Using Ohm's law, the current is $$I = \frac{10}{\frac{20}{3}} = \frac{30}{20} \, \mathrm{A} = 1.5 \, \mathrm{A}$$

The data given is $\omega = 210 \times \frac{2\pi}{60} \, \mathrm{rad} \, \mathrm{s}^{-1} = 7\pi \, \mathrm{rad} \, \mathrm{s}^{-1}$, $l = 0.2 \, \mathrm{m}$, $B = 0.2 \, \mathrm{T}$. The emf is given by $$\varepsilon = \frac{B \omega l^2}{2}$$ $$= 0.2 \times 7\pi \times \frac{(0.2)^2}{2}$$ $$= 88 \, \mathrm{mV}$$

The resistance in each branch is $6 \, \Omega$. They are in parallel. So, net resistance is $R' = \frac{6}{2} = 3 \, \Omega$. The total current flowing through the loop, $$I = \frac{V}{R'} = \frac{9 \, \mathrm{V}}{3 \, \Omega} = 3 \, \mathrm{A}$$ Thus, through each branch the current is $1.5 \, \mathrm{A}$. Hence, the voltage difference across the points is $$V_A - V_B = (4 \times 1.5) - (2 \times 1.5) = 3 \, \mathrm{V}$$

The fundamental frequency is $f = 50 \, \mathrm{Hz}$. The length of the string is $l = \frac{\lambda}{2}$. Linear mass density is $\mu = \frac{m}{l} = 20 \, \mathrm{g \, m^{-1}}$. Using $v = f \lambda$ $$\frac{v}{2l} = 50$$ $$\Rightarrow v = 100 \times l = \frac{100 \times 18}{20} = 90 \, \mathrm{m \, s^{-1}}$$

For $\lambda_1$ $$\frac{1}{\lambda_1} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{5}{36} \right) \cdots (i)$$ For $\lambda_2$ $$\frac{1}{\lambda_2} = RZ^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{7}{144} \right) \cdots (ii)$$ Dividing (ii) by (i) $$\frac{\lambda_1}{\lambda_2} = \frac{7}{\frac{144}{5}} = \frac{7}{20}$$ So, $$\frac{7}{4n} = \frac{7}{20}$$ $$\Rightarrow n = 5$$

Considering rotational axis as the diametrical axis for sphere and axis of cylinder. Then $$K_1^2 = \frac{2}{5} R^2 and K_2^2 = \frac{1}{2} R^2$$ Therefore, $$\frac{K_1}{K_2} = \sqrt{\frac{2/5}{1/2}} = \sqrt{\frac{4}{5}}$$ Thus, $$\frac{K_1}{K_2} = \frac{2}{\sqrt{5}}$$ Comparing it with the given value in the question, $$\frac{2}{\sqrt{5}} = \frac{2}{\sqrt{x}}$$ Therefore, $x = 5$

For minimum deviation in a prism, $$\mu = \frac{\sin \left( \frac{D_{min} + A}{2} \right)}{\sin \frac{A}{2}}$$ The value of minimum deviation is $$\Rightarrow \sqrt{2} = \frac{\sin \left( \frac{D_{min} + 60^\circ}{2} \right)}{\sin 30^\circ}$$ $$\Rightarrow \frac{1}{\sqrt{2}} = \sin \left( \frac{D_{min} + 60^\circ}{2} \right)$$ $$\Rightarrow 45^\circ = \left( \frac{D_{min} + 60^\circ}{2} \right)$$ $$\Rightarrow D_{min} = 30^\circ$$

The magnetic field due to the electron is given by $$B = \frac{\mu_0 q v \sin 90^\circ}{4 \pi r^2}$$ The given data is $v = 6.76 \times 10^6 \, \mathrm{m \, s^{-1}}$ $r = 0.52 \times 10^{-10} \, \mathrm{m}$ Thus, the magnetic field is $$B = \frac{10^{-7} \times (1.6 \times 10^{-19}) \times (6.76 \times 10^6)}{0.52 \times 0.52 \times 10^{-20}} = 40 \, \mathrm{T}$$

The pressure inside the bubble is given by $$\Delta P = P - P_0 = \rho gh + \frac{2T}{r}$$ The data given is $r = 10^{-3} \, \mathrm{m}$, $T = 0.075 \, \mathrm{N \, m^{-1}}$, $\rho = 1000 \, \mathrm{kg \, m^{-3}}$, $h = 10 \, \mathrm{cm}$. Therefore, the pressure in the bubble is $$\Delta P = (1000 \times 10 \times 10 \times 10^{-2}) + \frac{2 \times 0.075}{10^{-3}} = 1000 + 150 = 1150 \, \mathrm{Pa}$$

Work done is $W = \int F \, dx$. It is given that $F = 5x \, \mathrm{N}$. So, work done is $$W = \int_2^4 5x \, dx = \left. \frac{5x^2}{2} \right|_2^4$$ $$\Rightarrow W = \frac{5(16 - 4)}{2} = 30 \, \mathrm{J}$$

All the capacitors are in parallel combination and therefore potential drop across each capacitor would be the same. Charge on the capacitor can be written as, $Q = CV$. The charge on $C_1$ is $Q_1 = 2 \times 10 = 20 \, \mu \mathrm{C}$. The charge on $C_2$ is $Q_2 = x \times 10 = 10x \, \mu \mathrm{C}$. The charge on $C_3$ is $Q_3 = 3 \times 10 = 30 \, \mu \mathrm{C}$. By the given data, $$C_1 + C_2 + C_3 = 100 \, \mu \mathrm{C}$$ $$\Rightarrow 10(2 + x + 3) = 100$$ $$\Rightarrow x = 5$$

Orlon is a man-made polymer of acrylonitrile. Thus, its monomer is vinyl cyanide or Prop-2-ene-1-nitrile. Natural Rubber is obtained from latex which is a sap of a tree. Natural rubber is a polymer of isoprene. Teflon is made of a chemical compound named polytetrafluoroethylene (PTFE) which is a synthetic fluoropolymer. Nylon-6 is produced by ring-opening chain growth polymerisation of caprolactam in the presence of water vapour and an acid catalyst at the melt.

Hydroboration-oxidation of terminal alkenes give primary alcohols. Propene gives propanol. Propanol on oxidation with PCC gives propanal. Propanal undergo nucleophi magnesium bromide give 2-butanol.

Photochemical smog is a mixture of pollutants that are formed when nitrogen oxides and volatile organic compounds (VOCs) react to sunlight, creating a brown haze above cities. It tends to occur more often in summer, because that is when we have the most sunlight. The average temperature of Kashmir in January is estimated to be $-3.6^\circ \mathrm{C}$ in Srinagar, $-11.8^\circ \mathrm{C}$ in Pahalgam, and $-10.4^\circ \mathrm{C}$ in Gulmarg as their minimum temperatures.

Calcination is a process in which ore is heated in the absence of air or air might be supplied in a limited quantity. Roasting involves the heating of ore lower than its melting point in the presence of air or oxygen. $$2\mathrm{PbS} + 3\mathrm{O}_2 \xrightarrow{\Delta} 2\mathrm{PbO} + 2\mathrm{SO}_2$$ The above reaction is an example of roasting. Rest all the reactions are heated in the absence of oxygen.

In NF$_3$, nitrogen is having three bond pairs and a lone pair, hence, it has pyramidal shape. The bond order of N$_2$ = 3, and the bond order of O$_2$ = 2. Bond length is inversely related to bond order. Hence, bond length of N$_2$ is shorter than O$_2$. Isoelectronic species have identical bond order. Dipole moment of H$_2$O is more than that of H$_2$S due to higher electronegativity of O.

The reaction of aniline (aromatic amine) with nitrous acid results in the diazonium salt formation which is benzenediazonium chloride. When benzenediazonium chloride reacts with N,N-dimethylaniline at ice-cold condition, it gives p-dimethylaminoazobenzene. It is an example of coupling reaction.

In $\mathrm{H_4P_2O_7}$, each phosphorus undergo $\mathrm{sp^3}$ hybridisation. It has one P-O-P linkage in it. In $(\mathrm{HPO_3})_3$, each phosphorus undergo $\mathrm{sp^3}$ hybridisation. It has three P-O-P linkages in it. In $\mathrm{P_4O_{10}}$, each phosphorus undergo $\mathrm{sp^3}$ hybridisation. It has six P-O-P linkages in it.

According to Bohr's postulates the angular momentum in stationary orbit is quantised and is equal to $n h / 2 \pi$, where $n$ is the principal quantum number of the orbit. Bohr's model of the atom assumes fixed orbits and trajectories for the electron. Simultaneously known orbits and trajectories violate the Heisenberg Uncertainty Principle.

Reactivity of aromatic compounds towards electrophilic substitution depends on the electron density on benzene ring. Higher the electron density on benzene ring, more be the reactivity towards electrophilic substitution reaction. Hence, the correct decreasing order is $$NMe_2 > OCH_3 > CH_3 > H > CF_3$$

When concentration of $\mathrm{H}^+$ ion is more than $10^{-7} \, \mathrm{M}$, the proton concentration from water must be considered. Hence, the pH of $10^{-8} \, \mathrm{M}$ HCl solution is not equal to 8. $$\mathrm{H_2PO_4^-} \rightleftharpoons \mathrm{HPO_4^{2-}} + \mathrm{H^+}$$ $K_w$ increases with increase in temperature as dissociation of water increases with increase in temperature. When weak monoprotic acid is titrated against strong base, at half neutralization point, the solution becomes acidic buffer and its pH is given by $$\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[Salt]}{[Acid]} \right)$$ $$\Rightarrow \mathrm{pH} = \mathrm{p}K_a + 0 as [salt] = [acid]$$

In Be and Mg, the electrons are strongly bound. The energy required to excite these electrons is very high. Therefore, when the electron reverts back to its original position, the energy released does not fall in the visible region. Hence, no colour in the flame is seen.

The required product can be obtained by the following method. In this method nitro toluene is subjected to bromination, bromo group is substituted at ortho position to the methyl as it is more activating group. Then nitro group is reduced to amino group with iron metal in acid. The amine group undergoes diazotisation with nitrous acid. The diazonium group is replaced with chloro group in the presence of cuprous chloride. Finally, methyl group is oxidised in the presence of potassium permanganate.

Aldehydes and ketones react with water to give 1,1-geminal diols known as hydrates. In general, hydrates are not stable enough to be isolated as the equilibrium shifts back to starting materials (due to Le Chatelier's principle). But the following ketone forms stable diol due to the formation of hydrogen bonding as shown below.

In aqueous solutions carboxyl group can lose protons and the amino group can accept a proton thus giving rise to a dipolar known as zwitter ion. Due to this dipolar behaviour they have strong electrostatic interactions within them. For this reason melting point of amino acids is high. It has more than one basic groups amine and guanidine, hence it has two pKa values. It is crystalline solid. Since arginine is polar, it is not soluble in benzene.

The reaction between CO of syn gas and steam in presence of catalyst is called water gas shift reaction. $$\mathrm{CO + H_2O \rightarrow CO_2 + H_2O}$$ This reaction increases the yield of hydrogen obtained from coal gasification reaction as shown below. $$\mathrm{C + H_2O \rightarrow \underset{syn gas}{\mathrm{CO + H_2}}}$$

For a good quality cement the ratio of silica ($\mathrm{SiO_2}$) and alumina ($\mathrm{Al_2O_3}$) should be between 2.5 to 4.0. Similarly, the ratio of lime ($\mathrm{CaO}$) to the total oxide mixtures consisting of $\mathrm{SiO_2}$, $\mathrm{Al_2O_3}$ and $\mathrm{Fe_2O_3}$ should be roughly 2 : 1 : 1. If lime is in excess, the cement cracks during setting.

The principle of paper chromatography is partition. In paper chromatography there are two phases: one is the stationary phase and the other is the mobile phase. Here, water trapped in the paper acts as the stationary phase and solvent acts as the mobile phase.

Friedel-Crafts acylation is a reaction in which the addition of an acyl group to an aromatic ring takes place. The reaction will lead to the aromatic ring being transformed into a ketone.

Crystal field stabilisation energy of the complex can be calculated as follows, CFSE = number of $t_{2g}$ electrons $\($ $\times$ (-0.4) $\)$ + number of $e_g$ electrons $\($ $\times$ (0.6) $\)$ Complex $\hspace{1cm}$ CFSE $$[\mathrm{Ti(OH_2)_6}]^{3+} (t_{2g}^1 e_g^0) -0.4 \, \Delta_0$$ $$[\mathrm{Cr(H_2O)_6}]^{3+} (t_{2g}^3 e_g^0) -1.2 \, \Delta_0$$ $$[\mathrm{Mn(H_2O)_6}]^{3+} (t_{2g}^3 e_g^1) -0.6 \, \Delta_0$$ $$[\mathrm{Fe(H_2O)_6}]^{3+} (t_{2g}^3 e_g^2) 0$$

CsCl has body centered type structure in which $\mathrm{Cs}^+$ occupies at corner of a cube and $\mathrm{Cl}^-$ occupies the centre of the cube. $$2r_{\mathrm{Cs}^+} + 2r_{\mathrm{Cl}^-} = \sqrt{3}a (where a is the edge length of the cube)$$ $$r_{\mathrm{Cs}^+} + r_{\mathrm{Cl}^-} = \frac{\sqrt{3}}{2} a$$

The electronic configuration of $\mathrm{Co^{2+}}$ is $3d^7$. In an octahedral complex of $\mathrm{Co^{2+}}$ with $\mathrm{H_2O}$ ligands, the $d$ orbitals split into two sets of energy levels: $t_{2g}$ (lower energy) and $e_g$ (higher energy). In this case, since the complex is octahedral, we have six ligands surrounding the central cobalt ion. Each ligand donates a pair of electrons to the cobalt ion, occupying the available $d$ orbitals. So, $$\left[ \mathrm{Co\ (H_2O)_6} \right]^{2+} = d^7 (High spin) = t_{2g}^5 e_g^2$$ In the $t_{2g}$ set of orbitals, there will be 1 unpaired electron.

($\mathrm{Cr\ (H_2O)_5\ Cl}$)$\mathrm{Cl_2}$ $\rightarrow$ ($\mathrm{Cr\ (H_2O)_5\ Cl}$)^{+} + 2$\mathrm{Cl}$^{-} Number of ionisable $\mathrm{Cl}^{-}$ in $(\mathrm{Cr\ (H_2O)_5\ Cl})\mathrm{Cl_2}$ is 2. Molarity of $\mathrm{Cl}^{-}$ ions $= 0.01 \times 2 = 0.02\ \mathrm{M}$ Therefore, millimoles of $\mathrm{Cl}^{-}$ ions in the given solution $$= 20 \times 0.02$$ Therefore, millimoles of $\mathrm{AgNO_3}$ required $= 20 \times 0.2$ $$V \times 0.1 = 20 \times 0.02$$ $$V = 4\ \mathrm{mL}$$

In the reaction of $\mathrm{KMnO_4}$ and potassium iodide (KI) in an acidic medium, the $\mathrm{KMnO_4}$ acts as an oxidizing agent, and the iodide ions ($\mathrm{I^-}$) in KI act as reducing agents. $$\mathrm{KMnO_4 + KI + H^+ \rightarrow Mn^{+2} + I_2 + H_2O}$$ In $\mathrm{KMnO_4}$, the oxidation state of manganese is $+7$, which is the highest possible oxidation state for manganese in this compound. In the final product, $\mathrm{MnSO_4}$, the oxidation state of manganese is $+2$. Therefore, the total change in the oxidation state of manganese is: $$+7 (initial) - +2 (final) = 5$$ Hence the change in O. S. of Mn is $5$.

The chemical formula of Chromyl chloride is $\mathrm{CrO_2Cl_2}$. The oxidation state of chromium in chromyl chloride ($\mathrm{CrO_2Cl_2}$) is $+6$. Let the oxidation number of Chromium be $x$, the oxidation state of oxygen is $-2$, the oxidation state of chlorine is $-1$. Then, $$x + 2(-2) + 2(-1) = 0$$ $$x - 6 = 0$$ $$x = 6$$

Isoelectronic are those which have the same number of electrons. Let us write the number of electrons from the given species. Hence five species among the given are isoelectronic species.

To calculate the vapour pressure of the 30% (w/v) aqueous solution of glucose, we can use Raoult's law, which states that the vapour pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution. Weight of solution = $100 \times 1.2 = 120 \, \mathrm{gm}$ Weight of water = $120 - 30 = 90 \, \mathrm{gm}$ Now using formula $$\frac{P_0 - P}{P} = \frac{moles of glucose}{moles of water}$$ $$\frac{24 - P}{P} = \frac{30}{180} = \frac{3}{90}$$ $$24 \times 90 - P \times 90 = 3P$$ $$\Rightarrow P = 23.22$$

Coagulating value $= \frac{\text{Millimoles of electrolyte}}{\text{Volume of sol in litre}}$. $20 \, \mathrm{ml}$ of $0.5 \, \mathrm{M}$ NaCl contains NaCl $= 20 \times 0.5 = 10$ milli moles. $200 \, \mathrm{ml}$ of $\mathrm{As_2S_3}$ required NaCl for the coagulation $= 10$ milli moles. $\therefore 1000 \, \mathrm{ml}$ of $\mathrm{As_2S_3}$ required NaCl for the coagulation $= 10 \times \frac{1000}{200} = 50$ milli moles.

As per the question, to determine the activation energy of the catalyzed backward reaction, we need to use the Arrhenius equation and the given information. The Arrhenius equation is given by: $$k = A e^{(-E_a / RT)}$$ $$A e^{-\frac{300 \times 10^3}{600 \times R}} = A e^{-\frac{E_a}{300 \times R}}$$ $$\Rightarrow \frac{10^3}{2} = \frac{E_a}{300}$$ $$\Rightarrow E_a = 150 \times 10^3 \, \mathrm{J} \, \mathrm{mol}^{-1}$$ $$\Rightarrow E_a = 150 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$$ Therefore, the activation energy of the catalysed backward reaction is the energy of activation for the backward reaction, $E_b = E_a - \Delta H$. $$= 150 - 20 = 130 \, \mathrm{kJ} \, \mathrm{mol}^{-1}$$

Conductivity always decreases with the decrease in concentration both, for weak and strong electrolytes. This can be explained by the fact that the number of ions per unit volume that carry the current in a solution decreases on dilution. Molar conductivity of a strong and weak electrolyte increases with dilution. On dilution as volume of solution increases. Thus, on dilution, ions get more apart and mobility of ions increases which leads to increase in molar conductivity of the solution. Except (B) and (E), all statements are correct.

The equilibrium process can be written as follows, $\mathrm{NaCl(s) \rightleftharpoons NaCl(l)}$ \[ \Delta S = \frac{\Delta H}{\text{Temp}} \] \[ T = \frac{\Delta H}{\Delta S} \] \[ = \frac{30.4 \times 10^3}{28.4} \] \[ = 1070.4 \, \mathrm{K} \]