JEE Main 29 January 2025 Shift 1 question paper with solutions

By solving $x = y$ with the circle, we get $C(\sqrt{2}, \sqrt{2})$ and $D(-\sqrt{2}, -\sqrt{2})$. By solving $x + y = 1$ with the circle $x^2 + y^2 = 4$, we set $$A \left( \frac{1 + \sqrt{7}}{2}, \frac{1 - \sqrt{7}}{2} \right)$$ & $$B \left( \frac{1 - \sqrt{7}}{2}, \frac{1 + \sqrt{7}}{2} \right)$$ Therefore, the area of quadrilateral $ACBD$ is $$= 2 \times Area of \triangle BCD$$ $$= 2 \times \frac{1}{2} \begin{vmatrix} \sqrt{2} & \sqrt{2} & 1 \\ \frac{1 - \sqrt{7}}{2} & \frac{1 + \sqrt{7}}{2} & 1 \\ -\sqrt{2} & -\sqrt{2} & 1 \end{vmatrix}$$ $$= 2\sqrt{14}$$

Given the matrix: $$\begin{vmatrix} 1 + \sin^2 x & \cos^2 x & 4 \sin 4x \\ \sin^2 x & 1 + \cos^2 x & 4 \sin 4x \\ \sin^2 x & \cos^2 x & 1 + 4 \sin 4x \end{vmatrix}, x \in \mathbb{R}$$ Perform the row operations: $$R_2 \to R_2 - R_1 \& R_3 \to R_3 - R_1$$ The matrix becomes: $$\begin{vmatrix} 1 + \sin^2 x & \cos^2 x & 4 \sin 4x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$$ Expand about $R_1$, we get $$f(x) = 2 + 4 \sin 4x$$ Thus, $M = max value of f(x) = 6$ $M = min value of f(x) = -2$ Therefore, $M^4 - M^4 = 1280$

The parabolas are $$(x - 4)^2 + (y - 3)^2 = x^2 \ldots (i)$$ and $$(x - 4)^2 + (y - 3)^2 = y^2 \ldots$$ If point of intersection are $A(x_1, y_1)$ and $B(x_2, y_2)$. By solving (i) and (ii), we get $$x_1 + x_2 = 14 and x_1 x_2 = 25$$ $$(AB)^2 = 2 \left( (x_1 + x_2)^2 - 4x_1 x_2 \right) = 192$$

$\therefore$ centroid of $\triangle ABC =\left(\dfrac{9+3+5}{3},\dfrac{11+4+13}{3}\right) =\left(\dfrac{17}{3},\dfrac{28}{3}\right)$ Let image of centroid with respect to line mirror is $(h,k)$ $\therefore\; \left(\dfrac{k-\frac{28}{3}}{h-\frac{17}{3}}\right) \left(-\dfrac12\right) =-1 \qquad\cdots(1)$ $\left(\dfrac{h+\frac{17}{3}}{2}\right) + 6\left(\dfrac{k+\frac{28}{3}}{2}\right) =53 \qquad\cdots(2)$ Solving (1) \& (2) we get $h=3,\quad k=4$ $\therefore\;h^2+k^2+hk=37$

Given $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + 3\hat{k}$. We have $\vec{a} \times \vec{c} = \vec{c} \times \vec{b}$. Therefore, $\vec{a} \times \vec{c} + \vec{b} \times \vec{c} = 0$. This implies $(\vec{a} + \vec{b}) \times \vec{c} = 0$. Thus, $\vec{c} = \lambda(\vec{a} + \vec{b})$. So, $\vec{c} = \lambda(5\hat{i} - 6\hat{j} + 4\hat{k}) \ldots (1)$. The magnitude squared is $|\vec{c}|^2 = \lambda^2(25 + 36 + 16)$. Therefore, $|\vec{c}|^2 = 77\lambda^2$. The expression $(\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 168$ gives $\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{c} \cdot \vec{b} + |\vec{c}|^2 = 168$. Simplifying, $14 + \vec{c} \cdot (\vec{a} + \vec{b}) + 77\lambda^2 = 168$. Using equation (1), $\lambda|5\hat{i} - 6\hat{j} + 4\hat{k}|^2 + 77\lambda^2 = 154$. This simplifies to $77\lambda + 77\lambda^2 - 154 = 0$. Solving $\lambda^2 + \lambda - 2 = 0$, we find $\lambda = -2, 1$. Therefore, the maximum value of $|\vec{c}|^2$ occurs when $\lambda = -2$. Thus, $|\vec{c}|^2 = 77\lambda^2 = 77 \times 4 = 308$.

(i) number of numbers created using 1111133 = $\($ $\frac{7!}{5!2!}$ $\Rightarrow$ 21 $\)$ (ii) number of numbers created using 1111223 = $\($ $\frac{7!}{4!2!}$ $\Rightarrow$ 105 $\)$ (iii) number of numbers created using 1112222 = $\($ $\frac{7!}{4!3!}$ $\Rightarrow$ 35 $\)$ Total = 161

Given that $A$ is the area of rectangle $ABDE$ minus the area of region $EDC$. $$A \Rightarrow 4 - 2 \int_{0}^{1} \left(3 - x\right) - \left(\frac{x^2 + 3}{2}\right) \, dx$$ $$A \Rightarrow 4 - 2 \left\{ 3x - \frac{x^2}{2} - \frac{x^3}{6} - \frac{3}{2}x \right\}_{0}^{1}$$ $$A \Rightarrow 4 - 2 \left\{ 3 - \frac{1}{2} - \frac{1}{6} - \frac{3}{2} \right\} = \frac{7}{3}$$ So $6A = 14$

General term = $\binom{n}{r} \left\{ 7^{1/3} \right\}^{n-r} \left( 11^{1/12} \right)^r$ $$= \binom{n}{r} \left\{ 7 \right\}^{\frac{n-r}{3}} (11)^{r/12}$$ For integral terms, $r$ must be a multiple of 12. Therefore, $r = 12k$, $k \in \mathbb{W}$. Total values of $r = 183$. Hence $\max \, r = 12(182) = 2184$. Min value of $n = 2184$.

Consider $\frac{1}{\sqrt{x}} = \alpha$ where $x > 0$. $$\{9\alpha^2 - 9\alpha + 2\}\{2\alpha^2 - 7\alpha + 3\} = 0$$ $$(3\alpha - 2)(3\alpha - 1)(\alpha - 3)(2\alpha - 1) = 0$$ $$\alpha = \frac{1}{3}, \frac{2}{3}, 3, \frac{1}{2}$$ $$x = 9, 4, \frac{4}{9}, 1$$ So, no. of solutions = 4

$\dfrac{dy}{dx}-\dfrac{3\sin x}{\cos x\,\ln\cos x}\,y=-\dfrac{\sin x}{\cos x\,(\ln\cos x)^2}$ I.F.$=e^{-\int\dfrac{3\tan x}{\ln\cos x}\,dx}$ Let $\ln\cos x=t$ $-\tan x\,dx=dt$ $I.F.=e^{3\int\dfrac{dt}{t}}=e^{3\ln t}=t^3=(\ln\cosx)^3$ $\therefore$ Solution will be$y(\ln\cos x)^3=-\int(\tan x)(\ln\cosx)\,dx$ $y(\ln\cos x)^3=-\dfrac{(\ln\cos x)^2}{2}+c$ $\because\;y\!\left(\dfrac{\pi}{4}\right)=\dfrac{1{\ln2}$ $\Rightarrow\;c=0$ $\therefore\;y=-\dfrac{1}{2(\ln\cos x)}$ $y\!\left(\dfrac{\pi}{6}\right)=-\dfrac12\times\dfrac{1}{\ln\!\left(\cos\dfrac{\pi}{6}\right)}$ $=-\dfrac12\times\dfrac{1}{\ln\!\left(\dfrac{\sqrt3}{2}\right)}$ $=-\dfrac12\times\dfrac{1}{\ln\sqrt3-\ln2}$ $=\dfrac{1}{\ln4-\ln3}$

Given $\sec^2 x - \tan^2 x = 1$ (on replacing $y$ with $x$). Reflexive: $$\sec^2 x - \tan^2 y = 1$$ $$1 + \tan^2 x + 1 - \sec^2 y = 1$$ $$\sec^2 y - \tan^2 x = 1$$ Symmetric: $$\sec^2 x - \tan^2 y = 1$$ $$\sec^2 y - \tan^2 z = 1$$ Adding both: $$\sec^2 x - \tan^2 y + \sec^2 y - \tan^2 z = 1 + 1$$ $$\sec^2 x + 1 - \tan^2 z = 2$$ $$\sec^2 x - \tan^2 z = 1$$ Transitive: Hence equivalence relation.

Given $2ae = 4$. Therefore, $a = 2\sqrt{3}$. Thus, $1 - \frac{b^2}{12} = \frac{1}{3} \Rightarrow b^2 = 8$. $$\frac{2b^2}{a} \times \frac{2A^2}{B} = \frac{32}{\sqrt{3}}$$ Therefore, $$\frac{2 \times 8}{2\sqrt{3}} \times \frac{2A^2}{B} = \frac{32}{\sqrt{3}}$$ This implies $$\frac{A^2}{B} = 2 \Rightarrow A^2 = 2B$$ $$1 - \frac{A^2}{B} = \frac{1}{3}$$ Therefore, $$B = 3 \Rightarrow A^2 = 6$$ Equation 1: $\frac{x^2}{12} + \frac{y^2}{5} = 1 \ldots (i)$ Equation 2: $\frac{x^2}{6} + \frac{y^2}{9} = 1 \ldots (ii)$ On solving (i) and (ii), $$(x, y) = \left( \frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}} \right), \left( \frac{-\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}} \right), \left( \frac{\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}} \right), \left( \frac{-\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}} \right)$$ Four points are vertices of rectangle area $= \frac{24\sqrt{6}}{5}$.

Given $S_3 = 3a + 3d = 54$. This implies $a + d = 18$. $S_{20} = 10(2a + 19d)$ $$10(36 + 17d)$$ This implies $1600 < 10(36 + 17d) < 1800$. Therefore, $160 < 36 + 17d < 180$. This simplifies to $124 < 17d < 144$. Dividing throughout by 17 gives $\frac{7}{17} < d < \frac{8}{17}$. The common difference will be a natural number. Thus, $d = 8$ implies $a = 10$. Finally, $a_{11} = 10 + 10 \times 8 = 90$.

Given $L_1: \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} + 2\hat{j} + \hat{k})$. $$\Rightarrow \vec{r} = (\lambda - 1)\hat{i} + 2(\lambda + 1)\hat{j} + (\lambda + 1)\hat{k}$$ $L_2: \vec{r} = (\hat{j} + \hat{k}) + \mu (2\hat{i} + 7\hat{j} + 3\hat{k})$. $$\Rightarrow \vec{r} = 2\mu \hat{i} + (1 + 7\mu) \hat{j} + (1 + 3\mu) \hat{k}$$ For point of intersection equating respective components $$\Rightarrow \lambda - 1 = 2\mu$$ $$2(\lambda + 1) = 1 + 7\mu$$ $$\lambda + 1 = 1 + 3\mu$$ We get $$\Rightarrow \lambda = 3 and \mu = 1$$ $$\Rightarrow \vec{a} + \vec{b} = 3\hat{i} + 9\hat{j} + 4\hat{k}$$ $L_3: \vec{r} = 2\hat{i} + 8\hat{j} + 4\hat{k} + \alpha (3\hat{i} + 3\hat{j} + 4\hat{k})$. For $\alpha = 2$, $$\vec{r} = 8\hat{i} + 26\hat{j} + 12\hat{k}$$

The given expression is $\($ $\lim$_{n $\to$ $\infty$} $\sum$_{k=1}^{n} $\frac{k^3 + 6k^2 + 11k + 5}{(k+3)!}$ $\)$. This can be rewritten as: $$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^3 + 6k^2 + 11k + 6 - 1}{(k+3)!} $$ Which simplifies to: $$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{(k+1)(k+2)(k+3) - 1}{(k+3)!} $$ Further simplification gives: $$ \lim_{n \to \infty} \sum_{k=1}^{n} \left( \frac{(k+1)(k+2)(k+3)}{(k+3)!} - \frac{1}{(k+3)!} \right) $$ This can be expressed as: $$ \lim_{n \to \infty} \sum_{k=1}^{n} \left( \frac{1}{k!} - \frac{1}{(k+3)!} \right) $$ Evaluating the series, we have: $$ \lim_{k=1} \left( \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots + \frac{1}{n!} - \frac{1}{4!} - \frac{1}{5!} - \frac{1}{6!} - \cdots - \frac{1}{(n+3)!} \right) $$ This results in: $$ \frac{1}{1} + \frac{1}{2} + \frac{1}{6} = \frac{10}{6} = \frac{5}{3} $$

Given $$I = \int_{0}^{\frac{\pi}{4}} \left( \frac{\sin \theta + \cos \theta}{9 - 16 \sin 2\theta} \right) d\theta$$ Take $\sin \theta - \cos \theta = t$. Then $(\cos \theta + \sin \theta) d\theta = dt$ and $(\sin \theta - \cos \theta)^2 = t^2$. Thus, $\sin 2\theta = 1 - t^2$. When $\theta = 0 \rightarrow t = -1$ and $\theta = \frac{\pi}{4} \rightarrow t = 0$. Therefore, $$I = \int_{-1}^{0} \frac{dt}{9 + 16(1 - t^2)}$$ $$= \frac{1}{16} \int_{-1}^{0} \frac{dt}{25 - 16t^2}$$ $$= \frac{1}{4} \left[ \frac{1}{10} \log \left| \frac{5 - 4t}{5 + 4t} \right| \right]_{-1}^{0}$$ $$= \frac{1}{40} [0 + \log_e 9]$$ Thus, $$I = \frac{\log_e 9}{40}$$ Therefore, $$80I = 2 \log_e 9$$ $$80I = 4 \log_e 3$$

DR's of $L_3 = \vec{m} \times \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 2 & 1 & 1 \end{vmatrix}$ $$= -5\hat{i} - 3\hat{j} + \hat{k}$$ For $L_3$: $\($ $\frac{x - \alpha}{-5}$ = $\frac{y - \beta}{-3}$ = $\frac{z - \gamma}{1}$ = $\lambda$ $\)$ $A(\alpha - 5\lambda, \beta - 3\lambda, \gamma + \lambda)$ For $L_1$: $\($ $\frac{x - 1}{1}$ = $\frac{y - 2}{-1}$ = $\frac{z - 1}{2}$ = k $\)$ $B(k + 1, -k + 2, 2k + 1)$ Now $\alpha - 5\lambda = k + 1 \Rightarrow \alpha = 5\lambda + k + 1$ $\beta - 3\lambda = -k + 2 \Rightarrow \beta = 3\lambda - k + 2$ $\gamma + \lambda = 2k - 1 \Rightarrow \gamma = -\lambda + 2k - 1$ $|5\alpha - 11\beta - 8\gamma| = | - 25| = 25$

Given $\($ $\sum$_{l=1}^{10} (x_l - 2) = 30 $\)$ and $\($ $\sum$_{i=1}^{10} x_i = 50 $\)$. Therefore, Mean = 5. Variance = $\($ $\frac{4}{5}$ = $\frac{\sum x_l^2}{10}$ - ($\bar{x}$)^2 $\)$. $\[$ $\frac{4}{5}$ = $\frac{\sum x_l^2}{10}$ - 25 $\]$ $\[$ $\Rightarrow$ $\sum$ x_l^2 = 258 $\]$ Now, $\($ $\sum$_{l=1}^{10} (x_l - $\beta$)^2 = 98 $\)$ $\[$ $\sum$_{l=1}^{10} x_l^2 - 2$\beta$ $\sum$_{l=1}^{10} x_l + 10$\beta$^2 = 98 $\]$ $\[$ $\Rightarrow$ 258 - 2$\beta$(50) + 10$\beta$^2 = 98 $\]$ $\[$ $\Rightarrow$ 10$\beta$^2 - 100$\beta$ + 160 = 0 $\]$ $\[$ $\Rightarrow$ $\beta$^2 - 10$\beta$ + 16 = 0 $\]$ $\[$ $\Rightarrow$ $\beta$ = 8 as $\beta$ > 2 $\]$ Now, as per the question $\($ 2(x_1 - 1) + 4$\beta$, 2(x_2 - 1) + 4$\beta$, $\ldots$, 2(x_{10} - 1) + 4$\beta$ $\)$ Can be simplified as $\($ 2x_1 + 30, 2x_2 + 30, $\ldots$, 2x_{10} + 30 $\)$ $\($ $\mu$ = 2(5) + 30 = 40 $\)$ $\($ $\sigma$^2 = 2^2 $\left$( $\frac{4}{5}$ $\right$) = $\frac{16}{5}$ $\)$ $\($ $\beta$ $\mu$ = $\frac{8 \times 40}{\frac{16}{5}}$ = 100 $\)$

Since $AB = \sqrt{100} = 10$, therefore $|Z_1 - Z_2|_{\min} = 10 - 2 - 1 = 7$.

Given $|A| = \frac{11}{2}$. $$C_{11} = \sum_{k=1}^{2} a_{1k} \cdot A_{1k} = a_{11} A_{11} + a_{12} A_{12} = |A| = \frac{11}{2}$$ $$C_{12} = \sum_{k=1}^{2} a_{1k} \cdot A_{2k} = 0$$ $$C_{21} = \sum_{k=1}^{2} a_{2k} \cdot A_{1k} = 0$$ $$C_{22} = \sum_{k=1}^{2} a_{2k} \cdot A_{2k} = |A| = \frac{11}{2}$$ The matrix $C$ is given by: $$C = \begin{bmatrix} 11/2 & 0 \\ 0 & 11/2 \end{bmatrix}$$ The determinant $|C|$ is calculated as: $$|C| = \frac{121}{4}$$ Finally, $8|C|$ is: $$8|C| = 242$$

Given, $\int_0^1 f(\lambda x) d\lambda = a f(x)$ Let $\lambda x = u$ $d\lambda = \frac{1}{x} du$ From (1) $\frac{1}{x} \int_0^x f(u) du = a f(x)$ $$\Rightarrow \int_0^x f(u) du = ax f(x)$$ Differentiate both sides $$f(x) = a (x f'(x) + f(x))$$ $$\Rightarrow f(x) = ax f'(x) + a f(x)$$ $$\Rightarrow (1 - a) f(x) = ax f'(x)$$ $$\Rightarrow f'(x) = \frac{(1 - a)}{a} \cdot \frac{1}{x} f(x)$$ Integrate both sides w.r.t. $(x)$ $$\Rightarrow \int \frac{f'(x)}{f(x)} dx = \frac{(1 - a)}{a} \int \frac{1}{x} dx$$ $$\Rightarrow \ln f(x) = \left( \frac{1 - a}{a} \right) \ln x + c$$ Now at $x = 1, f(1) = 1$ $$\Rightarrow c = 0$$ Also given $f(16) = \frac{1}{8}$ $$\Rightarrow 1 = (16)^{\frac{1-a}{a}}$$ $$\Rightarrow 8 = 2^{\frac{4-4a}{a}}$$ $$\Rightarrow 2^{-3} = 2^{\frac{4-4a}{a}}$$ $$\Rightarrow -3 = \frac{4-4a}{a}$$ $$\Rightarrow -3a = 4 - 4a$$ $$\Rightarrow a = 4$$ $$\therefore f(x) = x^{-3/4}$$ $$f(x) = \frac{-3}{4} x^{-7/4}$$ Put $x = \frac{1}{16}$ $$f'\left(\frac{1}{16}\right) = \frac{-3}{4} \left(\frac{1}{16}\right)^{-7/4} = \frac{-3}{4} \cdot 2^{-4x\left(\frac{-7}{4}\right)} = -96$$ $$\therefore 16 - f'\left(\frac{1}{16}\right) \Rightarrow 16 - (-96) = 112$$

Given $A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$. Now finding characteristic equation $$\begin{vmatrix} 2 - \lambda & -1 \\ 1 & -\lambda \end{vmatrix} = 0$$ $$\Rightarrow (2 - \lambda)(-\lambda) - (-1)(1) = -2\lambda + \lambda^2 + 1 = 0$$ $$\Rightarrow \lambda^2 - 2\lambda + 1 = 0$$ $$\Rightarrow (\lambda - 1)^2 = 0$$ $$\Rightarrow \lambda = 1$$ Since $A$ satisfies $(A - I)^2 = 0$. Therefore, $A = I + N$ where $N = A - I$. $$N = \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix}$$ $$N^2 = 0$$ $$A^m = (I + N)^m = I + mN$$ $$A^m \cdot A^m = (I + mN)(I + mN) = I + 2mN + m^2N^2$$ Since $N^2 = 0$ $$\Rightarrow A^{m^2} = I + 2mN$$ Now putting in given condition $$I + m^2N + I + mN = 3I - A^{-6}$$ $$A^{-1} = \begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix}$$ $$A^{-6} = (A^{-1})^6 = I + (-6)N$$ Therefore, putting in (i) $$(m^2 + m)N = I - (I - 6N)$$ $$(m^2 + m)N = 6N$$ Since $N \neq 0$ $$\Rightarrow m^2 + m = 6$$ $$\Rightarrow m^2 + m - 6 = 0$$ $$\Rightarrow (m - 2)(m + 3) = 0$$ $$\Rightarrow m = 2, -3$$ Number of elements in $S$ is 2.

Given $$\lim_{x \to 0^+} \left( x \left( \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \ldots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1^2}{x^2} \right\rfloor + \left\lfloor \frac{2^2}{x^2} \right\rfloor + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \right) \geq 1$$ The expression $$\left( 1 + 2 + \ldots + p \right) - \left( 1^2 + 2^2 + \ldots + 9^2 \right) \geq 1$$ This simplifies to $$\frac{p(p+1)}{2} - \frac{9 \cdot 10 \cdot 19}{6} \geq 1$$ Thus, $$p(p+1) \geq 572$$ The least natural value of $$p$$ is 24.

Q6. (i) Single letter is used, then number of words = 5 (ii) Two distinct letters are used, then number of words $$^5C_2 \times \left( \frac{6!}{2!4!} \times 2 + \frac{6!}{3!3!} \right) = 10(30 + 20) = 500$$ (iii) Three distinct letters are used, then number of words $$^5C_3 \times \frac{6!}{2!2!2!} = 900$$ Total number of words = 1405

Given $\cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1)$. $2 \cos^{-1} x - \sin^{-1}(2x + 1) = \frac{3\pi}{2}$. Let $2\alpha - \beta = \frac{3\pi}{2}$ where $\cos^{-1} x = \alpha$, $\sin^{-1}(2x + 1) = \beta$. Then $2\alpha = \frac{3\pi}{2} + \beta$. We have $\cos 2\alpha = \sin \beta$. Thus, $2 \cos^2 \alpha - 1 = \sin \beta$. Substituting, $2x^2 - 1 = 2x + 1$. This simplifies to $x^2 - x - 1 = 0$. Solving, $\Rightarrow n = \frac{1 \pm \sqrt{5}}{2}$. Therefore, $$n = \begin{cases} \frac{1+\sqrt{5}}{2} & rejected \\ \frac{1-\sqrt{5}}{2} & \end{cases}$$ Thus, $4x^2 - 4x = 4$. Finally, $(2x - 1)^2 = 5$.

The current $I$ in the circuit is given by the formula: $$I = \frac{V}{\sqrt{R^2 + \omega^2 L^2}}$$ The voltage across the resistor $V_R$ is given by: $$V_R = \frac{R}{\sqrt{R^2 + \omega^2 L^2}} V$$

The maximum height $H_{Max}$ is given by $$H_{Max} = \frac{(u \sin \theta)^2}{2g}$$ For $(H_{max})_1$, we have $u^2 \sin^2(45^\circ - \alpha)$. For $(H_{max})_2$, we have $u^2 \sin^2(45^\circ + \alpha)$. This can be expressed as $$\left( \frac{1}{\sqrt{2}} \cos \alpha - \frac{1}{\sqrt{2}} \sin \alpha \right)^2$$ which simplifies to $$\left( \frac{1}{\sqrt{2}} \cos \alpha + \frac{1}{\sqrt{2}} \sin \alpha \right)^2$$ Finally, this results in $$\frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$$

Given $\tau = PE_0 \sin \theta$. If $\theta$ is small, $\tau = -(PE_0) \theta$. $$I = m \left( \frac{l}{2} \right)^2 \cdot 2 = \frac{ml^2}{2}$$ $$T = 2\pi \sqrt{\frac{ml^2}{2 \cdot PE_0}} = 2\pi \sqrt{\frac{ml^2}{2 \cdot q/E_0}}$$ $$T = 2\pi \sqrt{\frac{ml}{2qE_0}}$$

Q5. The dimension of angular momentum is $ML^2 T^{-1}$. The dimension of Planck's Constant is $ML^2 T^{-1}$. The dimension of torque is $ML^2 T^{-2}$. The dimension of energy is $ML^2 T^{-2}$. The dimension of surface tension is $MT^{-2}$. The dimension of impulse is $MLT^{-1}$. The dimension of pressure is $ML^{-1} T^{-2}$. The dimension of Young's modulus is $ML^{-1} T^{-2}$.

As $h$ increases, $g$ decreases, $T$ increases. $$T = 2\pi \sqrt{\frac{\ell}{g}}$$ $$g = \frac{g_0 R^2}{(R + h)^2}$$

Given $[LT^{-1}] = [A][T^2] = \frac{[B][T]}{[C] + [T]}$. We have $[C] = [T]$, $[A] = [LT^{-3}]$, $[B] = [LT^{-1}]$, and $[ABC] = [L^2 T^{-3}]$.

Given $\phi_1 = L_1 I_1 + M_{12} I_2$. The expression for $\varepsilon_1$ is given by $$\varepsilon_1 = -\frac{d\phi_1}{dt} = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_2}{dt}.$$

Given $\sin \theta_{1C} = \frac{n_1}{n_2}$ and $\sin \theta_{2C} = \frac{n_1}{n_3}$. Then, $$\sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2}$$ Substituting the values, $$\frac{n_1}{n_3} - \frac{n_1}{n_2} = \frac{1}{2}$$ Simplifying, $$\frac{n_1 n_2 - n_1 n_3}{n_3 n_2} = \frac{1}{2}$$ This gives, $$n_1 \left( \frac{2}{5} - 1 \right) = \frac{n_2}{2}$$ Solving further, $$\frac{n_1}{n_2} = -\frac{5}{6}$$ Thus, $$= \sin^{-1} \left( -\frac{5}{6} \right)$$

Maximum possible magnetic field is at the surface. $$B_{max} = \frac{\mu_0 I}{2 \pi a}$$ $$\frac{B_{max}}{2} = \frac{\mu_0 I}{4 \pi a}$$ It can be obtained inside as well as outside the wire. For inside, $$\frac{\mu_0 I}{4 \pi a} = \frac{\mu_0 I r}{2 \pi a^2}$$ $$\Rightarrow r = \frac{a}{2}$$ For outside, $$\frac{\mu_0 I}{4 \pi a} = \frac{\mu_0 I}{2 \pi r}$$ $$\Rightarrow r = 2a$$ Correct answer $\left[ \frac{a}{2}, 2a \right]$

Velocity of a just before hitting: $$u = \sqrt{2g \frac{R}{2}} = \sqrt{gR}$$ Just after collision, let velocity of A and B are $v_1$ and $v_2$ respectively. Therefore, by COM: $$mu = mv_1 + \frac{m}{2} v_2$$ $$2v_1 + v_2 = 2u \ldots (i)$$ $$e = 1 = \frac{v_2 - v_1}{u}$$ $$\Rightarrow v_2 - v_1 = u \ldots (ii)$$ From (i) -(ii) $$\Rightarrow 3v_1 = u \Rightarrow v_1 = \frac{u}{3} = \frac{1}{3} \sqrt{gR}$$

Negative potential will slow the electrons and if it is sufficient, it will make the photocurrent zero. $$eVs = hf - \phi_0$$

Given $\phi = BAN \cdot \cos(\omega t)$. The induced emf $\varepsilon$ is given by $$\varepsilon = -\frac{d\phi}{dt} = BA\omega N \cdot \sin(\omega t).$$ When $B$ is parallel to the plane, $\omega t = \frac{\pi}{2}$. Thus, $\phi = 0$, and $\varepsilon = BA\omega N$.

Given $$B = \frac{\rho g h}{\left( \frac{\Delta v}{v} \right)}$$ $$\frac{\Delta v}{v} \times 100 = \frac{\rho g h}{B} \times 100$$ $$\frac{1000 \times 10 \times 2.5 \times 10^3}{2 \times 10^9} \times 100\%$$ $$= 1.25\%$$

Given $$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}$$ $$\lambda^2 = \frac{h^2}{2m} \left( \frac{1}{k} \right)$$ $$Y = cx^2$$ Upward facing parabola passing through origin.

The truth table is given as follows: \begin{tabular}{|c|c|c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 1 \\ \hline \end{tabular} This is the truth table for an OR Gate.

Given $v = n \sqrt{gR}$. The initial velocity $v_0$ is given by $$v_0 = \sqrt{v^2 + 2g(2R)}$$ Substituting $v = n \sqrt{gR}$, we have $$v_0 = \sqrt{n^2 gR + 4gR}$$ Therefore, $$\frac{k_{bottom}}{k_{trog}} = \frac{v_0^2}{v^2} = \frac{n^2 + 4}{n^2}$$

Given the equation $(u + f)(v - f) = f^2$. The graph shows the relationship between $u$ and $v$. The axes are labeled $u$ and $v$ respectively.

Inside uniformly charged spherical shell, $E = 0$ Therefore, $A \rightarrow \mathrm{III}$ For uniformly charged infinite plate $$E = \frac{\sigma}{2\varepsilon_0}$$ Therefore, $B \rightarrow \mathrm{II}$ Outside of spherical shell $$E = \frac{Q}{4\pi\varepsilon_0 r_2} = \frac{\sigma R^2}{\varepsilon_0 r^2}$$ Therefore, $C \rightarrow \mathrm{IV}$ Between two plates $E = \frac{\sigma}{\varepsilon_0}$ Therefore, $D \rightarrow \mathrm{I}$

Work done in adiabatic process is given by $$\frac{nR \Delta T}{1 - \gamma}$$. So, it depends upon change in temperature.

EM wave carry both energy and momentum. Rest mass of photon is zero.

The torque $\vec{\tau}$ acting on the particle with respect to the origin can be calculated using the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$: $$\vec{\tau} = \vec{r} \times \vec{F}$$ Given the position vector $\vec{r} = (1, 1, 1) \, \mathrm{m}$ and the force vector $\vec{F} = \hat{i} - \hat{j} + \hat{k}$, we need to calculate the cross product: $$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix}$$ Calculating the determinant, we have: $$\vec{\tau} = \hat{i}(1 \cdot 1 - 1 \cdot (-1)) - \hat{j}(1 \cdot 1 - 1 \cdot 1) + \hat{k}(1 \cdot (-1) - 1 \cdot 1)$$ This simplifies to: $$\vec{\tau} = \hat{i}(1 + 1) - \hat{j}(1 - 1) + \hat{k}(-1 - 1)$$ $$\vec{\tau} = 2\hat{i} - 0\hat{j} - 2\hat{k}$$ The torque vector is $\vec{\tau} = 2\hat{i} - 2\hat{k}$. To find the magnitude of the torque in the $z$-direction, we look at the $\hat{k}$ component: $$\tau_z = -2$$ The magnitude of torque in the $z$-direction is: $$|\tau_z| = 2 \, \mathrm{Nm}$$ Thus, the magnitude of the torque in the $z$-direction is 2 Newton-meters.

Given $V = constant$, we have the relation $$\frac{P_1}{T_1} = \frac{P_2}{T_2}.$$ Given $P_2 = 2P_1$ and $T_2 = 2T_1$, we find $$T_2 = 2 \times 300 = 600 \, K.$$ Therefore, $T_2 = 327^\circ C$.

Given $n_1 \sin(90^\circ - \theta_1) = n_2 \sin r$. $$n_1 \times \frac{n_2}{2n_1} = n_2 \sin r$$ $$\sin r = \frac{1}{2}$$ $$r = 30^\circ$$ $$\tan r = \left(\frac{d/2}{t}\right)$$ $$t = \frac{d}{2 \tan r} = \frac{d \sqrt{3}}{2} = \frac{(4 \sqrt{3}) \sqrt{3}}{2}$$ $$= 6 \, \mathrm{cm}$$

Given $F = 100 \, \mathrm{N}$, $A_1 = 6 \, \mathrm{cm^2}$, and $A_2 = 1500 \, \mathrm{cm^2}$. Using the formula $\frac{F_1}{A_1} = \frac{F_2}{A_2}$, we have: $$\frac{100}{6} = \frac{F}{1500}$$ Solving for $F$ gives: $$F = \frac{50}{3} \times 1500$$ Calculating $F$: $$F = 50 \times 500 = 25 \times 10^3 \, \mathrm{N}$$ The work done $\omega$ is given by $\vec{F} \cdot \vec{S}$: $$\omega = 25 \times 10^3 \times \frac{20}{100}$$ Simplifying gives: $$= 5 \times 10^3 = 5 \, \mathrm{kJ}$$

Given $\vec{v}_b = 9 + 27 = 36 \, \mathrm{km/hr}$. $$\vec{v}_b = 36 \times \frac{1000}{3600} = 10 \, \mathrm{m/sec}$$ Time of flight $= \frac{2 \times 10}{10} = 2 \, \mathrm{sec}$ Range $= 10 \times 2 = 20 \, \mathrm{m} = 2000 \, \mathrm{cm}$

Total five nucleophiles are present: $\mathrm{NH_3}$, $\mathrm{PhSH}$, $(\mathrm{H_3C})_2\mathrm{S}$, $\mathrm{CH_2} = \mathrm{CH_2}$, $\overset{\ominus}{\mathrm{O}}\mathrm{H}$.

The element having strongest oxidising capacity will have highest value of standard reduction potential.

For a weak electrolyte, the variation of $\Lambda_m$ with $\sqrt{c}$ is shown in the graph. The graph depicts a curve that decreases as $\sqrt{c}$ increases.

For the reaction $\mathrm{AB_2(g) \rightleftharpoons AB(g) + \frac{1}{2} B_2(g)}$, at time $t = 0$, the initial pressure is $p_0$. At equilibrium, $t = t_{eq}$, the pressures are $p_0(1-x)$ for $\mathrm{AB_2}$, $p_0 x$ for $\mathrm{AB}$, and $\frac{p_0 x}{2}$ for $\mathrm{B_2}$. The total pressure $p$ is given by: $$p = p_0 x - p_0 x + p_0 x + \frac{p_0 x}{2}$$ Simplifying, we have: $$p = p_0 \left(1 + \frac{x}{2}\right)$$ Thus, $p_0$ is: $$p_0 = \frac{p}{\left(1 + \frac{x}{2}\right)}$$ The equilibrium constant $K_p$ is: $$K_p = \left( p_{AB} \right) \left( p_{B_2} \right)^{1/2}$$ Substituting the pressures, we get: $$K_p = \left( p_{0} x \right) \left( \frac{p_{0} x}{2} \right)^{1/2}$$ Simplifying further: $$K_p = \frac{p_0 (1-x)}{\left(1 + \frac{x}{2}\right)} \left( \frac{p}{1 + \frac{x}{2}} \times \frac{x}{2} \right)^{1/2}$$ Since $x \ll 1$, we approximate: $$K_p = p^{1/2} x^{3/2} \left(2\right)^{1/2}$$ Solving for $x$, we have: $$x = \sqrt[3]{\frac{2 K_p^2}{p}}$$

The structure in option (A) is a linear chain with an ethyl group at the third carbon and a methyl group at the fifth carbon. This corresponds to the name 3-Ethyl-5-methylheptane. The structure in option (B) is a different configuration.

Mass of substance is amount of matter present in it. Weight is force exerted by gravity on object.

Neglecting pairing energy I. $[\mathrm{Mn(CN)B}]^{3-} \Rightarrow \mathrm{Mn}^{3+}, \ t_2 \ g^4$, CFSE $= -0.4 \times 4 \Delta o = -1.6 \Delta$. II. $[\mathrm{Co(CN)8}]^{4-} \Rightarrow \mathrm{Co}^{2+}, \ t_2^6 e_g^1$, CFSE $= -0.4 \times 6 + 0.6 \times 1 = -1.8 \Delta$ III. $[\mathrm{Fe(CN)8}]^{4-} \Rightarrow \mathrm{Fe}^{2+}, \ t_{2g}^6 e_g^0$, CFSE $= -0.4 \times 6 = -2.4 \Delta_0$ IV. $[\mathrm{Fe(CN)8}]^{3-} \Rightarrow \mathrm{Fe}^{3+}, \ t_{2g}^5 e_g^0$, CFSE $= -0.4 \times 5 = -2 \Delta_0$ Order of stability III > IV > II > I

For (A) $[\mathrm{MnBr_4}]^{2-}$, $\mathrm{Mn}^{+2} \Rightarrow [\mathrm{Ar}] 3d^5$. In presence of ligand field: $$\Rightarrow [\mathrm{Ar}] \uparrow \uparrow \uparrow \uparrow \uparrow \square \square \square$$ This leads to $sp^3$ hybridization, paramagnetic in nature. For (B) $[\mathrm{FeF_6}]^{3-}$, $\mathrm{Fe}^{+3} \Rightarrow [\mathrm{Ar}] 3d^5$. In presence of ligand field: $$\Rightarrow [\mathrm{Ar}] \uparrow \uparrow \uparrow \uparrow \uparrow \square \square \square \square \square \square$$ This leads to $sp^3d^2$ hybridization, paramagnetic in nature. For (C) $[\mathrm{Co(C_2O_4)_3}]^{3-}$, $\mathrm{Co}^{+3} \Rightarrow [\mathrm{Ar}] 3d^6$. In presence of ligand field: $$\Rightarrow [\mathrm{Ar}] \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \square \square \square$$ This leads to $d^2sp^3$ hybridization, diamagnetic in nature. For (D) $[\mathrm{Ni(CO)_4}]$, $\mathrm{Ni}^{0} \Rightarrow [\mathrm{Ar}] 3d^8 4s^2$. In presence of ligand field: $$\Rightarrow [\mathrm{Ar}] \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \square \square$$ This leads to $sp^3$ hybridization, diamagnetic in nature.

Br at the para position of $\mathrm{NO_2}$ will undergo aromatic nucleophilic substitution by nucleophile $\mathrm{C_2H_5ONa}$.

Cathode: $\mathrm{Ag^+ (aq) + e^- \rightarrow Ag} \times 2$ Anode: $\mathrm{Mg \rightarrow Mg^{2+} (aq) + 2e^-}$ Cell reaction: $2\mathrm{Ag^+} + \mathrm{Mg} \rightarrow 2\mathrm{Ag} + \mathrm{Mg^{2+}}$ $Q = \frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^+}]^2}$ By Nernst equation $$E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q$$ $$E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln \left( \frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^+}]^2} \right)$$ $$= E^\circ_{cell} + \frac{RT}{2F} \ln \left( \frac{[\mathrm{Ag^+}]^2}{[\mathrm{Mg^{2+}}]} \right)$$

Melting point order: $\mathrm{Fe} > \mathrm{Mn}$, $\mathrm{Ru} > \mathrm{Tc}$, $\mathrm{Re} > \mathrm{Os}$.

For $\mathrm{AX_2}$ $$\Delta T_b = i \, K \, m$$ $$0.0156 = i \times 0.52 \times \frac{1.24}{124 \times 1}$$ $$3 = i$$ $$3 = 1 + 2\alpha$$ $$1 = \alpha$$ For $\mathrm{AY_2}$ $$\Delta T_b = i \, K_b \, m$$ $$0.0260 = i \times 0.52 \times \frac{25.4}{250 \times 2}$$ $$i \approx 1$$ Therefore, $\mathrm{AX_2}$ is completely ionised and $\mathrm{AY_2}$ is completely unionised.

Given $q_p = n \times c_p \times \Delta T$. $$500 = 0.5 \times \frac{5}{2} \times 8.3 \left( T_f - 298 \right)$$ Therefore, $T_f \approx 346.2 \, \mathrm{K}$. The ratio $\frac{\Delta H}{\Delta U} = \frac{C_p}{C_v} = \left( \frac{5}{3} \right)$. Thus, $\Delta U = \frac{3}{5} \times 500 = 300 \, \mathrm{J}$.

Given the rate equation $rate = k_2 [A] [B_2]$. $$\left( \frac{k_1}{k_{-1}} \right) = \left( \frac{[A]^2}{[A_2]} \right)$$ This implies $[A] = \sqrt{\frac{k_1}{k_{-1}}} \cdot \sqrt{[A_2]}$. Substituting in (1), we get $$Rate = k_2 \sqrt{\frac{k_1}{k_{-1}}} \cdot [A_2]^{\frac{1}{2}} \cdot [B_2]$$ Therefore, the order is $$\left( \frac{3}{2} \right) = 1.5$$

Bohr radius of hydrogen atom $\rightarrow a_0$ According to Bohr, the equation used to calculate the angular momentum of an electron in a hydrogen atom is $$mvr = \frac{nh}{2\pi} \ldots (i)$$ $m \rightarrow$ mass of electron $v \rightarrow$ velocity of electron $r \rightarrow$ radius of the orbit $n \rightarrow$ orbit number in which electron is present. Given that the electron is present in the second orbit, $n = 2$. The radius of the second orbit $r_2 = a_0 \times 2^2 = 4a_0$. General formula for radius of $n^{th}$ orbit, $$r_n = a_0 \times n^2$$ From (1) $$mvr = n \frac{h}{2\pi}$$ $$2\pi r = n \frac{h}{mv}$$ $\frac{h}{mv} = \lambda$ (de Broglie relationship, $\lambda \rightarrow$ de Broglie Wavelength) So, $2\pi r = n\lambda$. For the electron in the second orbit, $2\pi r_2 = n\lambda$. Substitute for $r_2$ $$2\pi \times 4a_0 = n\lambda$$ $$8\pi a_0 = n\lambda$$ Therefore, $\lambda = \frac{8\pi a_0}{n}$.

It is Clemmensen reduction, it will not reduce ester. Ester cannot be reduced by Clemmensen reduction.

The element having high value of Electron gain enthalpy (magnitude) will form a compound having higher ionic character so order of ionic character $$EB > EA > EC > ED$$

Amylose: It is a plant based starch it has $\alpha - C_1 - C_4$ glycosidic linkage. Cellulose: It has $\beta - C_1 - C_4$ glycosidic linkage. Glycogen: It has $\alpha - C_1 - C_4$ and glycosidic linkage (animal starch). Amylopectin: It is a plant based with $\alpha - C_1 - C_4$ and $C_1 - C_6$ glycosidic linkage.

Both compounds (A) and (B) are steam volatile due to intramolecular hydrogen bonding.

Given $r \propto q^{-}$ (for isoelectronic species) and $\propto \frac{1}{q^{+}}$. Therefore, Statement I is correct. Magnitude of electron gain enthalpy: $\mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I}$.

Benzyl Amine is most basic due to localised lone pair. Mole of benzyl Amine $\Rightarrow \frac{1}{107} = 0.00934$ mole 1 Mole of Benzyl amine consumed 1 mole of HCl So, Mole of HCl consumed $\rightarrow 0.00934$ mole Mass of HCl consumed $\rightarrow 0.00934 \times$ molar mass of HCl $$= 0.00934 \times 36.5$$ $$= 0.341 \, \mathrm{gm}$$ $$= 341 \, \mathrm{mg}$$

The insoluble product will be $\mathrm{Fe_2O_3}$. The molar mass is calculated as follows: molar mass $= 56 \times 2 + 16 \times 3$ $$= 112 + 48$$ $$= 160$$

Number of $\sigma$ bond $= 11$. Number of $\pi$ bond $= 4$. $$\sigma + \pi = 11 + 4 = 15$$

The reaction is given as $\mathrm{AB_2} \rightleftharpoons \mathrm{A^{2+}} + 2\mathrm{B^-}$. Initially, the concentrations are $1$, $0$, and $0$. At equilibrium, the concentrations are $1 - \alpha$, $\alpha$, and $2\alpha$. The expression for $i$ is: $$i = 1 + 2\alpha$$ Substituting $\alpha = 0.3$: $$i = 1 + 2 \times (0.3)$$ $$= 1.6$$ $$= 16 \times 10^{-1}$$

0.1 mole of compound (S) weight in gm = 0.1 $\times$ molar mass of compound (S) = 0.1 $\times$ 130 = 13 gm