JEE MAIN 2026 Previous Year Questions (PYQs) – Page 23 of 25

$A : (5,4,2)$

$\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$

$\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda$

Any general point $P$ on the line is $(2\lambda - 1, 3\lambda + 3, -\lambda + 1)$

Let the given point is $A(5,4,2)$

$\overrightarrow{AP} = (2\lambda - 6)\hat{i} + (3\lambda - 1)\hat{j} + (-\lambda - 1)\hat{k}$

$\because \overrightarrow{AP} \perp \text{Line } (L)$

$\overrightarrow{AP} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$

$2(2\lambda - 6) + 3(3\lambda - 1) - 1(-\lambda - 1) = 0$

$\Rightarrow \lambda = 1$

$\alpha = 1,; \beta = 6,; \gamma = 0$

Let the vector $\vec{u} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$

$\vec{u} = \hat{i} + 6\hat{j} + 0\hat{k}$

and $\vec{w} = 6\hat{i} + 2\hat{j} + 3\hat{k}$

So projection $= \frac{|\vec{u} \cdot \vec{w}|}{|\vec{w}|} = \frac{18}{7}$

Given circle

$x^2 + y^2 - 4x - 6y - 3 = 0$

$C(2,3)$ & $r = 4$

$\cos 30^\circ = \frac{r}{OR} = \frac{4}{OR}$

$\Rightarrow OR = \frac{8}{\sqrt{3}}$

Now

$OR^2 = (h - 2)^2 + (k - 3)^2$

$\Rightarrow 3(x^2 + y^2) - 12x - 18y - 25 = 0$

$(ax^2 + bx + c)\sum_{r=0}^{6} {}^6C_r(-2x)^r$

Coeff. of $x^2$:
$a \cdot {}^6C_0(-2)^0 + b \cdot {}^6C_1(-2)^1 + c \cdot {}^6C_2(-2)^2 = 0$

$\Rightarrow a - 12b + 60c = 0 \quad ...(1)$

Coeff. of $x^3$:
$a \cdot {}^6C_1(-2)^1 + b \cdot {}^6C_2(-2)^2 + c \cdot {}^6C_3(-2)^3 = 0$

$\Rightarrow -12a + 60b - 160c = 0 \quad ...(2)$

Coeff. of $x = -56$:
$b \cdot {}^6C_0(-2)^0 + c \cdot {}^6C_1(-2)^1 = -56$

$\Rightarrow b - 12c = -56 \quad ...(3)$

After solving (1), (2) & (3):

$a = 1300,; b = 100,; c = 3$

$\Rightarrow a + b + c = 1403$

$x^2 + x + 1 = 0$

$\Rightarrow x = \omega$ or $\omega^2$

$\therefore \alpha = \omega,; \beta = \omega^2$

$= (\omega + \omega^2)^4 + (\omega^2 + \omega)^4 + (\omega^4 + \omega^2)^4 + \cdots$

$= [1 + 1 + \cdots + 1] + [(1+1)^4 + (-1+1)^4 + \cdots]$

$= 17 + 128 = 145$

$= \frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}$

$= \frac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$

$= \frac{2\sin(30^\circ - 10^\circ)}{\sin 20^\circ}$

$= \frac{2\sin 20^\circ}{\sin 20^\circ} = 2 \cdot 2 = 4$

Let $|x - 1| = t$

$t^2 - 5t + 6 = 0$

$t = 2$ & $t = 3$

$|x - 1| = 2$ & $|x - 1| = 3$

$x - 1 = \pm 2,; x - 1 = \pm 3$

$x = 3, -1, 4, -2$

Sum of roots $= 3 + (-1) + 4 + (-2) = 4$

$h = \frac{4t}{5}$

$k = \frac{2t^2}{5} = \frac{2}{5}\left(\frac{5h}{4}\right)^2$

$8k = 5h^2$

$\Rightarrow 5x^2 = 8y$

$T = S_1$

$5(xx_1) - 4(y + y_1) = 5x_1^2 - 8y_1$

$5(xx_1) - 4(y + 2) = 5 - 8 \cdot 2$

$5x - 4y + 3 = 0$

Given quadratic equation has equal roots, thus

$D = 0 \Rightarrow (r(x))^2 = r'(x)\cdot f(x)$

$\frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}$

Integrate

$\ln(f'(x)) = \ln(f(x)) + \ln C \Rightarrow f'(x) = C f(x)$

Put $x = 0$

$1 = C \cdot 2 \Rightarrow C = \frac{1}{2}$

Now $2f'(x) = f(x)$

$\Rightarrow \frac{f'(x)}{f(x)} = 2$

Integrate

$\ln(f(x)) = 2x + d$

$\Rightarrow d = 0$

$\Rightarrow \ln(f(x)) = 2x \Rightarrow f(x) = e^{2x}$

Now let $g(x) = f(\ln x - x) = e^{2(\ln x - x)}$

$\therefore g'(x) = 2e^{2(\ln x - x)}\left(\frac{1}{x} - 1\right) \ge 0$

$\Rightarrow \frac{1 - x}{x} \ge 0$

$\Rightarrow x \in (0,1]$

$\Rightarrow \alpha = 0,; \beta = 1$

$\alpha + \beta = 1$

$a_{n+1} - \frac{1}{2}a_n = \frac{n^2 - 2n - 1}{n^2(n+1)^2} = \frac{2n^2 - (n+1)^3}{n^2(n+1)^2}$

$\Rightarrow a_{n+1} - \frac{1}{2}a_n = \frac{2}{(n+1)^2} - \frac{1}{n^2}$

$n = 1 \Rightarrow a_2 - \frac{1}{2}a_1 = \frac{2}{2^2} - \frac{1}{1^2}$

$2\left[a_3 - \frac{1}{2}a_2\right] = \frac{2}{3^2} - \frac{1}{2^2}$

$2^2\left[a_4 - \frac{1}{2}a_3\right] = \frac{2}{4^2} - \frac{1}{3^2}$

$\vdots$

$2^{n-1}\left[a_n - \frac{1}{2}a_{n-1}\right] = \frac{2}{n^2} - \frac{1}{(n-1)^2}$

Adding

$a_n = \frac{2}{n^2} - \frac{1}{2^{n-1}}$

$\Rightarrow \sum_{n=1}^{\infty} \left(a_n - \frac{2}{n^2}\right) = \sum_{n=1}^{\infty} \left(-\frac{1}{2^{n-1}}\right) = 2$

$S = {1,2,3,\ldots,50}$

$p = (6^m + 9^n)$ is divisible by $5$

No. of ways

$6^m = (5\lambda + 1)^m = 5k + 1$

$9^n = (10 - 1)^n = 10\mu - 1$ if $n$ is odd

$\Rightarrow n$ must be odd

$10\mu + 1$ if $n$ is even

$\Rightarrow$ No. of ways $= 50 \times 25 = 1250$

$q \Rightarrow (m+n)$ is square of a prime

$m+n = 4, 9, 25, 49$

No. of ways: $3, 8, 24, 48$

$q = 3 + 8 + 24 + 48 = 83$

$p + q = 1250 + 83 = 1333$