JEE MAIN 2026 Previous Year Questions (PYQs) – Page 22 of 25

$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$

$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$

$\vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}$

$\vec{a} \times \vec{c} = \vec{b}$

$(2c_3 - 2c_2)\hat{i} + (c_3 + 2c_1)\hat{j} - (c_2 + 2c_1)\hat{k} = 8\hat{i} + 7\hat{j} - 3\hat{k}$

$2c_3 - 2c_2 = 8,\quad c_3 + 2c_1 = 7,\quad c_2 + 2c_1 = 3$

$c_1 + c_2 + c_3 = 4,\quad c_1 + c_2 + c_3 = 4$

$c_1 = 2,\quad c_2 = -1,\quad c_3 = 3$

$|\vec{a} + \vec{c}|^2 = | \hat{i} + \hat{j} + 5\hat{k} |^2 = 27$

$ar \cdot ar^2 \cdot ar^3 = 64$

$a^3 r^6 = 64 \Rightarrow ar^2 = 4$

$a + ar + ar^2 = \frac{813}{7}$

$r^2 = 28$

$ar^2 + ar^4 + ar^6 = 4(1 + r^2 + r^4) = 4(1 + 28 + 784) = 3252$

$\frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{\tan^{-1}x}{1 + x^2}$

I.F. $= e^{\tan^{-1}x}$

$y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{\tan^{-1}x}{1 + x^2} dx$

$y \cdot e^{\tan^{-1}x} = \tan^{-1}x \cdot e^{\tan^{-1}x} - e^{\tan^{-1}x} + c$

$y(0) = 1 \Rightarrow c = 2$

$y(1) = \frac{2}{e^4} + \frac{\pi}{4} - 1$

$f(i) \ne i$, $f(x)$ is strictly increasing function

$f : A \to B$, where $A = {1,2,3,4,5,6}$

$B = {1,2,3,\ldots,9}$

$f : A \to B$ is equal to

Case–i: $f(1) = 2 \Rightarrow {}^7C_5 = 21$

Case–ii: $f(1) = 3 \Rightarrow {}^6C_5 = 6$

Case–iii: $f(1) = 4 \Rightarrow {}^5C_5 = 1$

No of function $A$ to $B = 21 + 6 + 1 = 28$

Let $T = \lim_{x \to 3} \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-3)^2}}$ ; $1^\infty$ form

$\Rightarrow T = e^{\lim \frac{18}{(x-3)^2} \cdot \frac{f(x+2) - f(3)}{f(3)}}$

$\Rightarrow T = e^{\lim \frac{18}{(x-3)^2} \cdot \frac{f(x+2) - f(3)}{18}}$

$\Rightarrow T = e^{\lim \frac{f(x+2) - f(3)}{(x-3)^2}}$ ; $0/0$ form apply L’Hospital

$\Rightarrow T = e^{\lim \frac{f'(x+2)}{2(x-3)}}$ ; $0/0$ form apply L’Hospital

$\Rightarrow T = e^{\lim \frac{f''(x+2)}{2}} = e^2$

$\Rightarrow \log_e (T) = 2$

Let $e_1$ be eccentricity of ellipse

$\Rightarrow e_1 = \sqrt{1 - \frac{16}{36}} = \sqrt{\frac{4}{9}} = \frac{\sqrt{5}}{3}$

So $ae_1 = 6 \cdot \frac{\sqrt{5}}{3} = 2\sqrt{5}$

Now $H : \frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$

$p.e = ae_1$

$p \cdot 5 = 2\sqrt{5}$

$p = \frac{2}{\sqrt{5}}$

$e^2 = 1 + \frac{q^2}{p^2}$

$25 = 1 + \frac{q^2}{p^2} \Rightarrow 25 = 1 + \frac{5q^2}{4}$

$q^2 = \frac{96}{5}$

So length of LR $= \frac{2q^2}{p} = \frac{96}{\sqrt{5}}$

$= 2\pi \int_0^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$ let $x + \frac{\pi}{6} = t \Rightarrow dx = dt$

$= 2\pi \int_{\pi/6}^{\pi/3} \frac{dt}{1 - \sin t}$

$= 2\pi \int_{\pi/6}^{\pi/3} \frac{1 + \sin t}{\cos^2 t} dt$

$= 2\pi \left[ \int_{\pi/6}^{\pi/3} \sec^2 t , dt + \int_{\pi/6}^{\pi/3} \sec t \tan t , dt \right]$

$= 2\pi \left[ (\tan t){\pi/6}^{\pi/3} + (\sec t){\pi/6}^{\pi/3} \right]$

$= 2\pi \left[ \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) + \left(2 - \frac{2}{\sqrt{3}}\right) \right]$

$= 2\pi \left[ \sqrt{3} + 2 - \sqrt{3} \right] = 4\pi$

Mean $(\bar{x}) = 8$ (Given)

$\Rightarrow \frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$

$\Rightarrow x + y = 14 \quad ...(1)$

Variance $(\sigma^2) = 16$ (Given)

$\Rightarrow 16 = \frac{2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2}{7} - 8^2$

$\Rightarrow x^2 + y^2 = 100 \quad ...(2)$

$(x + y)^2 = x^2 + y^2 + 2xy$

$\Rightarrow 14^2 = 100 + 2xy \Rightarrow xy = 48$

Since $x > y$

$\Rightarrow x = 8,; y = 6$

Now set $X = {1,2,3,4,6,5}$

Now we choose two numbers one after another without replacement

Total outcomes $= 6 \times 5 = 30$

We want the probability that smaller number $< 4$

$P(\text{smaller} < 4) = 1 - P(\text{smaller} \ge 4)$

$= 1 - \frac{6}{30} = \frac{4}{5}$