JEE MAIN 2026 Previous Year Questions (PYQs) – Page 16 of 25

$ 2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = 0 $ $ \Rightarrow (2\vec{a} + 3\vec{b}) \times \vec{c} = 0 \Rightarrow \vec{c} \parallel (2\vec{a} + 3\vec{b}) $ $ \vec{c} = \lambda (2\vec{a} + 3\vec{b}) $ $ = \lambda (7\hat{i} - 13\hat{j} + 19\hat{k}) $ Now $ (\vec{a} - \vec{b}) \cdot \vec{c} = -97 $ $ \Rightarrow \lambda (7 + 52 + 38) = -97 $ $ \Rightarrow \lambda = -1 $ Now $ \vec{c} = -7\hat{i} + 13\hat{j} - 19\hat{k} $ $ \vec{c} \times \vec{k} = -7\hat{j} + 13\hat{i} $ $ |\vec{c} \times \vec{k}|^2 = 7^2 + 13^2 = 218 $

$ f(0) = a $ RHL $ = \lim_{x \to 0^+} \frac{\sin x (1 - \cos x)}{x^3} = \frac{1}{2} $ LHL $ = \lim_{x \to 0^-} \left[ b^2 \sin\left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\sin x + \cos x) \cos x \right] \right) \right] = b^2 $ $ \therefore a = \frac{1}{2},; b^2 = \frac{1}{2} $ $ \Rightarrow a^2 + b^2 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} $

$ f(x) = e^x + \int_0^x y f(y),dy + xe^x \int_0^x f(y),dy $ $ f(x) = e^x + A + Bxe^x $ $ A = \int_0^x y f(y),dy = \int_0^x y(A + e^y + Bye^y),dy $ $ A = \frac{A}{2} + 0(-1) + B(e - 1) $ $ \frac{A}{2} + B(1 - e) = 1 $ $ B = \int_0^1 f(y),dy $ $ B = \int_0^1 (e^y + A + Bye^y),dy $ $ B = (e - 1) + A + B(0 - 1) $ $ B = e - 1 + A + B(0 - (-1)) $ $ B = e - 1 + A + B \Rightarrow A = 1 - e $ $ f(x) = e^x + A + Bxe^x $ $ f(0) = 1 + A + B\cdot 0 = 1 + A = 2 - e $ $ e + f(0) = 2 $

Let $ P = (2\cos\theta, 2\sin\theta) $ $ \therefore $ coordinates of $ Q = (4\cos\theta + 1,; 6\sin\theta + 3) $ $ \therefore $ locus of $ Q $ is $ \left( \frac{x - 1}{4} \right)^2 + \left( \frac{y - 3}{6} \right)^2 = 1 $ $ e^2 = 1 - \frac{16}{36} = \frac{5}{9} $ $ \Rightarrow \frac{5}{e^2} = 9 $

$ |z|^2 = 2^3 \cdot 5^2 \cdot 13 \cdot 17 $ $ \prod (1 + k^2) = (2)(5)(10)(17)(26) = 2 \cdot 5 \cdot 10 \cdot 17 \cdot 26 $ so $ n = 5 $

$ S = {a, b, c, d, e} $ $ P(S) $ contains 32 elements both set $ A $ and set $ B $ are subsets of $ P(S) $ Every element has 4 choices Favourable cases $ = 3^5 $ Total cases $ = 4^5 $ $ P = \frac{3^5}{4^5} = \frac{3^5}{2^{10}} $ $ p = 5,; q = 10 $ $ p + q = 15 $

$ \frac{\tan(x + 100^\circ)}{\tan x} = \tan(x + 50^\circ)\tan(x - 50^\circ) $ $ \frac{\sin(x + 100^\circ)\cos x}{\cos(x + 100^\circ)\sin x} = \frac{\sin(x + 50^\circ)\sin(x - 50^\circ)}{\cos(x + 50^\circ)\cos(x - 50^\circ)} $ Apply $ C $ & $ D $ $ \frac{\sin(2x + 100^\circ)}{\sin 100^\circ} = \frac{\cos 100^\circ}{-\cos 2x} $ $ 2\sin(2x + 100^\circ)\cos 2x + \sin 100^\circ + \sin 200^\circ = 0 $ $ \sin(4x + 100^\circ) + \sin 100^\circ + \sin 200^\circ = 0 $ $ \sin(4x + 100^\circ) = -2\sin 150^\circ \cos 50^\circ $ $ \sin(4x + 100^\circ) = -\cos 50^\circ = \sin(-40^\circ) $ $ 4x + 100^\circ = n\pi + (-1)^n(-40^\circ) $ $ x = \frac{n\pi + (-1)^n(40^\circ) - 100^\circ}{4} $ $ \Rightarrow x = 30^\circ,; 55^\circ,; 120^\circ,; 145^\circ \text{ in } (0,\pi) $ $ \therefore $ no. of solutions $ = 4 $

$ g(2) = 13 $ $ f(g(2)) = f(13) $ Now $ 4g(f(x)) = 3x^2 - 32x + 72 $ $ 4[3f(x)^2 + 2f(x) - 3] = 3x^2 - 32x + 72 $ Let $ f(x) = t $ $ 12t^2 + 8t - (3x^2 - 32x + 84) = 0 $ $ f(x) = \frac{-8 \pm \sqrt{64 + 48(3x^2 - 32x + 84)}}{24} $ $ f(x) = \frac{-8 \pm 4(3x - 16)}{24} $ $ f(x) = \frac{-8 + 4(3x - 16)}{24} $ $ \because f(0) = -3 \Rightarrow \text{we take +ve sign} $ $ f(x) = \frac{-8 + 4(3x - 16)}{24} $ $ f(13) = \frac{-8 + 4 \cdot 23}{24} = \frac{84}{24} = \frac{7}{2} $

$ T_k = (-1)^{k+1} \frac{k(k+1)}{k!} = (-1)^{k+1} \frac{k(k-1) + 2k}{k!} $ $ = (-1)^{k+1} \left[ \frac{k(k-1)}{k!} + \frac{2k}{k!} \right] $ $ = (-1)^{k+1} \left[ \frac{1}{(k-2)!} + \frac{2}{(k-1)!} \right] $ $ \Rightarrow \sum = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-2)!} + 2 \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{(k-1)!} $ $ = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} = \frac{1}{e} $

Equation of circle $ C_2 $ is

$ x^2 + y^2 - 5x - 5y = 0 $

its centre is $ \left( \frac{5}{2}, \frac{5}{2} \right) $

$ m_{AB} = -1 $

$ \therefore $ slope of required chord $ = 1 $

$ \therefore $ equation of required chord is $ x - y + 1 = 0 $

$ \therefore a = -1,; b = 2 $

$ \therefore a - b = -2 $