JEE MAIN 2026 Previous Year Questions (PYQs) – Page 17 of 25
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If $ \frac{\tan(A - B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1 $, $ A, B, C \in \left( 0, \frac{\pi}{2} \right) $, then
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Solution
$ \frac{\tan A - \tan B}{(1 + \tan A \tan B)\tan A} + \frac{1 + \cot^2 A}{1 + \cot^2 C} = 1 $
Put $\tan A = x,; \tan B = y,; \tan C = z$
$ \Rightarrow \frac{x - y}{(1 + xy)x} + \frac{(x^2 + 1)z^2}{x^2(z^2 + 1)} = 1 $
$ \Rightarrow x(x - y)(z^2 + 1) + z^2(1 + x^2)(1 + xy) = (1 + xy)x^2(1 + z^2) $
after solving we get
$ z^2 = xy \quad \therefore 1 + x^2 \ne 0 $
$ \therefore \tan^2 C = \tan A \cdot \tan B $
$ \Rightarrow \tan A,; \tan C,; \tan B$ are in G.P.
Let $ z $ be a complex number such that $ |z - 6| = 5 $ and $ |z + 2 - 6i| = 5 $. Then the value of $ z^3 + 3z^2 - 15z + 141 $ is equal to
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Solution
Center of first circle $ C_1(6,0),; r_1 = 5 $
Center of second circle $ C_2(-2,6),; r_2 = 5 $
$ \therefore C_1C_2 = r_1 + r_2 $
$ \therefore $ common point $ Z $ is mid point of $ C_1 $ & $ C_2 $
[Image used in solution — two circles diagram]
$ z = 2 + 3i $
$ z^2 = 4z - 13 $
$ z^3 = 3z - 52 $
$ z^3 + 3z^2 - 15z + 141 = 50 $
Let $ ABC $ be an equilateral triangle with orthocenter at the origin and the side $ BC $ on the line $ x + 2\sqrt{2}y = 4 $. If the co-ordinates of the vertex $ A $ are $ (\alpha, \beta) $, then the greatest integer less than or equal to $ |\alpha + \sqrt{2}\beta| $ is
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Solution
$ m_{BC} \cdot m_{AD} = -1 $
$ \Rightarrow \left( -\frac{1}{2\sqrt{2}} \right)\left( \frac{\beta}{\alpha} \right) = -1 $
$ \Rightarrow \beta = 2\sqrt{2}\alpha \quad ...(1)$
$ \therefore OD = \frac{|4|}{\sqrt{1 + 8}} = \frac{4}{3} \Rightarrow AO = \frac{8}{3} $
So $ AD = \frac{8}{3} + \frac{4}{3} = 4 $
$ \Rightarrow \frac{|\alpha + 2\sqrt{2}\beta - 4|}{3} = 4 \Rightarrow \alpha = \frac{16}{9} \text{ or } -\frac{8}{9} $
$ \because A(\alpha,\beta) $ & $(0,0)$ lies on same side of given line
$ \therefore (\alpha,\beta) = \left( -\frac{8}{9}, -\frac{16\sqrt{2}}{9} \right) $
$ \Rightarrow [\alpha + \sqrt{2}\beta] = \left[ \frac{-8 - 32}{9} \right] = 4 $
Let $ S = {1, 2, 3, 4, 5, 6, 7, 8, 9} $. Let $ x $ be the number of 9-digit numbers formed using the digits of the set $ S $ such that only one digit is repeated and it is repeated exactly twice. Let $ y $ be the number of 9-digit numbers formed using the digits of the set $ S $ such that only two digits are repeated and each of these is repeated exactly twice. Then,
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Solution
$ S = {1, 2, 3, \ldots, 9} $
$ x = {^9C_1} \cdot {^9C_2} \cdot \frac{9!}{2} = \frac{9 \times 8 \times 9!}{2} $
$ y = {^9C_2} \cdot {^7C_2} \cdot \frac{9!}{2! \times 2!} = \frac{9 \times 8}{2} \times \frac{7 \times 6}{2} \times \frac{9!}{2! \times 2!} $
$ \Rightarrow \frac{x}{y} = \frac{4}{21} $
$ \Rightarrow 21x = 4y $
Let $ S = { x^3 + ax^2 + bx + c : a, b, c \in \mathbb{N} \text{ and } a, b, c \le 20 } $ be a set of polynomials. Then the number of polynomials in $ S $, which are divisible by $ x^2 + 2 $, is
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Solution
$ x^3 + ax^2 + bx + c = (x^2 + 2)\left(x + \frac{c}{2}\right) $
$ \Rightarrow a = \frac{c}{2},; b = 2 $
$ \Rightarrow b = 2,; a = \frac{c}{2} \in {2, 4, \ldots, 20} $
Number of polynomials in $ S $ will be $ 10 $
A bag contains 10 balls out of which $ k $ are red and $ (10 - k) $ are black, where $ 0 \le k \le 10 $. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:
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Solution
Probability $ = \frac{{^1C_0} \cdot {^9C_3}}{\sum_{k=0}^{10} {^kC_0} \cdot {^{10-k}C_3}} $
$ = \frac{{^9C_3}}{{^0C_3} + {^1C_3} + {^2C_3} + \ldots + {^9C_3}} $
$ = \frac{{^9C_3}}{{^{11}C_4}} = \frac{14}{55} $
If $ \alpha, \beta $, where $ \alpha < \beta $, are the roots of the equation
$ \lambda x^2 - (\lambda + 3)x + 3 = 0 $ such that $ \frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3} $, then the sum of all possible values of $ \lambda $ is:
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Solution
$ \frac{\beta - \alpha}{\alpha \beta} = \frac{1}{3}, \quad \alpha + \beta = \frac{\lambda + 3}{\lambda}, \quad \alpha \beta = \frac{3}{\lambda} $
$ \beta - \alpha = \frac{\alpha \beta}{3} = \frac{1}{\lambda} $
on squaring
$ \alpha^2 + \beta^2 - 2\alpha \beta = \frac{1}{\lambda^2} \quad ...(1)$
$ \alpha^2 + \beta^2 + 2\alpha \beta = \frac{(\lambda + 3)^2}{\lambda^2} \quad ...(2)$
(2) $-$ (1)
$ 4\alpha \beta = \frac{(\lambda + 3)^2 - 1}{\lambda^2} $
$ \Rightarrow \frac{12}{\lambda} = \frac{\lambda^2 + 6\lambda + 8}{\lambda^2} $
$ \Rightarrow \lambda^2 - 6\lambda + 8 = 0 $
$ \Rightarrow \lambda = 0, 2, 4 $
Sum of possible values of $ \lambda $ is $ = 6 $
If $ \int \frac{1 - 5\cos^2 x}{\sin^5 x \cos^2 x} dx = f(x) + C $ where $ C $ is the constant of integration, then $ f\left( \frac{\pi}{6} \right) - f\left( \frac{\pi}{4} \right) $ is equal to
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Solution
$ \int \frac{dx}{\sin^5 x \cos^2 x} - 5 \int \frac{dx}{\sin^5 x} $
$ \int \frac{\sec^2 x dx}{\sin^5 x} - 5 \int \frac{dx}{\sin^5 x} $
By IBP
$ = \frac{\tan x}{\sin^5 x} - \int \frac{\cos x \tan x dx}{\sin^6 x} - 5 \int \frac{dx}{\sin^5 x} $
$ = \frac{\tan x}{\sin^5 x} + C $
$ f(x) = \frac{\tan x}{\sin^5 x} $
$ f\left( \frac{\pi}{6} \right) - f\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{3}} - (\sqrt{2})^5 = 4\sqrt{2} - \frac{32}{\sqrt{3}} $
$ = \frac{4}{\sqrt{3}}(8 - \sqrt{6}) $
Let $ f $ be a polynomial function such that
$ f(x^2 + 1) = x^4 + 5x^2 + 2 $, for all $ x \in \mathbb{R} $.
Then $ \int_0^3 f(x),dx $ is equal to
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Solution
$ f(x^2 + 1) = x^4 + 5x^2 + 2 $
put $ x^2 + 1 = t $
$ \Rightarrow f(t) = (t - 1)^2 + 5(t - 1) + 2 $
$ \Rightarrow f(t) = t^2 + 3t - 2 $
Now,
$ \int_0^3 f(t),dt = \int_0^3 (t^2 + 3t - 2),dt $
$ = \left[ \frac{t^3}{3} + \frac{3t^2}{2} - 2t \right]_0^3 $
$ = \frac{27}{3} + \frac{27}{2} - 6 = \frac{33}{2} $
The area of the region
$ R = {(x, y) : xy \le 8,; 1 \le y \le x^2,; x \ge 0 } $ is
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Solution
$ A = \int_1^2 (x^2 - 1),dx + \int_2^4 \left( \frac{8}{x} - 1 \right) dx $
$ = 8\log_e 4 - \frac{14}{3} = 16\log_e 2 - \frac{14}{3} $
$ = \frac{2}{3}(24\log_e 2 - 7) $
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