JEE MAIN 2026 Previous Year Questions (PYQs) – Page 15 of 25

$ 40^n = 2^{3n} \times 5^n $

$ E_2(60!) = \left[ \frac{60}{2} \right] + \left[ \frac{60}{2^2} \right] + \left[ \frac{60}{2^3} \right] + \left[ \frac{60}{2^4} \right] + \left[ \frac{60}{2^5} \right] $

$ = 30 + 15 + 7 + 3 + 1 = 56 $

$ E_5(60!) = \left[ \frac{60}{5} \right] + \left[ \frac{60}{5^2} \right] $

$ = 12 + 2 = 14 $

$ \therefore n = \min \left( \frac{56}{3}, 14 \right) = 14 $

$ \sqrt{2} = \frac{\left| \begin{matrix} -1 & 1 & 3 \ \alpha & -1 & -\alpha \ \alpha & 2 & 2\alpha \end{matrix} \right|}{\left| \begin{matrix} i & j & k \ \alpha & -1 & -\alpha \ \alpha & 2 & 2\alpha \end{matrix} \right|} $

$ \sqrt{2} = \frac{-1(-2\alpha + 2\alpha) - 1(2\alpha^2 + \alpha^2) + 3(2\alpha + \alpha)}{\sqrt{i(-2\alpha + 2\alpha)^2 + j(2\alpha^2 + \alpha^2)^2 + k(2\alpha + \alpha)^2}} $

$ \sqrt{2} = \frac{-3\alpha^2 + 9\alpha}{\sqrt{9\alpha^4 + 9\alpha^2}} $

$ \sqrt{2} = \frac{\alpha + 3}{\sqrt{\alpha^2 + 1}} $

$ \Rightarrow 2\alpha^2 + 2 = \alpha^2 + 9 - 6\alpha $

$ \alpha^2 + 6\alpha - 7 = 0 $

$ (\alpha + 7)(\alpha - 1) = 0 $

$ \alpha = -7,; 1 $

sum $ = -7 + 1 = -6 $

at $ \alpha = 0 \Rightarrow f(0) $

$x = 0,; x = 1,; y^2 = x$

$y = |0x - 5| - |1 - 0x| + 0x^2$

$y = 5 - 1 = 4$

$A_1 = \int_0^1 (4 - \sqrt{x}) dx$

$ = 4x - \frac{2}{3}x^{3/2} \Big|_0^1 $

$ = 4 - \frac{2}{3} = \frac{10}{3}$

at $ \alpha = 1 \Rightarrow f(1) $

$x = 0,; x = 1,; y^2 = x$

$y = |x - 5| - |1 - x| + x^2$

$x \in (0,1)$

$y = 5 - x - (1 - x) + x^2$

$y = 4 + x^2$

$A_2 = \int_0^1 \left( (4 + x^2) - \sqrt{x} \right) dx$

$ = 4x + \frac{x^3}{3} - \frac{2}{3}x^{3/2} \Big|_0^1 $

$ = 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}$

$f(0) + f(1) = |A_1 + A_2| = \left| \frac{10}{3} + \frac{11}{3} \right| = \left| \frac{21}{3} \right| = 7$

$ -1 \le \frac{1}{x^2 - 2x - 2} \le 1 $

$ \Rightarrow 1 + x^2 - 2x - 2 \ge 0 \Rightarrow \frac{(x-1)^2 - 2}{(x-1)^2 - 3} \ge 0 $

$ \Rightarrow \frac{(x-1-\sqrt{2})(x-1+\sqrt{2})}{(x-1-\sqrt{3})(x-1+\sqrt{3})} \ge 0 $

$ \Rightarrow x \in (-\infty, 1-\sqrt{3}] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [1+\sqrt{3}, \infty) \quad ...(1)$

$ \Rightarrow 1 - \frac{1}{x^2 - 2x - 2} \ge 0 \Rightarrow \frac{x^2 - 2x - 3}{x^2 - 2x - 2} \ge 0 $

$ \Rightarrow \frac{(x+1)(x-3)}{(x-1-\sqrt{3})(x-1+\sqrt{3})} \ge 0 $

$ \Rightarrow x \in (-\infty, -1] \cup [1-\sqrt{3}, 1+\sqrt{3}] \cup [3, \infty) \quad ...(2)$

$(1) \cap (2)$

$ \Rightarrow x \in (-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty) $

$ \therefore \alpha + \beta + \gamma + \delta = 4 $

$ \sum y_i = a \sum x_i + \sum b = a(1 + 2 + \cdots + 19) + 19b $

$ \sum y_i = \frac{a \cdot 19 \cdot 20}{2} + 19b $

$ \Rightarrow 30 = 10a + b \quad ...(1)$

Variance of $ X = \frac{\sum x_i^2}{19} - \left( \frac{\sum x_i}{19} \right)^2 $

$ = \frac{19 \cdot 20 \cdot 39}{19 \cdot 6} - (10)^2 = 30 $

Variance of $ Y = a^2 \times (\text{variance of } X)$

$ 750 = a^2 \times 30 \Rightarrow a^2 = 25 \Rightarrow a = \pm 5 $

if $ a = 5 \Rightarrow b = 30 - 50 = -20 $

if $ a = -5 \Rightarrow b = 30 + 50 = 80 $

sum of values of $ b = 80 - 20 = 60 $

$ f(x) = |\ln x| - |x - 1| $

$ = \begin{cases} \ln x - (x - 1), & x \ge 1 \ -\ln x + (x - 1), & 0 < x < 1 \end{cases} $

$ = \begin{cases} \ln x - x + 1, & x \ge 1 \ -\ln x + x - 1, & 0 < x < 1 \end{cases} $

$ f'(x) = \begin{cases} \frac{1}{x} - 1, & x \ge 1 \ -\frac{1}{x} + 1, & 0 < x < 1 \end{cases} $

$ f'(1^+) = f'(1^-) = 0 \Rightarrow f(x) $ is differentiable $ \forall x > 0 $

$ f'(x) < 0 ; \forall x > 1 $

$ f'(x) < 0 ; \forall 0 < x < 1 $

$ \Rightarrow f(x) $ is decreasing $ \forall x \in (0, \infty) $

$ \lim_{t \to x} \frac{2t f(x) - x^2 f(t)}{-1} = 3 $

$ x^2 f'(x) - 2x f(x) = 3 $

$ \frac{dy}{dx} - \frac{2y}{x} = \frac{3}{x^2} $

I.F. $ = e^{\int \frac{-2}{x} dx} = e^{-2\log x} = \frac{1}{x^2} $

$ \frac{y}{x^2} = \int \frac{3}{x^4} dx $

$ \frac{y}{x^2} = -\frac{1}{x^3} + c \Rightarrow y = cx^2 - \frac{1}{x} $

$ f(1) = 2 = c - 1 \Rightarrow c = 3 $

$ f(x) = 3x^2 - \frac{1}{x} $

$ f(2) = 12 - \frac{1}{2} $

$ 2f(2) = 23 $