Jee Main Math Class 12 Differentiation JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (c)

Explanation: $\frac{d^2 x}{d y^2}=\;\frac{d}{dy}\left(\frac{dx}{dy}\right)$ $=\;\frac{d}{dx}\left(\frac{dx}{dy}\right)\;\frac{dx}{dy}$ $=\;\frac{d}{dx}\left(\frac{1}{\frac{dy}{dx}}\right)\;\frac{dx}{dy}$ $=-\;\frac{1}{\left(\frac{dy}{dz}\right)^2}\cdot\;\frac{d^2 y}{d x^2},\;\frac{1}{\frac{dy}{dx}}$ $=-\;\frac{1}{\left(\frac{dy}{dx}\right)^3}\;\frac{d^2 y}{d x^2}$

Correct Answer: (d)

Explanation: $y = \left\{ x + \left(\sqrt{x^2-1}\right)^{15} \right\} + \left\{ x - \sqrt{x^2-1} \right\}^{15}$ $\frac{dy}{dx} = 15 (x+\sqrt{x^2-1})^{14} 15 (x-\sqrt{x^2-1})^{14} \left(1 - \frac{x}{\sqrt{x^2-1}}\right)$ $\frac{dy}{dx} = \frac{15}{\sqrt{x^2-1}} y$ ....(i) $\sqrt{x^2-1} \frac{dy}{dx} = 15y$ $\frac{x}{\sqrt{x^2-1}} \frac{dy}{dx} + \sqrt{x^2-1} \frac{d^2y}{dx^2} = 15 \frac{dy}{dx}$ $x\frac{dy}{dx} + (x^2-1)\frac{d^2y}{dx^2} = 15\sqrt{x^2-1} \cdot \frac{15}{\sqrt{x^2-1}} \cdot y = 225y$

Correct Answer: (c)

Explanation: $f'(0) = \lim\limits_{h \to 0} \; \frac{f(h)-f(0)}{h}$ $= \lim\limits_{h \to 0} \; \frac{(2^h+2^{-h}) \tan h \sqrt{\tan^{-1}(h^2-h+1)} - 0}{(7 h^2+3 h+1)^3 h}$ $= \sqrt{\pi}$

Correct Answer: (d)

Explanation: $\ln(y) = 3 \sin^{-1} x$ $\frac{1}{y} \cdot y' = 3 \left( \frac{1}{\sqrt{1-x^2}} \right)$ $\Rightarrow y' = \frac{3y}{\sqrt{1-x^2}} \text{ at } x = \frac{1}{2}$ $\Rightarrow y' = \frac{3 e^{3 \left( \frac{\pi}{6} \right)}}{\frac{\sqrt{3}}{2}} = 2 \sqrt{3} e^{\frac{\pi}{2}}$ $\Rightarrow y'' = 3 \left( \frac{ \sqrt{1-x^2} y' - y \frac{1}{2 \sqrt{1-x^2}} (-2 x) }{ (1-x^2) } \right)$ $\Rightarrow (1-x^2) y'' = 3 \left( 3 y + \frac{x y}{\sqrt{1-x^2}} \right)$ $\downarrow \text{ at } x = \frac{1}{2}, y = e^{3 \sin^{-1} \left( \frac{1}{2} \right)} = e^{3 \left( \frac{\pi}{6} \right)} = e^{\frac{\pi}{2}}$ $= 3 e^{\frac{\pi}{2}} \left( 3 + \frac{1}{\sqrt{3}} \right)$ $(1-x^2) y'' - x y' \big|_{x = \frac{1}{2}}$ $= 3 e^{\frac{\pi}{2}} \left( 3 + \frac{1}{\sqrt{3}} \right) - \frac{1}{2} \left( 2 \sqrt{3} e^{\frac{\pi}{2}} \right) = 9 e^{\frac{\pi}{2}}$

Correct Answer: (b)

Explanation: Concept: Treat $f'(1)$ , $f''(2)$ , and $f'''(3)$ as constants. Differentiate the polynomial and substitute the given points to form equations. Explanation: Let $a = f'(1)$ , $b = f''(2)$ , $c = f'''(3)$ . Then $f(x) = x^3 + a x^2 + 2b x + c$ . Differentiate: $f'(x) = 3x^2 + 2a x + 2b$ . $f''(x) = 6x + 2a$ . $f'''(x) = 6$ . Apply conditions: $f'(1)=a \Rightarrow a = 3 + 2a + 2b \Rightarrow a + 2b = -3$ …(1) $f''(2)=b \Rightarrow b = 12 + 2a \Rightarrow 2a - b = -12$ …(2) $f'''(3)=c \Rightarrow c = 6$ . Solve (1) and (2): From (2), $b = 12 + 2a$ . Substitute in (1): $a + 2(12+2a) = -3 \Rightarrow 5a = -27 \Rightarrow a = -\frac{27}{5}$ . Then $b = 12 + 2\left(-\frac{27}{5}\right) = \frac{60}{5} - \frac{54}{5} = \frac{6}{5}$ . Now $f'(5) = 3(25) + 2a(5) + 2b = 75 + 10a + 2b$ . Plug values: $75 + 10\left(-\frac{27}{5}\right) + 2\left(\frac{6}{5}\right) = 75 - 54 + \frac{12}{5} = 21 + \frac{12}{5} = \frac{105}{5} + \frac{12}{5} = \frac{117}{5}$ . Answer: $\frac{117}{5}$ (Option B).