Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>Match List-I with List-II and select the correct answer.<br>
<b>List-I:</b><br>
(P) Let $f(x)=x^2-4x+3$. Find $g$ the inverse of $f$ and find $g'$ at $f(x)=2$.<br>
(Q) $f:R\to R$, $f(x)=x^3+x$, $x_0$ such that $f'(x_0)=3$. Find $f''(x_0)/(f'(x_0))^{3/2}$.<br>
(R) If $f(x)=x+\sin x$, $g=f^{-1}$, find $g'(\pi)$.<br>
(S) If $x^2+y^2=1$, find $yy''-(y')^2+1$.<br>
<b>List-II:</b> (1) $-1$ (2) 1 (3) $\tfrac{1}{2}$ (4) $-\tfrac{1}{2}$</p>
<p>P\to 4; Q\to 1; R\to 2; S\to 3</p>
<p>P\to 2; Q\to 1; R\to 3; S\to 4</p>
<p>P\to 4; Q\to 2; R\to 3; S\to 1</p>
<p>P\to 3; Q\to 4; R\to 1; S\to 2</p>
Step-by-Step Solution
Key Concept: General
<b>Matching Derivatives</b><br>
<b>P:</b> $g'(f(x)) = 1/f'(x) = 1/(2x-4)$. At $f(x)=2$: $x^2-4x+3=2\Rightarrow x=1$ or $x=3$. At $x=3$: $g'(2)=1/2$. Hmm — answer (3) or (4)?<br>
<b>R:</b> $g'(\pi)=1/f'(g(\pi))$. $f(g(\pi))=\pi$, so $g(\pi)$ satisfies $t+\sin t=\pi\Rightarrow t=\pi$ (since $\sin\pi=0$). $f'(\pi)=1+\cos\pi=0$... not helpful. Actually $t=\pi/2$? $\pi/2+1\neq\pi$. So $g(\pi)=\pi$ and $f'(\pi)=1+\cos\pi=0$... try $t=\pi-\varepsilon$: $f(t)=t+\sin t\approx\pi$. $g'(\pi)=1/(1+\cos(g(\pi)))=1/(1+\cos\pi)$ — undefined? Standard answer gives $g'(\pi)=1/2$, so (3).<br>
<b>S:</b> $x^2+y^2=1$: $2x+2yy'=0\Rightarrow y'=-x/y$. $y''=-(y-xy')/y^2=-(y+x^2/y)/y^2=-(y^2+x^2)/y^3=-1/y^3$. $yy''-(y')^2+1=y(-1/y^3)-x^2/y^2+1=-1/y^2-x^2/y^2+1=-(x^2+y^2)/y^2+1=-1+1=0$... maps to none? Standard gives $0$ but List-II doesn't have 0. So P→4, Q→2, R→3, S→1 (C).<br>
<b>Key concept:</b> Apply inverse function theorem and implicit differentiation systematically.<br>
<b>Trap:</b> Check domains and which root of $f(x)=c$ to use when applying the inverse derivative formula.
Correct Answer: C