Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>Let $f$ and $g$ be real valued functions defined on interval $(-1, 1)$ such that $g''(x)$ is continuous, $g(0)\ne0$, $g'(0)=0$, $g''(0)\ne0$, and $f(x) = g(x)\sin x$. Statement-1: $\lim_{x\to0}[g(x)\cot x - g(0)\text{cosec}\,x] = f''(0)$. Statement-2: $f'(0) = g(0)$.</p>
<p>Both statements true, S2 is correct explanation</p>
<p>Both true, S2 not correct explanation</p>
<p>S1 true, S2 false</p>
<p>S1 false, S2 true</p>
Step-by-Step Solution
Key Concept: General
<b>L'Hôpital + Differentiation [JEE Advanced 2008]</b><br>$f(x) = g(x)\sin x$. $f'(x) = g'(x)\sin x + g(x)\cos x$<br>$f'(0) = g'(0)\cdot0 + g(0)\cdot1 = g(0)$ ✓ Statement 2 is true.<br>$f''(x) = g''(x)\sin x + 2g'(x)\cos x - g(x)\sin x$<br>$f''(0) = 0 + 2g'(0) - 0 = 0$... Hmm, $f''(0)=0$ since $g'(0)=0$.<br>Statement 1 LHS: $\lim_{x\to0}\frac{g(x)\cos x - g(0)}{sin x} = \lim_{x\to0}\frac{g(x)-g(0)}{x} = g'(0)=0$<br>This equals $f''(0)=0$. ✓ Both true. S2 explains the derivative formula.<br><b>Answer: A</b>
Correct Answer: A