IIT-JEE 2008 · Paper 1 · Mathematics Q11

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MathematicsAssertion and reason+3 / -1Question 11 of 69

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: $\displaystyle\lim_{x\to0}\left[g(x)\cot x-g(0)\operatorname{cosec}x\right]=f''(0)$ .

and

STATEMENT-2: $f'(0)=g(0)$ .

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

Answer and worked solution

Official answer: A

Statement 2. $f'(x)=g'(x)\sin x+g(x)\cos x$ , so $f'(0)=g(0)$ . True.

Statement 1. Using $g(x)\cos x=f'(x)-g'(x)\sin x$ and $g(0)=f'(0)$ :
$g(x)\cot x-g(0)\operatorname{cosec}x=\frac{g(x)\cos x-g(0)}{\sin x}=\frac{f'(x)-f'(0)}{\sin x}-g'(x).$
As $x\to0$ this tends to $f''(0)-g'(0)=f''(0)$ . True.

Explanation. The proof of Statement 1 uses $f'(0)=g(0)$ , so Statement 2 explains it.

(A)

Topics:Limits Higher-order derivatives

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IIT-JEE 2008 · Paper 1, Mathematics question 11