Practicing Success
Let f and g be real valued functions defined on interval $(-1,1)$ such that $g''(x)$ is continuous, $g(0)=0, g''(0) \neq 0$, and $f(x)=g(x) \sin x$, $g'(0)=0$. Statement-1: $\lim\limits_{x \rightarrow 0}\{g(x) \cot x-g(0) ~cosec x \}=f''(0)$ 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. |
Statement-1 is True, Statement-2 is True; statement-2 is not a correct explanation for Statement-1. |
$\lim\limits_{x \rightarrow 0}\{g(x) \cot x-g(0) ~cosec x\}$ $=\lim\limits_{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x}$ ($\frac{0}{0}$ form) $=\lim\limits_{x \rightarrow 0} \frac{g'(x) \cos x-g(x) \sin x}{\cos x}$ [Using L' Hospital's Rule] = 0 [∵ g'(0) = 0 and g(0) = 0] Now, f(x) = g(x) sin x $\Rightarrow f'(x)=g'(x) \sin x+g(x) \cos x$ ......(i) $\Rightarrow f''(x)=g''(x) \sin x+2 g'(x) \cos x-g(x) \sin x$ $\Rightarrow f''(0) = 0$ ∴ $\lim\limits_{x \rightarrow 0}\{g(x) \cot x-g(0) ~cosec x\}=f''(0)$ [∵ f''(0) = 0] So, statement - 1 is true. From (i), we have f'(0) = g(0) So, statement - 2 is true. But, statement - 2 is not a correct explanation for statement - 1. |