Let f be defined by f(x) = begin{cases} frac{ | vedclass

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(B) For $x 
eq 0$,$f(x) = \frac{	an x}{x}$. Using the Taylor series expansion for $	an x$ near $x = 0$:$	an x = x + \frac{x^3}{3} + \frac{2x^5}{15

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Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is a correct explanation for Statement-$1$.

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(B) For $x 
eq 0$,$f(x) = \frac{	an x}{x}$. Using the Taylor series expansion for $	an x$ near $x = 0$:$	an x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$So,$f(x) = \frac{x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots}{x} = 1 + \frac{x^2}{3} + \frac{2x^4}{15} + \dots$Since $f(0) = 1$,the function is continuous at $x = 0$.For $x$ near $0$,$f(x) \approx 1 + \frac{x^2}{3}$.Since the coefficient of $x^2$ is positive $(1/3 > 0)$,$f(x)$ has a local minimum at $x = 0$. Thus,Statement-$1$ is true.Now,$f'(x) = \frac{d}{dx} (1 + \frac{x^2}{3} + \dots) = \frac{2x}{3} + \dots$At $x = 0$,$f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{(1 + h^2/3 + \dots) - 1}{h} = \lim_{h 	o 0} \frac{h}{3} = 0$.Thus,Statement-$2$ is true.Since $f'(0) = 0$ and $f''(0) = 2/3 > 0$,the second derivative test confirms $x = 0$ is a local minimum. Thus,Statement-$2$ is a correct explanation for Statement-$1$.

Let $f$ be defined by $f(x) = \begin{cases} \frac{\tan x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$.
Statement-$1$: $x = 0$ is a point of local minima for $f$.
Statement-$2$: $f'(0) = 0$.

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Let $f$ be defined by $f(x) = \begin{cases} \frac{\tan x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$.Statement-$1$: $x = 0$ is a point of local minima for $f$.Statement-$2$: $f'(0) = 0$.