If f ( x ) is differentiable and strictly increasing function, then the value of lim x 0 f ( x 2 ) - f ( x ) f ( x ) - f ( 0 ) is [2004]

Question

If $f (x)$ is differentiable and strictly increasing function, then the value of $\lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ is [2004]

Smart Achievers · Accepted Answer

(3)

Let $L = \lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ $[using L.H. Rule]$

$[∵ f' (a) > 0 as f being strictly increasing]$

$L = \lim_{x \to 0} \frac{f' (x^{2}) \cdot 2 x - f' (x)}{f' (x)}$

$= \lim_{x \to 0} \frac{f' (x^{2}) \cdot 2 x}{f' (x)} - 1$

$= 0 - 1$

$= - 1$

Solution

Q.
If $f (x)$ is differentiable and strictly increasing function, then the value of $\lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ is [2004]

1 1

2 0

3 - 1

4 2

Ans.
(3)

Let $L = \lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ $[using L.H. Rule]$

$[∵ f' (a) > 0 as f being strictly increasing]$

$L = \lim_{x \to 0} \frac{f' (x^{2}) \cdot 2 x - f' (x)}{f' (x)}$

$= \lim_{x \to 0} \frac{f' (x^{2}) \cdot 2 x}{f' (x)} - 1$

$= 0 - 1$

$= - 1$

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Solution

Q. If f(x) is differentiable and strictly increasing function, then the value of limx→0f(x2)-f(x)f(x)-f(0) is [2004]

1 1

2 0

3 - 1

4 2

Ans. (3) Let L=limx→0f(x2)-f(x)f(x)-f(0) [using L.H. Rule] [∵ f'(a)>0 as f being strictly increasing] L=limx→0f'(x2)·2x-f'(x)f'(x) =limx→0f'(x2)·2xf'(x)-1 =0-1 =-1

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Q.
If $f (x)$ is differentiable and strictly increasing function, then the value of $\lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ is [2004]

Ans.
(3)

Let $L = \lim_{x \to 0} \frac{f (x^{2}) - f (x)}{f (x) - f (0)}$ $[using L.H. Rule]$

$[∵ f' (a) > 0 as f being strictly increasing]$

$L = \lim_{x \to 0} \frac{f' (x^{2}) \cdot 2 x - f' (x)}{f' (x)}$

$= \lim_{x \to 0} \frac{f' (x^{2}) \cdot 2 x}{f' (x)} - 1$

$= 0 - 1$

$= - 1$