Let f- be a function such that f2 - 4 and f-2 - 1- Then- the value of limx- 2x2f2-4fxx-2 is equal to

Lim_x→ 2x^2f(2) 4f(x)x 2 nbsp; [ 00form ] By L'hospital rule, we have =lim _x→ 22x· f(2) 4f'(x) =4f(2) 4· f'(2) = 4 ·4 4·1 = 12 ...

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Standard XII

Mathematics

Variable Separable Method

Let f:ℝ→ℝ be ...

Question

Let $f : R \to R$ be a function such that $f (2) = 4$ and $f^{'} (2) = 1.$ Then, the value of $lim x \to 2 \frac{x^{2} f (2) - 4 f (x)}{x - 2}$ is equal to

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Solution

The correct option is D $12$
$lim x \to 2 \frac{x^{2} f (2) - 4 f (x)}{x - 2}$ $[\frac{0}{0} form]$
By L'hospital rule, we have
$= lim x \to 2 2 x \cdot f (2) - 4 f^{'} (x)$
$= 4 f (2) - 4 \cdot f^{'} (2)$
$= 4 \cdot 4 - 4 \cdot 1$
$= 12$

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Let f:R→R be a function such that f(2)=4 and f′(2)=1. Then, the value of limx→2x2f(2)−4f(x)x−2 is equal to

The correct option is D 12limx→2x2f(2)−4f(x)x−2 [00form] By L'hospital rule, we have =limx→22x⋅f(2)−4f′(x) =4f(2)−4⋅f′(2) =4⋅4−4⋅1 =12

Let $f : R \to R$ be a function such that $f (2) = 4$ and $f^{'} (2) = 1.$ Then, the value of $lim x \to 2 \frac{x^{2} f (2) - 4 f (x)}{x - 2}$ is equal to

The correct option is D $12$
$lim x \to 2 \frac{x^{2} f (2) - 4 f (x)}{x - 2}$ $[\frac{0}{0} form]$
By L'hospital rule, we have
$= lim x \to 2 2 x \cdot f (2) - 4 f^{'} (x)$
$= 4 f (2) - 4 \cdot f^{'} (2)$
$= 4 \cdot 4 - 4 \cdot 1$
$= 12$