If G (x) = - √ 25 - x2, then lim x → 1 ( G (x) - G (1)/ x - 1) has the value

Question

If G (x) = - √ 25 - x2, then lim x → 1 ( G (x) - G (1)/ x - 1) has the value (A) (1/√24) (B) (1/5) (C)  - √24 (D) None of these

Tardigrade Admin · Accepted Answer

Given, G (x) = - √ 25 - x2 ∴ lim x → 1 ( G (x) - G (1)/ x - 1) = lim x → 1 ( G ; (x) - 0/ 1 - 0) [using L' Hospital's rule] = G' (1) = (1/√ 24) Bigg [ ∵ G (x) = - √ 25 - x2 ⇒ G' (x) = ( 2x/ 2 √ 25 - x2 ) Bigg ]

Q. If G (x) = - $25 - x^{2}, t h e n l i m_{x \to 1} \frac{G ( x ) - G ( 1 )}{x - 1}$ has the value

$\frac{1}{24}$

$\frac{1}{5}$

$- 24$

None of these

Given, G (x) = - $25 - x^{2}$
$∴ l i m_{x \to 1} \frac{G ( x ) - G ( 1 )}{x - 1} = l i m_{x \to 1} \frac{G ; ( x ) - 0}{1 - 0}$
[using L' Hospital's rule]
= G' (1) = $\frac{1}{2 4}$
$[∵ G (x) = - 25 - x^{2} \Rightarrow G^{'} (x) = \frac{2 x}{2 25 - x ^{2}}]$

Q. If G (x) = - 25−x2​,thenlimx→1​x−1G(x)−G(1)​ has the value

24​1​

51​

−24​