If G(x) = -√(25 - x^2), then lim_x → 1 (G(x) - G(1))/(x - 1) has the value

If G(x) = -√(25 - x^2), then lim_x → 1 (G(x) - G(1))/(x -… — JEE Main 1983 Mathematics PYQ

1983

G(x) = -\sqrt{25 - x^2}

, then

\lim_{x \to 1} \frac{G(x) - G(1)}{x - 1}

has the value

A.
$\frac{1}{\sqrt{24}}$
B.
$\frac{1}{5}$
C.
$-\sqrt{24}$
D.
None of these

Show answer and solution

Correct option: A

Solution

The limit is just G'(1) by definition of the derivative. Differentiating G(x) = -(25 - x²)^(1/2) gives G'(x) = x/√(25 - x²). At x = 1, G'(1) = 1/√(25 - 1) = 1/√24.

[1983]

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