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प्रश्न
f(x) = `sqrt(4 - x^2)` in [– 2, 2]
उत्तर
We have, `sqrt(4 - x^2) = (4 - x^2)^(1/2)`
Since (4 – x2) and square root function are continuous and differentiable in their domain, given function f(x) is also continuous and differentiable in [– 2, 2]
Also f(–2) = f(2) = 0
So, conditions of Rolle's theorem are satisfied.
Hence, there exists a real number c ∈ (–2, 2) such that f'(c) = 0.
Now f'(x) = `1/2(4 - x^2)^((-1)/2)(-2x)`
= `- x/sqrt(4 - x^2)`
So, f'(c) = 0
⇒ `"c"/sqrt(4 - "c"^2)` = 0
⇒ c = 0 ∈ (–2, 2)
Hence Rolle's theorem has been verfired.
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