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प्रश्न
Discuss the applicability of Rolle’s theorem on the function given by f(x) = `{{:(x^2 + 1",", "if" 0 ≤ x ≤ 1),(3 - x",", "if" 1 ≤ x ≤ 2):}`
उत्तर
We have, f(x) = `{{:(x^2 + 1",", "if" 0 ≤ x ≤ 1),(3 - x",", "if" 1 ≤ x ≤ 2):}`
We know that polynomial function is everywhere continuous and differentiability.
So, f(x) is continuous and differentiable at all points except possibly at x = 1.
Now `lim_(x -> 1^-) (x^2 + 1)` = 1 + 1 = 2
And `lim_(x -> 1^+) (3 - x)` = 3 – 1 = 2
Also f(1) = 12 + 1 = 2
So, f(x) is continuous at x = 1
Also f'(x) = `{{:(2x",", "if" 0 < x < 1),(-x",", "if" 1 < x 2):}`
f'(1) = 2(1) = 2
And f'(1) = –1
Thus f'(1) ≠ f'(1).
So, f(x) is not differentiable at x = 1
Hence, Rolle's theorem is not applicable on the interval [0, 2].
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