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प्रश्न
Given an interval [a, b] that satisfies hypothesis of Rolle's theorem for the function f(x) = x4 + x2 – 2. It is known that a = – 1. Find the value of b.
उत्तर
f(x) = x4 + x2 – 2
Since the hypothesis of Rolle's theorem are satisfied by f in the interval [a, b], we have
f(a) = f(b), where a = – 1
Now, f(a) = f(– 1)
= (– 1)4 + (– 1)2 – 2
= 1 + 1 – 2
= 0
and f(b) = b4 + b2 – 2
∴ f(a) = f(b) gives 0 = b4 + b2 – 2
i.e. b4 + b2 – 2 = 0.
Since b = 1 satisfies this equation, b = 1 is one of the roots of this equation.
Also, b = – 1 satisfies the equation.
But, if b = – 1, then a = – 1 gives [a, b ] = [– 1, – 1] which is not possible.
∴ b ≠ – 1.
Hence, b = 1.
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