﻿ Between any two real roots of the equation ex sin x – 1 = 0, the equation excos x + 1 = 0 has : Kaysons Education

Between Any Two Real Roots Of The Equation ex sin x – 1 = 0, The Equation excos x + 1 = 0 Has

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Question

Solution

Correct option is

At least one root

be two roots of the equation ex sin x– 1 = 0. Such that α < β. Then,

Clearly, f (x) is continuous on [α, β] and differentiable on (α, β).

∴ by Rolle’s theorem there exists c Ïµ (α, β) such that f’(c) = 0

SIMILAR QUESTIONS

Q1

Find c of the Lagrange’s mean value theorem for which

Q2

Let f (x) and g (x) be differentiable for 0 ≤ x ≤ 2 such that (0) = 2, g(0) = 1 and f (2) = 8. Let there exists a real number c in [0, 2] such that f’(c) = 3g’(c) then the value of g(2) must be:

Q3

If f (x) = loge x and g(x) = x2 and c Ïµ (4, 5), then  is equal to:

Q5

In [0, 1] lagrange’s mean value theorem is not applicable to

Q6

Let f (x) satisfy the requirement of lagrange’s mean value theorem in [0, 2]. If f (0) and

Q7

Let f : [2, 7] and [0, ) be a continuous and differentiable function.

Then the value of (f (7) – f (2))  is (where c Ïµ (2, 7))

Q8

The equation sin x + x cos x = 0 has at least one root in the interval

Q9

Let f (x) = ax5 + bx4 + cx3 + dx2 + ex, where abcde Ïµ R and f (x) = 0 has a positive root α, then

Q10

f (x) is a polynomial of degree 4 with real coefficients such that f (x) = 0 is satisfied by x = 1, 2, 3 only, then f’(1). f’(2). f’(3) is equal to: