JEE Challenger
More from Application of Derivatives
MathMedAdv 2007 • Paper 2

Existence of Stationary Points of Cosine in Subintervals

This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Let f(x)=2+cos⁡xf(x) = 2 + \cos x for all real xx.

STATEMENT-1 :\textbf{STATEMENT-1 :} For each real tt, there exists a point cc in [t,t+π][t, t + \pi] such that f′(c)=0f'(c) = 0.

because\textbf{because}

STATEMENT-2 :\textbf{STATEMENT-2 :} f(t)=f(t+2π)f(t) = f(t + 2\pi) for each real tt.

Options

A

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Correct
C

Statement-1 is True, Statement-2 is False

D

Statement-1 is False, Statement-2 is True

Topics & Concepts

Application of DerivativesRolles TheoremTrigonometric FunctionsRoots of Derivative

Step-by-Step Solution

  1. Check Statement-1:
    The derivative is f′(x)=−sin⁡xf'(x) = -\sin x. The zeros of f′(x)f'(x) occur at x=kπx = k\pi for k∈Zk \in \mathbb{Z}. The distance between any two consecutive zeros is π\pi. Therefore, any closed interval [t,t+π][t, t + \pi] of length π\pi on the real line must contain at least one integer multiple of π\pi. Hence, there always exists c∈[t,t+π]c \in [t, t + \pi] such that f′(c)=0f'(c) = 0. Statement-1 is True.

  2. Check Statement-2:
    Since cosine has period 2π2\pi, f(t+2π)=2+cos⁡(t+2π)=2+cos⁡t=f(t)f(t + 2\pi) = 2 + \cos(t + 2\pi) = 2 + \cos t = f(t). Statement-2 is True.

  3. Explanation Check:
    Periodicity with period 2π2\pi guarantees by Rolle's Theorem that f′(c)=0f'(c) = 0 in an interval of length 2π2\pi, but does not in itself explain why a zero exists in every subinterval of smaller length π\pi. Thus, Statement-2 is NOT the correct explanation for Statement-1.

Correct Answer: B

PreviousNext