Let \(f(x)=2+\cos x\) for all real \(x\). STATEMENT-1 : For each real \(t\), there exists a point \(c\) in \([t, t+\pi]\) such that \(f^{\prime}(c)=0\). because STATEMENT-2 : \(f(t)=f(t+2 \pi)\) for each real \(t\).

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

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

(C) Statement-1 is True, Statement-2 is False

(D) Statement-1 is False, Statement-2 is True

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Application of Derivatives

Let

f (x) = 2 + cos x

for all real

x

.

STATEMENT-1 : For each real

t

, there exists a point

c

[t, t + π]

such that

f^{'} (c) = 0

.

because

STATEMENT-2 :

f (t) = f (t + 2 π)

for each real

t

(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

Answer: (B)

For Statement-1:

f (x) = 2 + cos x

is continuous and differentiable everywhere.

f^{'} (x) = - sin x

.

In any interval

[t, t + π]

, the sine function completes half a period and must equal zero at least once. Therefore, there exists

c \in [t, t + π]

such that

f^{'} (c) = - sin c = 0

. Statement-1 is TRUE.

For Statement-2:

f (t + 2 π) = 2 + cos (t + 2 π) = 2 + cos t = f (t)

. Statement-2 is TRUE.

However, Statement-2 (periodicity with period

2 π

) does not directly explain Statement-1 (existence of a critical point in an interval of length

π

). The correct explanation would involve Rolle's theorem or the intermediate value theorem applied to

f^{'}

. Therefore, Statement-2 is NOT a correct explanation for Statement-1.

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option