Let f: R → R be a continuous function defined by f(x) = 1/e^x + 2e^-x .Statement-1: f(c) = 1/3 , for some c R .Statement-2: 0 < f(x) ≤ 1/2√(2) , for a - Mathematics MCQ AIEEE 2026 | ExamTest.live

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Let f : R \to R be a continuous function defined by f (x) = \frac{1}{e ^{x} + 2 e ^{- x}} .Statement-1: f (c) = \frac{1}{3}, for some c \in R .Statement-2: 0 < f (x) \leq \frac{1}{2 2}, for all x \in R .

A $Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1.$
B $Statement-1 is true, Statement-2 is false.$
C $Statement-1 is false, Statement-2 is true.$
D $Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1.$

Solution & Step-by-step Explanation

Let g (x) = e^{x} + 2 e^{- x} . By AM-GM, \frac{e ^{x} + 2 e ^{- x}}{2} \geq e^{x} \cdot 2 e^{- x} = 2 .So e^{x} + 2 e^{- x} \geq 22 .Since g (x) is continuous and lim_{x \to \pm \infty} g (x) = \infty, its range is [22, \infty) .Thus, the range of f (x) = 1/ g (x) is (0, \frac{1}{2 2}] . 22 \approx 2.828 . \frac{1}{2 2} \approx 0.353 .Since 1/3 \approx 0.333 lies in (0, 0.353], there exists some c such that f (c) = 1/3 .Both statements are true and Statement-2 explains why Statement-1 is possible.

Try it yourself before checking the explanation above.

Let f : R \to R be a continuous function defined by f (x) = \frac{1}{e ^{x} + 2 e ^{- x}} .Statement-1: f (c) = \frac{1}{3}, for some c \in R .Statement-2: 0 < f (x) \leq \frac{1}{2 2}, for all x \in R .

Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1.

Statement-1 is true, Statement-2 is false.

Statement-1 is false, Statement-2 is true.

Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1.

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