GATE 2025 EC Q.51: solution and answer

GATE 2025 · ECQ.51 · 2 marks · MSQ

Consider a non-negative function $f (x)$ which is continuous and bounded over the interval [2, 8]. Let $M$ and $m$ denote, respectively, the maximum and the minimum values of $f (x)$ over the interval.

Among the combinations of $α$ and $β$ given below, choose the one(s) for which the inequality

β \leq \int_{2}^{8} f (x) d x \leq α

is guaranteed to hold.

(A)
$β = 5 m, α = 7 M$
(B)
$β = 6 m, α = 5 M$
(C)
$β = 7 m, α = 6 M$
(D)
$β = 7 m, α = 5 M$

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Answer and solution

ANSWER

Only option (A), $β = 5 m, α = 7 M$ , is guaranteed. From $m \leq f (x) \leq M$ we get $6 m \leq \int_{2}^{8} f (x) d x \leq 6 M$ . Because $m, M \geq 0$ , this gives $5 m \leq \int_{2}^{8} f (x) d x \leq 7 M$ . Options (B), (C) and (D) all fail for $f (x) = 1$ .

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

Bounds on the integrand: $f$ is continuous on the closed interval $[2, 8]$ , so its maximum $M$ and minimum $m$ exist, and $m \leq f (x) \leq M$ for every $x$ in the interval.
Integrate over $[2, 8]$ , whose length is $8 - 2 = 6$ . Integration preserves inequalities, so $6 m \leq \int_{2}^{8} f (x) d x \leq 6 M .$
Both bounds are attained: for a constant function $f (x) = c \geq 0$ we have $m = M = c$ , and the integral is exactly $6 c$ . So a combination $β = p m, α = q M$ works for every admissible $f$ only if $p \leq 6$ and $q \geq 6$ .
Non-negativity matters: $f \geq 0$ gives $m \geq 0$ and $M \geq 0$ , so $5 m \leq 6 m \leq 7 m$ and $5 M \leq 6 M \leq 7 M$ . (If $m$ could be negative, $5 m$ would be larger than $6 m$ and the argument below would fail.)
(A) $β = 5 m \leq 6 m \leq \int_{2}^{8} f (x) d x \leq 6 M \leq 7 M = α$ . The inequality holds for every admissible $f$ , so it is guaranteed.
(B) $α = 5 M$ : take $f (x) = 1$ , so $M = 1$ . The integral is $6$ , but $α = 5 < 6$ . Fails.
(C) $β = 7 m$ : take $f (x) = 1$ , so $m = 1$ . The integral is $6$ , but $β = 7 > 6$ . Fails.
(D) With $f (x) = 1$ , both $β = 7 > 6$ and $α = 5 < 6$ . Fails. Only (A) is guaranteed.

KEY IDEA

Bound the integral using the interval length: $(b - a) m \leq \int_{a}^{b} f (x) d x \leq (b - a) M$ , which here is $6 m \leq \int_{2}^{8} f \leq 6 M$ . A constant function reaches both bounds, so a lower bound $p m$ needs $p \leq 6$ and an upper bound $q M$ needs $q \geq 6$ . The step from $6 m$ down to $5 m$ and from $6 M$ up to $7 M$ is valid only because $f \geq 0$ .

WATCH OUT FOR

Using the upper limit $8$ (or the lower limit $2$ ) as the multiplier instead of the interval length $8 - 2 = 6$ . Another mistake is dropping the non-negativity condition: if $f$ could be negative, then $5 m > 6 m$ when $m < 0$ , and even (A) fails. For example, $f (x) = - 1$ gives an integral of $- 6$ , which is less than $5 m = - 5$ . A third mistake is marking (C) because $α = 6 M$ is a valid upper bound, while missing that $β = 7 m$ breaks the lower bound. Both halves must hold.

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GATE 2025 EC Q.51: solution and answer

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