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GO Classes | IIITH PGEE 2026 Mock Test 12 | Question: 43

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By the Mean Value Theorem (MVT), for $a=3$ and $b=6$, there exists $c \in(3,6)$ such that:

$$
f^{\prime}(c)=\frac{f(6)-f(3)}{6-3}=\frac{f(6)-3}{3}
$$

Since $f^{\prime}(x) \geq 2$ for $3 \leq x \leq 6$, we have:

$$
\frac{f(6)-3}{3} \geq 2
$$

Multiplying both sides by $3 :$

$$
f(6)-3 \geq 6 \quad \Rightarrow \quad f(6) \geq 9.
$$

Thus, the smallest possible value of $f(6)$ is $9.$
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