Definite Integration
Differentiation Under Integral
Grade 12
Question:
<p>Let \(f:[0,2]\to\mathbb{R}\) be continuous with \(f(t)+f(2-t)\ge 2\) for all \(t\). If \(g(x)=\int_0^x f(t)\,dt\), show \(g(2)\ge 2\). [JEE Advanced 2014]</p>
g(2) ≥ 2
g(2) ≤ 2
g(2) = 1
g(2) = 0
Step-by-Step Solution
Key Concept: g(2) = \int_0^2 f(t)dt. Use t\to 2-t: g(2) = \int_0^2 f(2-t)dt. Add: 2g(2) = \int_0^2[f(t)+f(2-t)]dt \geq \int_0^2 2 dt = 4. So g(2) \geq 2.
<div class='solution'>
<p>$g(2)=\int_0^2 f(t)\,dt$. Substitute $t\to 2-t$: $g(2)=\int_0^2 f(2-t)\,dt$.</p>
<p>Add: $2g(2)=\int_0^2[f(t)+f(2-t)]\,dt\ge\int_0^2 2\,dt=4$.</p>
<p>$\therefore g(2)\ge 2$. ✓</p>
</div>
Correct Answer: A