JEE Advanced 2015 · Paper 2 · Mathematics Q56

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MathematicsMultiple correct+4 / -2Question 56 of 60

Let $f'(x)=\dfrac{192x^3}{2+\sin^4\pi x}$ for all $x\in\mathbb R$ with $f\left(\dfrac12\right)=0$ . If $m\le\displaystyle\int_{1/2}^1f(x)\,dx\le M$ , then the possible values of $m$ and $M$ are

m=13,\ M=24

m=\dfrac14,\ M=\dfrac12

m=-11,\ M=0

m=1,\ M=12

Answer and worked solution

Official answer: D

Bound the derivative. For $x>0$ , $64x^3\le f'(x)\le96x^3$ .

Bound f. With $f(\frac12)=0$ , integrate from $\frac12$ :
$16x^4-1\le f(x)\le24x^4-\frac32\quad\text{on }[\tfrac12,1].$

Bound the integral.
$2.6\le\int_{1/2}^1f\,dx\le3.9.$
Only $m=1$ , $M=12$ is consistent with this.

(D)

Topics:Properties of definite integrals

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JEE Advanced 2015 · Paper 2, Mathematics question 56