IIT-JEE 2008 · Paper 1 · Mathematics Q19

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MathematicsSingle correct+4 / -1Question 19 of 69

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Paragraph for Question Nos. 18 to 20

Consider the functions defined implicitly by the equation $y^3-3y+x=0$ on various intervals in the real line.

If $x\in(-\infty,-2)\cup(2,\infty)$ , the equation implicitly defines a unique real valued differentiable function $y=f(x)$ .

If $x\in(-2,2)$ , the equation implicitly defines a unique real valued differentiable function $y=g(x)$ satisfying $g(0)=0$ .

The area of the region bounded by the curve $y=f(x)$ , the $x$ -axis, and the lines $x=a$ and $x=b$ , where $-\infty<a<b<-2$ , is

\displaystyle\int_a^b\frac{x}{3\left((f(x))^2-1\right)}dx+bf(b)-af(a)

\displaystyle-\int_a^b\frac{x}{3\left((f(x))^2-1\right)}dx+bf(b)-af(a)

\displaystyle\int_a^b\frac{x}{3\left((f(x))^2-1\right)}dx-bf(b)+af(a)

\displaystyle-\int_a^b\frac{x}{3\left((f(x))^2-1\right)}dx-bf(b)+af(a)

Answer and worked solution

Official answer: A

Sign of $f$ . For $x<-2$ , $f^3-3f=-x>2$ , so $f>2>0$ , and the area is $\int_a^bf(x)\,dx$ .

Integrate by parts.
$\int_a^bf\,dx=\Big[xf\Big]_a^b-\int_a^bxf'\,dx=bf(b)-af(a)+\int_a^b\frac{x}{3\left(f^2-1\right)}dx,$
using $f'=-\dfrac1{3(f^2-1)}$ .

(A)

Topics:Area under a curve Integration by parts

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IIT-JEE 2008 · Paper 1, Mathematics question 19