IIT-JEE 2010 · Paper 2 · Mathematics Q32

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MathematicsSingle correct+3 / -1Question 32 of 57

Passage

Paragraph for Questions 31 to 33

Consider the polynomial $f(x)=1+2x+3x^2+4x^3.$ Let $s$ be the sum of all distinct real roots of $f(x)$ and let $t=|s|$ .

The area bounded by the curve $y=f(x)$ and the lines $x=0$ , $y=0$ and $x=t$ , lies in the interval

\left(\dfrac34,3\right)

\left(\dfrac{21}{64},\dfrac{11}{16}\right)

(9,10)

\left(0,\dfrac{21}{64}\right)

Answer and worked solution

Official answer: A

Area. On $[0,t]$ , $f>0$ , so the area is
$\int_0^tf\,dx=t+t^2+t^3+t^4,$
an increasing function of $t$ .

Bounds. $t\in\left(\frac12,\frac34\right)$ : at $t=\frac12$ the area is $0.9375$ , and at $t=\frac34$ it is about $2.05$ . So the area lies in $\left(\frac34,3\right)$ .

(A)

Topics:Area under a curve

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IIT-JEE 2010 · Paper 2, Mathematics question 32