IIT-JEE 2010 · Paper 2 · Mathematics Q31

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MathematicsSingle correct+3 / -1Question 31 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 real number $s$ lies in the interval

\left(-\dfrac14,0\right)

\left(-11,-\dfrac34\right)

\left(-\dfrac34,-\dfrac12\right)

\left(0,\dfrac14\right)

Answer and worked solution

Official answer: C

One real root. $f'(x)=2+6x+12x^2$ has discriminant $36-96<0$ , so $f'>0$ and $f$ is increasing: exactly one real root, which is $s$ .

Locate it. $f\left(-\frac12\right)=1-1+\frac34-\frac12=\frac14>0$ and $f\left(-\frac34\right)=1-\frac32+\frac{27}{16}-\frac{27}{16}=-\frac12<0$ . So $s\in\left(-\frac34,-\frac12\right)$ .

(C)

Topics:Theory of equations (higher degree)Monotonicity

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IIT-JEE 2010 · Paper 2, Mathematics question 31