JEE Advanced 2010 Paper 2Question 1
The real number $s$ lies in the interval
Quadratic Equation · Higher Degree Polynomial Equations
Solution
Concept. $f'(x)>0$ for all $x$, so $f$ has exactly one real root; locate it by sign changes. Formulas used. (1) $f'>0$ on $\mathbb R$ $\Rightarrow$ $f$ is strictly increasing, so it has at most one zeroStep 1 - Derivative. $f'(x)=2+6x+12x^2$ has discriminant $36-96<0$ and positive leading coefficient, so $f'>0$. By (1), $s$ is the only real root. Step 2 - Values. $$f\left(-\tfrac12\right)=1-1+\tfrac34-\tfrac12=\tfrac14>0,$$ $$f\left(-\tfrac34\right)=1-\tfrac32+\tfrac{27}{16}-\tfrac{27}{16}=-\tfrac12<0.$$ Step 3 - Conclude. $s\in\left(-\frac34,-\frac12\right)$. Answer: (C)