If the solve the problem


If $4 a+2 b+c=0$, then the equation $3 a x^{2}+2 b x+c=0$ has at least one real root Iying in the interval

(a) (0, 1)
(b) (1, 2)
(c) (0, 2)
(d) none of these


(c) (0, 2)


$f(x)=a x^{3}+b x^{2}+c x+d$             ......(1)



$f(2)=8 a+4 b+2 c+d$

$=2(4 a+2 b+c)+d$

$=d$              $(\because(4 a+2 b+c)=0)$

is continuous in the closed interval [0, 2] and is derivable in the open interval (0, 2).

Also, f(0) = f(2)

By Rolle's Theorem,

$f^{\prime}(\alpha)=0 \quad$ for $0<\alpha<2$

Now, $f^{\prime}(x)=3 a x^{2}+2 b x+c$

$\Rightarrow f^{\prime}(\alpha)=3 a \alpha^{2}+2 b \alpha+c=0$

Equation (1) has atleast one root in the interval $(0,2)$.

Thus, $f^{\prime}(\mathrm{x})$ must have root in the interval $(0,2)$.

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