Question

State whether true or false:

The equation $3x^2+px-1=0$ has at least one real root in the interval $(-1,1)$.

• A. True
• B. False

Answer 1

The correct option is A True

Using Rolle's theorem,we have

Condition:$1-f\left(x\right)$ is continuous in $[a,b]$

Condition:$2-f\left(x\right)$ is differentiable in $(a,b)$

Condition:$3-f\left(a\right)=f\left(b\right)$

Given:$f\left(x\right)=3x^2+px-1=0$

Discriminant$=\Delta =p^2-4\times 3\times -1=p^2+12$

Condition:$3-f\left(x\right)$ is continuous in $[-1,1]$

$f(-1)=f\left(1\right)\Rightarrow 3+p-1=3-p-1$

$\Rightarrow 2p=0$ or $p=0$

$\Rightarrow f\left(x\right)=3x^2-1$

Thus there exists a point $c\in (-1,1)$ for which $f^{\prime }\left(c\right)=0$

$\Rightarrow f^{\prime }\left(x\right)=6x$

$\Rightarrow f^{\prime }\left(c\right)=6c=0$

$\Rightarrow c=0$

Hence there exists atleast one value of $c$ in $(-1,1)$ for which $f^{\prime }\left(c\right)=0$

Hence the given statement is true.