Question

Find the values of $c$ that satisfy the Rolle's theorem for integrals on $[-2,1]$.
$f\left(t\right)=2t-t^3-t^2$

• A. $c=\frac{1+\sqrt{7}}{-3}$
• B. $c=\frac{1-\sqrt{7}}{-3}$
• C. $c=\frac{-1+\sqrt{7}}{-3}$
• D. $c=\frac{-1-\sqrt{7}}{-3}$

Answer 1

Answer: (A), (B)

The correct options are
A $c=\frac{1-\sqrt{7}}{-3}$
B $c=\frac{1+\sqrt{7}}{-3}$

Given : $f\left(t\right)={2t-t}^3-t^2(-2,1)$

According to rolle's theorem $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{(b-a)}$

$(2-{3c}^2-2c)=\frac{f\left(1\right)-f(-2)}{(1-(-2\left)\right)}=\frac{(2-1-1)-(-4+8-4)}{3}(2-{3c}^2-2c)=0({3c}^2+2c-2)=0c=\frac{-2\pm \sqrt{4+4\times 2\times 3}}{6}=\frac{-2\pm 2\sqrt{7}}{6}=\frac{-1\pm \sqrt{7}}{3}c=\frac{1+\sqrt{7}}{-3},\frac{+1-\sqrt{7}}{-3}$

Hence the correct answers are $c=\frac{1+\sqrt{7}}{-3},\frac{1-\sqrt{7}}{-3}$