Question

Number of values of x satisfying the equation ${\int }_{-1}^x(8t^2+\frac{28}{3}t+4)dt=\frac{\left(\frac{3}{2}\right)x+1}{{\log }_{x+1}\sqrt{x+1}}.$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

${\int }_{-1}^x({8t}^2+\frac{28}{3}t+4)dt=\frac{\left(\frac{3}{2}\right)x+1}{{\log }_{x+1}\sqrt{x+1}}$
$\Rightarrow {[\frac{{8t}^3}{3}+\frac{28}{3}t+4]}_{-1}^x=\frac{\frac{3}{2}x+1}{{\log }_{x+1}\sqrt{x+1}}$
$\Rightarrow \frac{{8x}^3}{3}+\frac{{14x}^2}{3}+4x-\frac{8}{3}+\frac{14}{3}-4=\frac{\frac{3}{2}x+1}{{\log }_{x+1}\sqrt{x+1}}$
$\Rightarrow \frac{{8x}^3+{14x}^3+12x-6}{3}=\frac{3x+2}{2{\log }_{x+1}\sqrt{x+1}}$
Only $x=1$ satisfy this equation.