Question

Assertion :g(x)= max{f(t) : $\{f\left(f\right):0\le t\le x\},$ then ${\int }_0^{1}g\left(x\right)dx=\frac{29}{24}$ Reason: f(x) is increasing in $(0,\frac{1}{2})$ and decreasing in $(\frac{1}{2},1)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$f\left(x\right)=-x^2+x+1$
$f^{\prime }\left(x\right)=1-2x$
$f^{\prime }\left(x\right)>0\Rightarrow 1-2x>0\Rightarrow x<\frac{1}{2}$
$f^{\prime }\left(x\right)<0\Rightarrow 1-2x<0\Rightarrow x>\frac{1}{2}$
$\Rightarrow f\left(x\right)$ is increasing in $(0,\frac{1}{2})$ and decreasing in $(\frac{1}{2},1)$

now $g\left(x\right)=max\left\{f\right(x):0\le t\le x\}$
$=\{x-x^2+10\le x<\frac{1}{2}$
$=\{\frac{5}{4}\frac{1}{2}\le x\le 1$
${\int }_0^{1}g\left(x\right)dx={\int }_0^{1/2}(x-x^2+1)dx+{\int }_{1/2}^{1}5/4dx=\frac{29}{24}$