Question

Let $a+b=4,a<2$ and $g\left(x\right)$ be a monotonically increasing function of $x$. Then $f\left(a\right)={\int }_0^{a}g\left(x\right)dx+{\int }_0^{b}g\left(x\right)dx$

• A. Increases with increase in $(b-a)$
• B. Decreases with increase in $(b-a)$
• C. Increases with decrease in $(b-a)$
• D. None of these

Answer 1

Answer: (A)

Increases with increase in $(b-a)$

We have, $a+b=4\Rightarrow b=4-a$ and $b-a=4-2a=t$(say)

Now, $f\left(a\right)={\int }_0^{a}g\left(x\right)dx+{\int }_0^{a}g\left(x\right)dx={\int }_0^{a}g\left(x\right)dx+{\int }_0^{4-a}g\left(x\right)dx$

$\Rightarrow \frac{df\left(a\right)}{da}=g\left(a\right)-g(4-a)$

As $a<2$ and $g\left(x\right)$ is increasing

$\Rightarrow 4-a>a\Rightarrow g\left(a\right)-g(4-a)<0\Rightarrow \frac{df\left(a\right)}{da}<0$

Now, $\frac{df\left(a\right)}{da}=\frac{df\left(a\right)}{dt}.\frac{dt}{da}=-2.\frac{df\left(a\right)}{dt}\Rightarrow \frac{df\left(a\right)}{dt}>0.$

Thus, $f\left(a\right)$ is an increasing funciton of $t.$

Hence, the given expression increases with increase in $(b-a).$