Question

Let $f$ be a continuous function on $[a,b]$. If $F\left(x\right)=({\int }_a^{x}f\left(t\right)dt-{\int }_x^{b}f\left(t\right)dt)(2x-(a+b\left)\right)$, then there exist some $cϵ(a,b)$ such that

• A. ${\int }_a^{c}f\left(t\right)dt={\int }_c^{b}f\left(t\right)dt$
• B. ${\int }_a^{c}f\left(t\right)dt-{\int }_c^{b}f\left(t\right)dt=f\left(c\right)(a+b-2c)$
• C. ${\int }_a^{c}f\left(t\right)dt-{\int }_c^{b}f\left(t\right)dt=f\left(c\right)[2c-(a+b\left)\right]$
• D. ${\int }_a^{c}f\left(t\right)dt+{\int }_c^{b}f\left(t\right)dt=f\left(c\right)[2c-(a+b\left)\right]$

Answer 1

The correct option is B ${\int }_a^{c}f\left(t\right)dt-{\int }_c^{b}f\left(t\right)dt=f\left(c\right)(a+b-2c)$

$F\left(x\right)=({\int }_a^{x}f\left(t\right)dt-{\int }_x^{b}f\left(t\right)dt)(2x-(a+b\left)\right)$
Since, $f\left(x\right)$ is continuous on $[a,b]$ , so $F\left(x\right)$ is also continuous on $[a,b]$.
Also , $F\left(x\right)$ is differentiable on $(a,b)$
$F\left(a\right)=-(a-b){\int }_a^{b}f\left(t\right)dt$
$\Rightarrow F\left(a\right)=(b-a){\int }_a^{b}f\left(t\right)dt$
Also $F\left(b\right)=(b-a){\int }_a^{b}f\left(t\right)dt$
$\Rightarrow F\left(a\right)=F\left(b\right)$
Hence, Rolle's theorem is applicable to $F\left(x\right)$ on $[a,b]$, so there exists c$\in$ (a,b) such that
$F^{\prime }\left(c\right)=0$
Now , $F^{\prime }\left(x\right)=2({\int }_a^{x}f\left(t\right)dt-{\int }_x^{b}f\left(t\right)dt)+(2x-(a+b\left)\right)\left(f\right(x)+f(x\left)\right)=0$
$\Rightarrow 2({\int }_a^{c}f\left(t\right)dt-{\int }_x^{b}f\left(t\right)dt)+2f\left(c\right)(2c-(a+b\left)\right)=0$
$\Rightarrow {\int }_a^{c}f\left(t\right)dt-{\int }_c^{b}f\left(t\right)dt=f\left(c\right)(a+b-2c)$