Question

Let $g\left(y\right)={\int }_0^{y}f\left(t\right)dt$, when f is such that

(i) $-\frac{1}{2}\le f\left(t\right)\le 0\forall 0\le t\le 1$

(ii) $\frac{1}{2}\le f\left(t\right)\le 1\forall 1\le t\le 3$

(iii) $f\left(t\right)\le 1\forall 3\le t\le 4$

then $g\left(4\right)$ satisfies the inequality?

• A. $0\le g\left(4\right)\le 2$
• B. $\frac{1}{2}\le g\left(4\right)\le 3$
• C. $g\left(4\right)\le 3$
• D. None of these

Answer 1

The correct option is C $g\left(4\right)\le 3$

$g\left(4\right)={\int }_0^{4}f\left(t\right)dt={\int }_0^{1}f\left(t\right)dt+{\int }_1^{3}f\left(t\right)dt+{\int }_3^{4}f\left(t\right)dt$

Now using given conditions,

$-\frac{1}{2}\times 1\le {\int }_0^{1}f\left(t\right)dt\le 0\times 1=0$ ....(A)

$\frac{1}{2}\times 2\le {\int }_1^{3}f\left(t\right)dt\le 1\times 2=2$ ....(B)

${\int }_3^{4}f\left(4\right)dt\le 1\times 1=1$ .....(C)

Using $m(b-a)\le {\int }_a^{b}f\left(x\right)dx\le m(b-a)$

Where m is least value of $f\left(x\right)$ and $M$ is the greater value of $f\left(x\right)$ on $[a,b]$

By adding (A),(B)&(C)we get

${\int }_0^{4}f\left(t\right)dt+{\int }_1^{3}f\left(t\right)dt+{\int }_3^{4}f\left(t\right)dt\le 0+2+1$

i.e $g\left(4\right)\le 3$