Question

If $\underset{t\to a}{lim}\frac{{\int }_a^{t}f\left(t\right)dt-\left(\frac{t-a}{2}\right)(f\left(t\right)+f\left(a\right))}{{(t-a)}^3}=0$ then maximum degree of $f\left(x\right)$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (D)

$1$

Substituting $t=a+h$
$\therefore$ $t-a=h$ as $t\to a$, $h\to 0$
$\therefore$ Required limit $=\underset{h\to 0}{lim}\frac{{\int }_a^{a+h}f\left(t\right)dt-\frac{h}{2}[f(a+h)+f\left(a\right)]}{h^3}$ ($\frac{0}{0}$ form)
$\therefore$ $0=\underset{h\to 0}{lim}\frac{f(a+h)-\frac{1}{2}[f(a+h)+f\left(a\right)]-\frac{h}{2}{f}^{\prime }(a+h)}{3h^2}$
$0=\underset{h\to 0}{lim}\frac{f^{\prime }(a+h)-\frac{1}{2}{f}^{\prime }(a+h)-\frac{1}{2}{f}^{\prime }(a+h)-\frac{h}{2}{f}^{\prime \prime }(a+h)}{6h}$
$0=\underset{h\to 0}{lim}\frac{0-h{f}^{\prime \prime }(a+h)}{2\times 6h}$,
$0=-\frac{1}{12}{f}^{\prime \prime }(a+0)$
$\Rightarrow$ $0=f^{\prime \prime }\left(a\right)$ $\Rightarrow$ $f^{\prime }\left(x\right)=$ constant$=c$ (say)
$\Rightarrow$ $f\left(x\right)=cx+d$, whose degree is $1$
Hence, option $D$ is correct.