Q66 / 120 · JEE Advanced 2006 · Mathematics

Paragraph: Read the following passage and answer the questions. For every function $f(x)$ which is twice differentiable, these will be good approximation of $\int_a^b f(x) d x \cong\left(\frac{b-a}{2}\right)\{f(a)+f(b)\}$. Now, if we take $c=\frac{a+b}{2}$, then using above again, we get $\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x \cong \frac{b-a}{4}\{f(a)+f(b)+2 f(c)\}$ and so on. We get approximation for value of $\int_a^b f(x) d x$.Question: If $\lim _{t \rightarrow a} \frac{\int_a^t f(x) d x-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a)^3}=0$, then degree of polynomial function $f(x)$ at most is

  • A
    0
  • Correct answer
    B
    1
  • C
    3
  • D
    2

Solution

$$ \begin{aligned} \lim _{t \rightarrow a} \frac{\int_a^t f(t) d t-\frac{(t-a)}{2}(f(t)+f(a))}{(t-a)^3} & =0 \\ \Rightarrow \quad \lim _{h \rightarrow 0} \frac{\int_a^{a+h} f(t) d t-\frac{h}{2}(f(a+h)+f(a))}{h^3} & =0 \\ \Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(a+h)-\frac{1}{2}(f(a+h)+f(a))-\frac{h}{2}\left(f^{\prime}(a+h)\right)}{3 h^2} & =0 \end{aligned} $$ $f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)$ $\Rightarrow \quad \lim _{h \rightarrow 0} \frac{-\frac{1}{2} f^{\prime}(a+h)-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h}=0$ $\Rightarrow \quad \lim _{h \rightarrow 0} \frac{-\frac{h}{2} f^{\prime}(a+h)}{6 h}=0$ $\Rightarrow \quad f^{\prime \prime}(a)=0, \forall a \in R$ $\Rightarrow(x)$ must have maximum degree 1.

Hence, the correct answer is option: 'B'.

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