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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Mathematics
Chapter
Limits

JEE Advanced 2006 · Mathematics · Limits

Paragraph: Read the following passage and answer the questions. For...

Paragraph: Read the following passage and answer the questions. For every function f(x)f(x) which is twice differentiable, these will be good approximation of ∫abf(x)dx≅(b−a2){f(a)+f(b)}\int_a^b f(x) d x \cong\left(\frac{b-a}{2}\right)\{f(a)+f(b)\}. Now, if we take c=a+b2c=\frac{a+b}{2}, then using above again, we get ∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx≅b−a4{f(a)+f(b)+2f(c)}\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 ∫abf(x)dx\int_a^b f(x) d x.Question: If lim⁡t→a∫atf(x)dx−(t−a)2{f(t)+f(a)}(t−a)3=0\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)f(x) at most is

Options

  1. A0
  2. B1
  3. C3
  4. D2

Correct answer

B. 1

Step-by-step solution

lim⁡t→a∫atf(t)dt−(t−a)2(f(t)+f(a))(t−a)3=0⇒lim⁡h→0∫aa+hf(t)dt−h2(f(a+h)+f(a))h3=0⇒lim⁡h→0f(a+h)−12(f(a+h)+f(a))−h2(f′(a+h))3h2=0\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.
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