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Question Id : 4306 | Context :NIMCET 2017

Question

Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If lim, then f(2) is
🎥 Video solution / Text Solution of this question is given below:
Given it has extremum values at x=1 and x=2
⇒f′(1)=0  and  f′(2)=0
Given f(x) is a fourth degree polynomial 
Let  f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e
Given 
\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3
\lim _{{x}\rightarrow0}\lbrack1+\frac{a{x}^4+b{x}^3+c{x}^2+\mathrm{d}x+e}{{x}^2}\rbrack=3
\lim _{{x}\rightarrow0}\lbrack1+a{x}^2+bx+c+\frac{d}{x}+\frac{e}{{x}^2}\rbrack=3
For limit to have finite value, value of 'd' and 'e' must be 0
⇒d=0  & e=0
Substituting x=0 in limit 
⇒ c+1=3
⇒ c=2
f^{\prime}(x)=4a{x}^3+3b{x}^2+2cx+d
x=1 and x=2 are extreme values,
⇒f^{\prime}(1)=0 and $f^{\prime}(2)=0
⇒ 4a+3b+4=0 and 32a+12b+8=0 
By solving these equations
we get, a=\frac{1}{2} and b=-2
So,
f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}
⇒f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)
⇒f(2)=2^{2}(2-4+2)
⇒f(2)=0

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