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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If $\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$, then f(2) is
Solution
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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