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## NIMCET Previous Year Questions (PYQs)

#### NIMCET Limit PYQ

NIMCET PYQ
$\lim _{{x}\rightarrow1}\frac{{x}^4-1}{x-1}=\lim _{{x}\rightarrow k}\frac{{x}^3-{k}^2}{{x}^2-{k}^2}=$, then find k

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NIMCET PYQ
Let $f(x)=\frac{x^2-1}{|x|-1}$. Then the value of $lim_{x\to-1} f(x)$ is

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NIMCET PYQ
is equal to

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NIMCET PYQ
Which of the following is NOT true?

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NIMCET PYQ
For $a\in R$ (the set of al real numbers), $a \ne 1$, $\lim _{{n}\rightarrow\infty}\frac{({1}^a+{2}^a+{\ldots+{n}^a})}{{(n+1)}^{a-1}\lbrack(na+1)(na+b)\ldots(na+n)\rbrack}=\frac{1}{60}$ . Then one of the value of $a$ is

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NIMCET PYQ
$\lim_{x\to \infty} (\frac{x+7}{x+2})^{x+5}$ equal to

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NIMCET PYQ
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

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#### 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,$f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$⇒$f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)$⇒$f(2)=2^{2}(2-4+2)$⇒$f(2)=0$NIMCET PYQ NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ #### Solution Function is the form of therefore using by L'Hospital rule Again apply L'Hospital Rule, Putting x = 0, we get NIMCET PYQ Find NIMCET Previous Year PYQNIMCET NIMCET 2020 PYQ #### Solution NIMCET PYQ If$f(x)=\lim _{{x}\rightarrow0}\, \frac{{6}^x-{3}^x-{2}^x+1}{\log _e9(1-\cos x)}$is a real number then$\lim _{{x}\rightarrow0}\, f(x)\$

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