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NIMCET Previous Year Questions (PYQs)
NIMCET Limit PYQ
NIMCET PYQ
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NIMCET Mathematics PYQNIMCET Limit Continuity Differentiability (Disha JEE) PYQ
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NIMCET Mathematics PYQNIMCET Limit Continuity Differentiability (Disha JEE) PYQ
Solution
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NIMCET Previous Year PYQNIMCET NIMCET 2023 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 Previous Year PYQNIMCET NIMCET 2023 PYQ
Solution
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NIMCET Previous Year PYQNIMCET NIMCET 2023 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 Previous Year PYQNIMCET NIMCET 2023 PYQ
Solution
is equal to 
NIMCET PYQ
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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ
The value of the limit $$\lim _{{x}\rightarrow0}\Bigg{(}\frac{{1}^x+{2}^x+{3}^x+{4}^x}{4}{\Bigg{)}}^{1/x}$$ is
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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ
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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ
Which of the following is NOT true?
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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ
Solution
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NIMCET Previous Year PYQNIMCET NIMCET 2022 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 Previous Year PYQNIMCET NIMCET 2022 PYQ
Solution
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NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ
$\lim_{x\to \infty} (\frac{x+7}{x+2})^{x+5}$ equal to
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NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ
Solution
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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ
⇒f′(1)=0 and f′(2)=0
Given f(x) is a fourth degree polynomial
Let $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$
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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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ
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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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ
therefore using by L'Hospital rule
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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
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NIMCET Previous Year PYQNIMCET NIMCET 2023 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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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ
Solution
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