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## NIMCET Previous Year Questions (PYQs)

NIMCET PYQ
If the equation $|x^2 – 6x + 8| = a$ has four real solution then find the value of $a$?

NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

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NIMCET PYQ
Between any two real roots of the equation $e^x sin x = 1$, the equation $e^x cos x = –1$ has

NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

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NIMCET PYQ

Not Available

NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

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NIMCET PYQ
Let C denote the set of all tuples (x,y) which satisfy $x^2 -2^y=0$ where x and y are natural numbers. What is the cardinality of C?

NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ

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NIMCET PYQ
If $\alpha , \beta$ are the roots of $x^2-x-1=0$ and $A_n=\alpha^n+\beta^n$, the Arithmetic mean of $A_{n-1}$ and $A_n$ is

NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

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NIMCET PYQ
If the roots of the quadratic equation $x^2+px+q=0$ are tan 30° and tan 15° respectively, then the value of 2 + p - q is

NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

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NIMCET PYQ
The quadratic equation whose roots are  is

NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ

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NIMCET PYQ
For what value of p, the polynomial  $x^4-3x^3+2px^2-6$ is exactly divisible by $(x-1)$

NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

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NIMCET PYQ
α, β are the roots of the an equation $x^2- 2x cosθ + 1 = 0$, then the equation having roots αn and βn is

NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ

#### Solution

NIMCET PYQ
The equation (x-a)3+(x-b)3+(x-c)3 = 0 has

NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ

#### Solution

Let f(x) = (x – a)3 + (x – b)3 + (x – c)3.
Then f'(x) = 3{(x – a)2 + (x – b)2 + (x –c)2}
clearly , f'(x) > 0 for all x.
so, f'(x) = 0 has no real roots.
Hence, f(x) = 0 has two imaginary and one real root

NIMCET PYQ
Let  and  be the roots f the equation  and  are the roots of the equation , then the value of r,

NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ

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NIMCET PYQ
If α≠β and $\alpha^2=5\alpha-3,\beta^2=5\beta-3$, then the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is

NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

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NIMCET PYQ
Roots of equation are $ax^2-2bx+c=0$ are n and m , then the value of $\frac{b}{an^2+c}+\frac{b}{am^2+c}$ is

NIMCET Previous Year PYQNIMCET NIMCET 2020 PYQ

#### Solution

NIMCET PYQ
If a + b + c = 0, then the value of

NIMCET Previous Year PYQNIMCET NIMCET 2020 PYQ

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