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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Quadratic Equations PYQ


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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2023 PYQ

Solution


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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2023 PYQ

Solution

Number of Roots

Given:

\( e^x \sin x = 1 \) has two real roots → say \( x_1 \) and \( x_2 \)

Apply Rolle’s Theorem:

Since \( f(x) = e^x \sin x \) is continuous and differentiable, and \( f(x_1) = f(x_2) \), ⇒ There exists \( c \in (x_1, x_2) \) such that \( f'(c) = 0 \)

Compute:

\[ f'(x) = e^x(\sin x + \cos x) = 0 \Rightarrow \tan x = -1 \] At this point, \[ e^x \cos x = -1 \]

\[ \boxed{\text{At least one root}} \]



MCA 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?





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2024 PYQ

Solution


MCA 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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2022 PYQ

Solution


MCA 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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2022 PYQ

Solution


MCA NIMCET PYQ
The quadratic equation whose roots are  is





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2018 PYQ

Solution


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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2021 PYQ

Solution


MCA 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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2017 PYQ

Solution



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





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MCA NIMCET Previous Year PYQMCA NIMCET 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

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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2018 PYQ

Solution



MCA 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 





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2021 PYQ

Solution


MCA NIMCET PYQ
If $\alpha$ and $\beta$ are the roots of the equation $2x^{2}+ 2px + p^{2} = 0$, where $p$ is a non-zero real number, and $\alpha^{4}$ and $\beta^{4}$ are the roots of $x^{2} - rx + s = 0$, then the roots of $2x^{2} - 4p^{2}x + 4p^{4} - 2r = 0$ are:





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2014 PYQ

Solution


MCA NIMCET PYQ
If x and y are positive real numbers satisfying the system of equations $x^{2}+y\sqrt{xy}=336$ and $y^{2}+x\sqrt{xy}=112$, then x + y is:





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2014 PYQ

Solution


MCA NIMCET PYQ
The value of k for which the equation $(k-2)x^{2}+8x+k+4=0$ has both real, distinct and negative roots is





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2015 PYQ

Solution


MCA 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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2020 PYQ

Solution


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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2020 PYQ

Solution


MCA NIMCET PYQ
$a, b, c$ are positive integers such that $a^{2}+2b^{2}-2bc=100$ and $2ab-c^{2}=100$. Then the value of $\frac{a+b}{c}$ is





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2015 PYQ

Solution


MCA NIMCET PYQ
If $x^{2} + 2ax + 10 - 3a > 0$ for all x ∈ R, then





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2014 PYQ

Solution



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