Jamia Millia Islamia MCA Inequation Previous Year Questions (PYQs)
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Showing 4 questions
If $x$, when divided by 4, leaves remainder 3,
then find the remainder when $(2020 + x)^{2022}$ is divided by 8.
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Solution
Then $2020 + x = 2020 + 4k + 3 = 4k + 2023$
Now $2020 \equiv 4 \pmod{8} $
$\Rightarrow 2020 + x \equiv 4 + 3 = 7 \pmod{8}$
So $(2020 + x)^{2022} \equiv 7^{2022} \pmod{8}$
Since $7 \equiv -1 \pmod{8}$,
$(-1)^{2022} = 1$.
Hence remainder = 1.
If $|\alpha^2| = 4$ and $-3 \le \lambda \le 2$, then the range of $|\lambda \alpha^2|$ is……
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Solution
Given $|\alpha^2| = 4$ and $-3 \le \lambda \le 2$.
Then $|\lambda \alpha^2| = 4|\lambda|$.
Minimum value at $\lambda = 0$ → 0.
Maximum value at $\lambda = -3$ → $4 \times 3 = 12$.
Hence, range is $[0, 12]$.
$\boxed{\text{Answer: (C) } [0, 12]}$
What is the number of non-zero integral solutions of the equation
$\,|x| + |1 - x| = 6\,$?
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Solution
Case 1: $x \ge 1 \Rightarrow |x|=x,\ |1-x|=x-1$
$\Rightarrow x + (x-1) = 6 \Rightarrow 2x = 7 \Rightarrow x=\tfrac{7}{2}$ (not integer).
Case 2: $x < 1 \Rightarrow |x|=-x,\ |1-x|=1-x$
$\Rightarrow -x + (1-x) = 6 \Rightarrow -2x = 5 \Rightarrow x=-\tfrac{5}{2}$ (not integer).
Hence, there are **no** non-zero integral solutions.
The area enclosed by $3|x| + 4|y| \le 12$ is
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Solution
Area of rhombus $= \dfrac{1}{2} \times 8 \times 6 = 24$.
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