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Jamia Millia Islamia Previous Year Questions (PYQs)

Jamia Millia Islamia Inequation PYQ


Jamia Millia Islamia PYQ
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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

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.

Jamia Millia Islamia PYQ
If $|\alpha^2| = 4$ and $-3 \le \lambda \le 2$, then the range of $|\lambda \alpha^2|$ is……





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

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 $[0, 12]$. $\boxed{\text{Answer: (C) } [0, 12]}$

Jamia Millia Islamia PYQ
What is the number of non-zero integral solutions of the equation $\,|x| + |1 - x| = 6\,$?





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

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.

Jamia Millia Islamia PYQ
The area enclosed by $3|x| + 4|y| \le 12$ is





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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

Area of rhombus $= \dfrac{1}{2} \times 8 \times 6 = 24$.


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