Jamia Millia Islamia MCA 2017 Previous Year Questions (PYQs) – Page 1 of 10
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If 20th term of an A.P. is 30 and its 30th term is 20, then its 10th term is
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Solution
Let the first term be $a$ and common difference $d$.
$ a + 19d = 30 $
$ a + 29d = 20 $
Subtract → $10d = -10 \Rightarrow d = -1$
Then $a = 49$
10th term = $a + 9d = 49 - 9 = 40$
Let sum of $n$ terms of an A.P. is $2n(n-1)$, then sum of their squares is
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Solution
Sum $S_n = 2n(n-1)$ → $a = 2$, $d = 4$
Sum of squares = $\sum (a + (r-1)d)^2$
$= n[a^2 + (n-1)(a+d)^2 + …]$
Simplifying gives $\dfrac{8n(n+1)(2n+1)}{3}$
For what value of $x$, the $\log_2(5·2^x+1)$, $\log_4(2^{1-x}+1)$, and 1 are in A.P.?
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Solution
Let $a=\log_2(5·2^x+1)$, $b=\log_4(2^{1-x}+1)$, $c=1$
For A.P.: $2b=a+c$
Simplify using $\log_4 y = \frac{1}{2}\log_2 y$
After algebra → $x = 1 - \log_2 5$
If the ratio of sum of $m$ terms and $n$ terms of an A.P. be $m^2:n^2$, then the ratio of its $m$th and $n$th terms will be
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Solution
$S_m/S_n = m^2/n^2$
We know $S_n = \dfrac{n}{2}[2a+(n-1)d]$
$\Rightarrow \dfrac{m[2a+(m-1)d]}{n[2a+(n-1)d]} = \dfrac{m^2}{n^2}$
Simplify → $\dfrac{2a+(m-1)d}{2a+(n-1)d} = \dfrac{m}{n}$
$\Rightarrow \text{ratio of } t_m : t_n = (2m-1):(2n-1)$
The value of $9^{1/3} × 9^{1/9} × 9^{1/27} × ... ∞$ is
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Solution
$\log$ both sides: $\log y = \frac{1}{3}\log 9 + \frac{1}{9}\log 9 + …$
Geometric series sum = $\frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}$
$\Rightarrow \log y = \frac{1}{2}\log 9 = \log 3$
$\Rightarrow y = 3$
If $\alpha$ and $\beta$ are roots of $x^2 + px + q = 0$, then value of $\alpha^2 + \alpha\beta + \beta^2$ is
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Solution
$\alpha+\beta=-p$, $\alpha\beta=q$
$\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = p^2 - 2q$
So $\alpha^2 + \alpha\beta + \beta^2 = p^2 - q$
If the roots of $x^2 - bx + c = 0$ are two consecutive numbers, then $b^2 - 4c$ is equal to
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Solution
Let roots be $r$ and $r+1$
Then $b = r+(r+1) = 2r+1$, $c = r(r+1)$
$\Rightarrow b^2 - 4c = (2r+1)^2 - 4r(r+1) = 1$
The number of real roots of equation $(x-1)^2 + (x-2)^2 + (x-3)^2 = 0$ is
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Solution
Each term is non-negative. Sum of squares = 0 ⇒ each = 0
Impossible since $x$ cannot be simultaneously 1, 2, and 3.
So, no real root.
If the equations $x^2 + 2x + 3\lambda = 0$ and $2x^2 + 3x + 5\lambda = 0$ have a non-zero common root, then $\lambda$ is equal to
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Solution
Let the common root be $x$.
From first: $x^2 + 2x + 3\lambda = 0$
From second: $2x^2 + 3x + 5\lambda = 0$
Multiply (1) by 2 and subtract:
$(2x^2 + 4x + 6\lambda) - (2x^2 + 3x + 5\lambda) = 0$
$\Rightarrow x + \lambda = 0 \Rightarrow x = -\lambda$
If $P_n = {}^nP_r$ and $C_r = {}^nC_{r-1}$, then $(n, r)$ is
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Solution
${}^nP_r = \dfrac{n!}{(n-r)!}$ and ${}^nC_{r-1} = \dfrac{n!}{(r-1)!(n-r+1)!}$
Equating: $\dfrac{n!}{(n-r)!} = \dfrac{n!}{(r-1)!(n-r+1)!}$
Simplify → $(r-1)! = (n-r+1)!/(n-r)! = (n-r+1)$
$\Rightarrow r-1 = n-r+1 \Rightarrow n = 2r - 2$
Try small integers: for $r=2$, $n=3$ works.
Hence $(n, r) = (3, 2)$
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