AMU MCA 2016 Previous Year Questions (PYQs) – Page 1 of 5
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The value of the integral $\int_{0}^{1} \frac{e^{5\log_e x} - e^{4\log_e x}}{e^{\log_e x^3} - e^{\log_e x^2}} , dx$ is
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Solution
Use $e^{\log_e x^n} = x^n$
So integral becomes:
$\int_{0}^{1} \frac{x^5 - x^4}{x^3 - x^2} , dx$
Factor:
$= \int_{0}^{1} \frac{x^4(x-1)}{x^2(x-1)} , dx$
Cancel $(x-1)$:
$= \int_{0}^{1} x^2 , dx$
$= \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}$
The area bounded by the circle $x^2 + y^2 = 4$ and the line $x = y\sqrt{3}$ in the first quadrant (in sq units) is
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Solution
Circle: $x^2 + y^2 = 4 \Rightarrow r = 2$
Line: $x = \sqrt{3}y \Rightarrow \frac{x}{y} = \sqrt{3} \Rightarrow \tan\theta = \frac{y}{x} = \frac{1}{\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}$
Required region = sector of angle $\frac{\pi}{6}$
Area $= \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 4 \cdot \frac{\pi}{6} = \frac{\pi}{3}$
$\int \frac{x^3}{1 + x^4} , dx$ equals
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Solution
Let $t = 1 + x^4 \Rightarrow dt = 4x^3 dx$
$\int \frac{x^3}{1 + x^4} dx = \frac{1}{4} \int \frac{dt}{t}$
$= \frac{1}{4}\log(1 + x^4) + C$
The value of the integral $\int_{-2}^{0} \frac{dx}{\sqrt{12 - x^2 - 4x}}$ is
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Solution
$12 - x^2 - 4x = 16 - (x+2)^2$
Let $u = x+2$
Limits: $x=-2 \to u=0$, $x=0 \to u=2$
$\int_{0}^{2} \frac{du}{\sqrt{16 - u^2}}$
$= \sin^{-1}\left(\frac{u}{4}\right)\Big|_0^2$
$= \sin^{-1}\left(\frac{2}{4}\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$
$\int \sqrt{16 - 9x^2} , dx$ equals
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Solution
Use standard formula:
$\int \sqrt{a^2 - b^2 x^2} dx = \frac{x}{2}\sqrt{a^2 - b^2 x^2} + \frac{a^2}{2b}\sin^{-1}\left(\frac{bx}{a}\right) + C$
Here $a=4,\ b=3$
$= \frac{x}{2}\sqrt{16 - 9x^2} + \frac{16}{6}\sin^{-1}\left(\frac{3x}{4}\right)$
$= \frac{x}{2}\sqrt{16 - 9x^2} + \frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right) + C$
The sum of two numbers is $10$. Their product will be maximum when they are
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Solution
Let numbers be $x$ and $10-x$
Product $= x(10-x) = 10x - x^2$
Maximum at vertex: $x = \frac{10}{2} = 5$
Numbers: $5,5$
The maximum value of $\frac{\log x}{x}$ is
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Solution
Let $f(x)=\frac{\log x}{x}$
$f'(x)=\frac{1-\log x}{x^2}$
Set $f'(x)=0 \Rightarrow \log x=1 \Rightarrow x=e$
Maximum value $= \frac{\log e}{e}=\frac{1}{e}$
The function $f(x)=\cos^2 x$ is strictly decreasing on
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Solution
$f'(x)=2\cos x(-\sin x)=-\sin 2x$
On $(0,\frac{\pi}{2})$, $\sin 2x > 0 \Rightarrow f'(x)<0$
So decreasing on $(0,\frac{\pi}{2})$
Consider the following propositions:
$p$: I take medicine
$q$: I can sleep
Then, the compound statement $\sim p \to \sim q$ means
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Solution
$\sim p \to \sim q$ means
"If not $p$, then not $q$"
If $\left[\matrix{\alpha & \beta \cr \gamma & -\alpha}\right]$ is to be square root of the two rowed unit matrix, then $\alpha, \beta$ and $\gamma$ should satisfy
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Solution
Let $A = \left[\matrix{\alpha & \beta \cr \gamma & -\alpha}\right]$
$A^2 = I$
Multiply:
$A^2 = \left[\matrix{\alpha^2 + \beta\gamma & 0 \cr 0 & \alpha^2 + \beta\gamma}\right]$
Equating with identity:
$\alpha^2 + \beta\gamma = 1$
$\Rightarrow 1 - \alpha^2 - \beta\gamma = 0$
Final Answer: $\boxed{1 - \alpha^2 - \beta\gamma = 0}$
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