AMU MCA 2017 Previous Year Questions (PYQs) – Page 1 of 5
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For any three sets $A, B$ and $C$ the set
$(A \cup B \cup C)\cap (A \cap B' \cap C')' \cap C'$ is equal to
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Solution
$ (A \cup B \cup C)\cap (A \cap B' \cap C')' \cap C' $
$ = (A \cup B \cup C)\cap (A' \cup B \cup C)\cap C' $
$ = [(A \cup B \cup C)\cap A'] \cup [(A \cup B \cup C)\cap (B \cup C)] \cap C' $
$ = (B \cup C)\cap C' $
$ = (B \cap C') \cup (C \cap C') $
$ = B \cap C' $
Let $P={\theta;\ \sin\theta - \cos\theta = \sqrt{2}\cos\theta}$ and
$Q={\theta;\ \sin\theta + \cos\theta = \sqrt{2}\sin\theta}$ be two sets. Then,
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Solution
For $P$
$ \sin\theta - \cos\theta = \sqrt{2}\cos\theta $
$ \Rightarrow \tan\theta - 1 = \sqrt{2} $
$ \Rightarrow \tan\theta = \sqrt{2} + 1 $
For $Q$
$ \sin\theta + \cos\theta = \sqrt{2}\sin\theta $
$ \Rightarrow 1 + \cot\theta = \sqrt{2} $
$ \Rightarrow \cot\theta = \sqrt{2} - 1 $
$ \Rightarrow \tan\theta = \sqrt{2} + 1 $
$ \therefore P = Q $
The function $f(x)=x-[x]$ where $[,]$ denotes the greatest integer function is
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Solution
Let $c$ be an integer.
For LHL,
$ \lim_{x \to c^-} f(x)=\lim_{h\to0^+}[(c-h)-[c-h]] $
Since $[c-h]=c-1$,
$ =\lim_{h\to0}(c-h-(c-1)) $
$ =\lim_{h\to0}(1-h)=1 $
For RHL,
$ \lim_{x \to c^+} f(x)=\lim_{h\to0^+}[(c+h)-[c+h]] $
Since $[c+h]=c$,
$ =\lim_{h\to0}(c+h-c) $
$ =\lim_{h\to0}h=0 $
Thus, $ \text{LHL} \ne \text{RHL} $
$\therefore f(x)$ is discontinuous at all integers.
Hence, $f(x)$ is continuous at non-integer points only.
If $f(x)=x e^{x(1-x)}$, then $f(x)$ is
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Solution
$ f(x)=x e^{x(1-x)} $
$ f'(x)=e^{x(1-x)}+x e^{x(1-x)}(1-2x) $
$ =e^{x(1-x)}[1+x-2x^2] $
$ e^{x(1-x)}\neq0 $
$ \Rightarrow 1+x-2x^2=0 $
$ \Rightarrow (1-x)(2x+1)=0 $
$ \Rightarrow x=1,\ -\frac{1}{2} $
For $ x\in\left[-\frac{1}{2},1\right],\ f'(x)>0 $
$ \therefore f(x)$ is increasing on $\left[-\frac{1}{2},1\right]$
The range of the function $f(x)=\frac{2+x}{2-x},\ x\ne2$ is
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Solution
Let $ y=\frac{2+x}{2-x} $
$ \Rightarrow 2y-xy=2+x $
$ \Rightarrow 2y-2=x+xy $
$ \Rightarrow 2y-2=x(1+y) $
$ \Rightarrow x=\frac{2y-2}{1+y} $
For real $x$, denominator $\neq0$
$ \Rightarrow 1+y\neq0 $
$ \Rightarrow y\neq-1 $
$ \therefore \text{Range}=R-{-1} $
If $\sin(\alpha+\beta)=1,\ \sin(\alpha-\beta)=\frac{1}{2},\ \alpha,\beta\in\left[0,\frac{\pi}{2}\right]$ then the value of $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$ is
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Solution
$ \sin(\alpha+\beta)=1 $
$ \Rightarrow \alpha+\beta=\frac{\pi}{2} $
$ \sin(\alpha-\beta)=\frac{1}{2} $
$ \Rightarrow \alpha-\beta=\frac{\pi}{6} $
Solving,
$ \alpha=\frac{\pi}{3},\ \beta=\frac{\pi}{6} $
$ \tan(\alpha+2\beta)\tan(2\alpha+\beta) $
$ =\tan\left(\frac{\pi}{3}+\frac{\pi}{3}\right)\tan\left(\frac{2\pi}{3}+\frac{\pi}{6}\right) $
$ =\tan\left(\frac{2\pi}{3}\right)\tan\left(\frac{5\pi}{6}\right) $
$ =(-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)=1 $
Using the information from the figure, $\cos^{-1} x$ is equal to
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Solution
From the figure,
$ BC=\sqrt{AB^2-AC^2}=\sqrt{1-x^2} $
$ \Rightarrow \cos\alpha=\sqrt{1-x^2},\ \sin\alpha=x $
$ \Rightarrow \cos\beta=x,\ \sin\beta=\sqrt{1-x^2} $
$ \Rightarrow \sin^{-1}x=\alpha,\ \cos^{-1}x=\beta $
In right triangle,
$ \alpha+\beta=\frac{\pi}{2} $
$ \Rightarrow \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2} $
$ \Rightarrow \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x $
Let $f(x)$ be a function defined by
$f(x)=\begin{cases}4x-5,\ x\le2 \ x-\lambda,\ x>2 \end{cases}$
If $\lim_{x\to2} f(x)$ exists, then the value of $\lambda$ is
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Solution
For limit to exist,
$ \lim_{x\to2^-} f(x)=\lim_{x\to2^+} f(x) $
$ \lim_{x\to2^-} f(x)=4(2)-5=3 $
$ \lim_{x\to2^+} f(x)=2-\lambda $
Equating,
$ 3=2-\lambda $
$ \Rightarrow \lambda=-1 $
The tangent of the angle between the lines whose intercepts on the axes are respectively $a,-b$ and $b,-a$ is
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Solution
Equations of lines:
$ \frac{x}{a}-\frac{y}{b}=1 $
$ \frac{x}{b}-\frac{y}{a}=1 $
Slopes:
$ m_1=\frac{b}{a},\quad m_2=\frac{a}{b} $
Angle between lines:
$ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| $
$ =\left|\frac{\frac{b}{a}-\frac{a}{b}}{1+\frac{b}{a}\cdot\frac{a}{b}}\right| $
$ =\left|\frac{\frac{b^2-a^2}{ab}}{2}\right| $
$ =\left|\frac{b^2-a^2}{2ab}\right| $
$ \therefore \tan\theta=\pm\frac{b^2-a^2}{2ab} $
The equation of the circle whose radius is $5$ and which touches the circle
$x^2+y^2-2x-4y-20=0$ externally at the point $(5,5)$ is
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Solution
Given circle:
$ x^2+y^2-2x-4y-20=0 $
$ \Rightarrow (x-1)^2+(y-2)^2=25 $
Centre $=(1,2)$, radius $=5$
New circle also radius $=5$
Since they touch externally, centres lie on line joining and midpoint is $(5,5)$
$ \Rightarrow \frac{h+1}{2}=5,\ \frac{k+2}{2}=5 $
$ \Rightarrow h=9,\ k=8 $
Equation:
$ (x-9)^2+(y-8)^2=25 $
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