AMU MCA Function Previous Year Questions (PYQs) – Page 1 of 2
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Showing 14 questions
Let $f(x,y,z) = e^{\sqrt{1 - x^2 - y^2}}$. Then the range of $f$ is
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Solution
For real values:
$1 - x^2 - y^2 \ge 0$
$x^2 + y^2 \le 1$
Minimum value of $\sqrt{1 - x^2 - y^2}$ is $0$ (when $x^2+y^2=1$)
Maximum value is $1$ (when $x=y=0$)
So exponent varies from $0$ to $1$
$e^0 = 1$
$e^1 = e$
Hence range is $[1, e]$
If $a, b$ be two fixed positive integers such that $f(a + x)= b + \left[b^3 + 1 - 3b^2 f(x) + 3b{f(x) - (f(x))^3}\right]^{1/3}$ for all real $x$, then $f(x)$ is a periodic function with period
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Solution
If $f(x)=\sin^2 x$ and the composite function $g(f(x))=|\sin x|$, then the function $g(x)$ is equal to
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Solution
The curves $x=\log(y+e)$ and $y=\log\left(\dfrac{1}{x}\right)$
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Solution
Let $K$ be a set of real number and $f:K \to R$ such that for all $x$ and $y$,
$ |f(x) - f(y)| \le |x - y|^5 $. If $f(3)=7$, then the value of $f(9)$ is equal to
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Solution
Let $ f(x) = x^3 + x $, then the equation
$ \frac{2}{y - f(2)} + \frac{3}{y - f(3)} + \frac{4}{y - f(4)} = 0 $ has
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If $g(x)=x^2+x-2$ and $\frac{1}{2}(g \circ f)(x)=2x^2-5x+2$, then $f(x)$ is equal to
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If $R$ is the set of real numbers and $f:R\to R$ is a function defined by $f(x)=\sin x$, then $f^{-1}([-1,1])$ is
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Solution
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.
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