AMU MCA Function Previous Year Questions (PYQs) – Page 2 of 2
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Showing 14 questions
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} $
Let $f(x,y) = x^3 + y^3$ for all $(x,y) \in \mathbb{R}^2$. Then
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Solution
$f_x = 3x^2$
$f_y = 3y^2$
At (0,0) → critical point
Second derivatives:
$f_{xx} = 6x$, $f_{yy} = 6y$
At (0,0) → 0
Function changes sign around origin → saddle point
If $S$ is the set of all real numbers except $-1$ and $*$ is defined by
$a * b = a + b + ab$, then the inverse of $2 * 3$ is
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Solution
First compute:
$2 * 3 = 2 + 3 + (2)(3) = 11$
Inverse element $x$ satisfies:
$11 * x = 0$ (identity element is $0$)
$11 + x + 11x = 0$
$x(12) = -11$
$x = -\frac{11}{12}$
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