Jamia Millia Islamia MCA Maxima And Minima Previous Year Questions (PYQs)
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Showing 5 questions
Maximum value of $\left(\dfrac{1}{x}\right)^x$ is:
1
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4
Solution
Let
$\displaystyle y = \left(\dfrac{1}{x}\right)^x = e^{x\ln(1/x)} = e^{-x\ln x}$
Take $\ln$ on both sides:
$\ln y = -x\ln x$
Differentiate w.r.t $x$:
$\dfrac{1}{y}\dfrac{dy}{dx} = -(\ln x + 1)$
$\Rightarrow \dfrac{dy}{dx} = -y(\ln x + 1)$
For maximum or minimum, set $\dfrac{dy}{dx}=0$:
$\ln x + 1 = 0 \Rightarrow x = \dfrac{1}{e}$
Now,
$y_{\max} = \left(\dfrac{1}{1/e}\right)^{1/e} = e^{1/e}$
Minimum value of $a^{x} + \dfrac{a}{a^{a x}}$
(where $a > 0$, $a \neq 1$, and $x \in \mathbb{R}$) is:
1
2
3
4
Solution
**Solution:**
Let $y = a^{x} + \dfrac{a}{a^{a x}} = a^{x} + a^{1 - a x}$
Let $t = a^{x}$, $t > 0$.
Then $y = t + \dfrac{a}{t^{a}}$
Differentiate:
\[
\dfrac{dy}{dt} = 1 - a^2 t^{-a-1} = 0 \Rightarrow t^{a+1} = a^2 \Rightarrow t = a^{\tfrac{2}{a+1}}.
\]
Substitute back:
\[
y_{\min} = a^{\tfrac{2}{a+1}} + a^{1 - a\tfrac{2}{a+1}} = 2\sqrt{a}.
\]
$\boxed{\text{Answer: (A) }2\sqrt{a}}$
The absolute maximum value of $y = x^3 - 3x + 2$ in $0 \le x \le 2$ is
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4
Solution
$y' = 3x^2 - 3 = 0 \Rightarrow x = 1$
Now,
$y(0) = 2$, $y(1) = 0$, $y(2) = 8 - 6 + 2 = 4$
Hence, maximum value = 4.
The points of extremum of the function
$f(x) = \displaystyle \int_0^x e^{t^2} (1 - t^2)\,dt$ are:
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2
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4
Solution
$f'(x) = e^{x^2}(1 - x^2)$.
Setting $f'(x) = 0 \Rightarrow 1 - x^2 = 0 \Rightarrow x = \pm 1$.
If $y = a\log|x| + bx^2 + x$ has extreme values at $x=-1$ and $x=-2$, then
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Solution
From $y'=a/x + 2bx + 1=0$ → solving gives $a=2,\ b=-\tfrac{1}{2}$.
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