NIMCET Maxima And Minima Previous Year Questions (PYQs) – Page 1 of 2
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Showing 16 questions
The minimum value of $px + qy$ when $xy=r^2$ and $p,q,x,y$ are positive numbers is
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Solution
Use AM-GM or substitution $y=\dfrac{r^2}{x}$:
$px + q\dfrac{r^2}{x}$
Min occurs when derivatives equal:
$px = q\dfrac{r^2}{x}$
$\Rightarrow x^2 = \dfrac{qr^2}{p}$
Compute minimum:
$2r\sqrt{pq}$
Suppose that the temperature at a point (x,y), on a metal plate is $T(x,y)=4x^2-4xy+y^2$, An ant, walking on the plate, traverses a circle of radius 5 centered at the origin. What is the highest temperature encountered by the ant?
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Solution
A particle is at rest at the origin. It moves
along the x −axis with an acceleration $x-x^2$
, where x is the distance of the
particle at time t. The particle next comes to
rest after it has covered a distance
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Solution
Consider the function $$f(x)=\begin{cases}{-{x}^3+3{x}^2+1,} & {if\, x\leq2} \\ {\cos x,} & {if\, 2{\lt}x\leq4} \\ {{e}^{-x},} & {if\, x{\gt}4}\end{cases}$$ Which of the following statements about f(x) is true:
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Solution
The function $f(x)=2\sin x+\sin 2x,\ x\in[0,2\pi]$ has absolute maximum and minimum at
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Solution
$f'(x)=2\cos x+2\cos 2x$
$=2(2\cos^2x+\cos x-1)$
Critical points:
$\cos x=\frac{1}{2},-1$
$x=\frac{\pi}{3},\pi,\frac{5\pi}{3}$
Check values → max at $\frac{\pi}{3}$, min at $\frac{5\pi}{3}$
The maximum value of
$(\cos\alpha_1)(\cos\alpha_2)\cdots(\cos\alpha_n)$
where $0\le \alpha_1,\alpha_2,\ldots,\alpha_n\le\pi$ and
$(\cot\alpha_1)(\cot\alpha_2)\cdots(\cot\alpha_n)=1$ is
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Solution
By AM–GM, maximum occurs when
$\alpha_1=\alpha_2=\cdots=\alpha_n=\frac{\pi}{4}$
Then
$\cos\alpha_i=\frac{1}{\sqrt2}$
Product $=\left(\frac{1}{\sqrt2}\right)^n=\frac{1}{2^{n/2}}$
The curve $y=\frac{x}{1+x\tan x}$ attains maxima
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Solution
Given the curve
$y = \dfrac{x}{1 + x \tan x}$
Differentiate using quotient rule:
$y' = \dfrac{(1 + x \tan x) - x(\tan x + x \sec^2 x)}{(1 + x \tan x)^2}$
Simplify the numerator:
$N = 1 + x \tan x - x \tan x - x^2 \sec^2 x$
$N = 1 - x^2 \sec^2 x$
Set $N = 0$ for maxima/minima:
$1 - x^2 \sec^2 x = 0$
$x^2 \sec^2 x = 1$
$x \sec x = \pm 1$
Hence the curve attains maxima when:
$x \sec x = 1$
$x=\cos x$
If $y = a \log x + b x^2 + x$ has its extremum value at $x = -1$ and $x = 2$, then
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Solution
$y' = \frac{a}{x} + 2bx + 1$
Since $y'=0$ at $x=-1$ and $2$:
At $x= -1$:
$-!a + (-2b) + 1 = 0$
$\Rightarrow -a -2b + 1 = 0$ … (i)
At $x= 2$:
$\frac{a}{2} + 4b + 1 = 0$
$\Rightarrow a + 8b + 2 = 0$ … (ii)
Solve (i) and (ii):
From (i): $a = 1 - 2b$
Put into (ii):
$(1 - 2b) + 8b + 2 = 0$
$1 - 2b + 8b + 2 = 0$
$6b + 3 = 0$
$b = -\frac12$
Then
$a = 1 - 2b = 1 - 2(-\frac12) = 1 + 1 = 2$
So
$a = 2,\ b = -\frac12$
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