Mathematics

Maxima and Minima

191 Questions

Maxima and minima problems involve finding the highest and lowest values of mathematical functions within given intervals. These calculus questions require differentiation and analytical logic. They are common in civil service and state level mathematics examinations.

Extreme function valuesComplex number minimizationMinimum variables calculationMaximum matrix analysisCalculus optimization

Maxima and Minima Questions

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

The maximum value of $\left| z \right| $ when $z$ satisfies the condition $\displaystyle \left| z+\dfrac { 2 }{ z }  \right| =2$ is

  1. $\sqrt { 3 } -1$
  2. $\sqrt { 3 } +1$
  3. $\sqrt { 3 } $
  4. $\sqrt { 2 } +\sqrt { 3 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have $\displaystyle \left| z \right| =\left| z+\frac { 2 }{ z } -\frac { 2 }{ z }  \right| \le \left| z+\frac { 2 }{ z }  \right| +\frac { 2 }{ \left| z \right|  } $

$\displaystyle \Rightarrow \left| z \right| \le 2+\frac { 2 }{ \left| z \right|  } \Rightarrow { \left| z \right|  }^{ 2 }\le 2\left| z \right| +2\ \Rightarrow { \left| z \right|  }^{ 2 }-2\left| z \right| +1\le 1+2\Rightarrow { \left( \left| z \right| -1 \right)  }^{ 2 }\le 3\ \Rightarrow -\sqrt { 3 } \le \left| z \right| -1\le \sqrt { 3 } \Rightarrow 1-\sqrt { 3 } \le \left| z \right| \le 1+\sqrt { 3 } $
That is , the maximum value of $\left| z \right| $ is $1+\sqrt { 3 } $.

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If the complex number z satisfies the condition |z| $\geq$ 3, then the least value of $\displaystyle \left | z + \frac{1}{z} \right |$ is equal to.

  1. $2$
  2. $\dfrac{4}{3}$
  3. $1$
  4. $\dfrac{8}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
By using triangle inequality:  $||z _1-|z _2||\le |z _1+z _2|\le |z _1|+|z _2|$

We have    $|z+\dfrac{1}{z}|\leq |z|+|\dfrac{1}{z}|$

Now Given that
$|z|\geq 3$

$\therefore |z+\dfrac{1}{z}|\leq |3|+|\dfrac{1}{3}|$

$\Rightarrow |z+\dfrac{1}{z}|\leq 3-\dfrac{1}{3}$

$\Rightarrow |z+\dfrac{1}{z}|\leq \dfrac{8}{3}$
Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $\left | z-i \right |\leq 2$ and $z _{0}=13+5i$, then the maximum value of $\left | iz+z _{0} \right |$ is

  1. $12$
  2. $15$
  3. $13$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\left | iz+z _{0} \right |=\left | iz +1 + z _{0} -1\right |$
$\left | iz+z _{0} \right |=\left | iz -i^{2} + z _{0} -1\right |$
$=|{i}({z}-{i})+13+5{i}-1|$
$\leq|{i}||{z}-{i}|+|12+5i|\leq 1\times2+13\le15$

Hence, option B.
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The least integral value of $a$ for which the graphs of the functions $y = 2ax + 1$ and $\displaystyle y=(a-6)x^{2}-2$ do not intersect is:

  1. -6

  2. -5

  3. 3

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For no intersection of graphs of functions,  $ y = 2ax + 1$ and $ y = (a-6)x^2 -2$, There should not any common points between two curves.


Putting the value of $y$ from equation of line into equation of given parabola, we get,

$\Rightarrow (2ax + 1) = (a-6)x^2 - 2$

$\Rightarrow (a-6)x^2  - (2a)x -3 = 0$ ...$(1)$

Equation $(1)$ is a quadratic equation in $x$. 

For no intersection of both given functions, the equation $(1)$ must not have any real solutions.

A quadratic equation have no real roots if the value of it's discriminant is less than zero.

Hence $D = b^2 - 4ac < 0 $

$\Rightarrow D = ((-2a)^2) - 4 \times (a-6) \times (-3) < 0$

$\Rightarrow  D = 4a^2 +12a -72 < 0$

$\Rightarrow (a +6)(a -3) <0$

Hence Value of $a$ for the graphs of given functions do not intersect lies between $(-6 ,3)$

So the least integral value will be $(-5)$. Correct answer is $A$.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

The value of $a$ for which the function $f(x)=a\ \sin x+\dfrac{1}{3}\sin 3x$ has an extremum at $x=\dfrac{\pi}{3}$ is

  1. $1$
  2. $-1$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$f\left( x \right) = a\sin x + \dfrac{1}{3}\sin 3x$
$ \Rightarrow f'\left( x \right) = a\cos x + \dfrac{1}{3}\cos 3x \times 3$
$ \Rightarrow f'\left( x \right) = a\cos x + \cos 3x$
For extremum at ${\dfrac{\pi }{3}}$
$f'\left( {\dfrac{\pi }{3}} \right) = 0$
$ \Rightarrow a\cos \left( {\dfrac{\pi }{3}} \right) + \cos 3\left( {\dfrac{\pi }{3}} \right) = 0$
$ \Rightarrow \dfrac{a}{2} - 1 = 0$
$\Rightarrow a = 2$
Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

If $p$ and $q$ are positive real numbers such that ${p}^{2}+{q}^{2}=1$, then the maximum value of $(p+q)$ is

  1. $2$
  2. $\cfrac{1}{2}$
  3. $\cfrac{1}{\sqrt{2}}$
  4. $\sqrt{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$AM\geq GM\implies \dfrac{p+q}{2}\geq \sqrt{pq}$

squaring on both sides 
$(p+q)^{2}\geq {4}p{q}$
$p^{2}+q^{2}+2{p}{q}\geq 4{p}{q}$
$1\geq 2{p}{q}\implies  {p}{q}\leq \dfrac{1}{2}$
$(p+q)^{2}=1+2{p}{q}\leq 1+1$
$(p+q)^{2}\leq 2\implies p+q\in[-\sqrt{2},\sqrt{2}]$
The maximum value of $p+q$ is $\sqrt{2}$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let $A = (3,-4), B = (1,2)$ .Let $P = (2k-1,2k+1)$ be  a variable point  such that PA+PB is the minimum. then $k$ is

  1. $\dfrac 79$
  2. $0$
  3. $\dfrac 78$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To minimize PA + PB, P must lie on the line segment AB. The slope of AB is (2 - (-4)) / (1 - 3) = 6 / -2 = -3. The equation of line AB is y - 2 = -3(x - 1) => y = -3x + 5. Substituting P(2k-1, 2k+1): 2k+1 = -3(2k-1) + 5 => 2k+1 = -6k + 3 + 5 => 8k = 7 => k = 7/8.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let x and y be two varibles such that $\displaystyle x> 0$ and $xy=1$. Find the minimum value of $x+y$.

  1. $ 2 $
  2. $ \dfrac {1}{2}$
  3. $ \dfrac {2}{3}$
  4. $ 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let,  $\displaystyle z= x+y= x+\dfrac{1}{x}$
for minimum value of $z$
$\cfrac{dz}{dx}=0\Rightarrow 1-\cfrac{1}{x^2}=0\Rightarrow x=\pm 1$
but given $x>0, \Rightarrow x=1$
Hence minimum value of $z$ is 2.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

If 'x' is real, then maximum value of $\dfrac{3x^2+9x+17}{3x^2+9x+7}$ is - 

  1. $41$
  2. $1$
  3. $\dfrac{17}{7}$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Maximise: $\cfrac{3{ x }^{ 2 }+9x+17}{3{ x }^{ 2 }+9x+7}$

Now we can see that coefficient of ${x}^{2}$ and $x$ are same in ${N}^{x}$ and ${D}^{x}$ so
$\Rightarrow$ $\cfrac{3{ x }^{ 2 }+9x+10+7}{3{ x }^{ 2 }+9x+7}$
$y=1+\cfrac{10}{3{ x }^{ 2 }+9x+7}$
We want to maximise $y$ so we need to minimize $3{ x }^{ 2 }+3x+7$
$y=1+\cfrac{10}{min(3{ x }^{ 2 }+9x+7)}$
$y=1+40=41$ ($\because$ we know min value of quadratic is $\cfrac{-D}{ya}$)

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

If $F(x)=2x^3-21\,x^2+36x-20$, then 

  1. f has maxima at x=1

  2. f has minima at x=1

  3. f has maximum value -128

  4. f has minimum value -3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider given the function,

$F\left( x \right)=2{{x}^{3}}-21{{x}^{2}}+36x-20$      ……(1)

Differentiate with respect to x,

${{F}^{'}}\left( x \right)=6{{x}^{2}}-42x+36$          ……..(2)


For maxima and minima,

$ F\left( x \right)=0 $

$ 6{{x}^{2}}-42x+36=0 $

$ {{x}^{2}}-7x+6=0 $

$ {{x}^{2}}-6x-x+6=0 $

$ x\left( x-6 \right)-1\left( x-6 \right)=0 $

$ \left( x-6 \right)\left( x-1 \right)=0 $

$ x=1,6 $


Differentiate equation 2nd with respect to x,

${{F}^{''}}\left( x \right)=12x-42$

At $x=1\Rightarrow {{F}^{''}}\left( x \right)<0$

Hence, F(x) Is maximum.


At $x=6\Rightarrow F\left( x \right)>0$

Hence, function F(x) is minimum.

 

Hence, this is the answer.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

If $\displaystyle xy=a^{2}$ and $\displaystyle S=b^{2}x+c^{2}y$ where a,b and c are constants then the minimum value of S is 

  1. $abc$
  2. $\displaystyle bc\sqrt{a}$
  3. $2abc$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $x y = a^2$ and $S = b^2x + c^2y$
$\Rightarrow S = b^2 x + c^2a^2/x$
$\Rightarrow \dfrac{dS}{dx} = b^2 - c^2a^2/x^2$
For maximum or minimum value of $S$
$ \dfrac{dS}{dx} = 0 = b^2 - c^2a^2/x^2 \Rightarrow x =\pm  ac/b$
Now $\dfrac{dS}{dx} = 2 c^2a^2/x^3$
Clearly at $x =  ac/b$,  $\dfrac{dS}{dx} = 2 b^3/ac > 0 $ (Assuming that $ b^3/ac>0$)
Hence minimum value of $S$ is $= b^2(ac/b)+c^2(b/ac)= 2abc$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

If $\displaystyle \theta +\phi =\frac{\pi }{3}$ then $\displaystyle  \sin \theta \cdot\sin \phi$ has a maximum value at $\displaystyle \theta$ =

  1. $\displaystyle \dfrac{\pi }{6}$
  2. $\displaystyle \dfrac{2\pi }{3}$
  3. $\displaystyle \dfrac{\pi }{4}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $y = \sin\theta.\sin\phi = \sin\theta.\sin(\dfrac{\pi}{3}-\theta)$
For maximum value of $y$ 
$\dfrac{dy}{dx} = 0 = \cos\theta.\sin(\dfrac{\pi}{3}-\theta) - \sin\theta.\cos(\dfrac{\pi}{3}-\theta) = \sin(2\theta -\dfrac{\pi}{3})$
$\Rightarrow \theta = \dfrac{\pi}{6}$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Find the two positive numbers $x$ & $y$ such that their sum is $60$ and $\displaystyle xy^{3}$ is maximum

  1. $15$ & $45$
  2. $30$ & $30$
  3. $20$ & $40$
  4. $10$ & $50$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let one number be $x$
Hence the other number will be $(60-x)$.
Let 
$K=x^{3}.(60-x)$
Differentiating $K$ with respect to $x$, we get 
$\dfrac{dK}{dx}$
$=3x^{2}(60-x)-x^{3}=0$
Or 
$x^{2}[180-3x-x]=0$
Or 
$x=0$ and $x=\dfrac{180}{4}=45$.
Now its given that the numbers are positive.
Hence $x=0$ is ruled out.
Thus we get $x=45$.
Hence
$y=15$.
Therefore the numbers are $45,15$.