Mathematics

Maxima and Minima

129 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 complex number $z$ satisfies the condition $\left|\displaystyle {z}-\frac{25}{z}\right|=24$. Then the maximum distance from the origin to the point '$z$' in the argand plane is

  1. 20

  2. 25

  3. 30

  4. 35
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$|z| = |z - \frac{25}{z} + \frac{25}{z}| \leq 24 + \frac{25}{|z|}$
$\therefore$ $|z|^2 - 24|z| - 25 \leq 0.$
$\therefore$ $(|z| - 25)(|z| + 1) \leq 0., \Rightarrow |z| \leq 25.$
Hence maximum distance of z from origin is 25.

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


If $\left |z-\displaystyle \frac{6}{z}\right|=2$, then the greatest value of $|z|$ is

  1. $\sqrt{7}-1$
  2. $\sqrt{7}+1$
  3. $\sqrt{7}$
  4. $\displaystyle \frac{\sqrt{7}}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

we have,
$|z|=\left |z-\dfrac{6}{z}+\dfrac{6}{z} \right |\leq \left |z-\dfrac{6}{z} \right |+\left |\dfrac{6}{z} \right|$
$|z|\leqslant 2+\left |\dfrac{6}{z} \right|$
$\Rightarrow |z|^{2}-2 |z|-6\leqslant 0$
Hence, 

$ {1 - \sqrt{7}} \leq |z| \leq {1+\sqrt 7} $
$0 < |z| \leq {1+\sqrt 7} $
Hence, option B is correct

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

If $|z+4|\leq 3$, then the maximum value of $|{z}+1|$ is

  1. $0$
  2. $4$
  3. $10$
  4. $6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The condition |z + 4| <= 3 describes a disk centered at -4 with radius 3. We want to maximize |z - (-1)|, which is the distance from z to -1. The maximum distance from a point in the disk to -1 occurs at the point furthest from -1, which is -4 - 3 = -7. The distance from -7 to -1 is |-7 - (-1)| = 6.

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

A point $'z'$ moves on the curve $|z - 4 - 3i| = 2$ in an argand plane. The maximum and minimum values of $|z|$ are

  1. $2, 1$
  2. $6, 5$
  3. $4, 3$
  4. $7, 3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let $w = 4 + 3i$. 
We can write, $|z| = |(z-w) + w|$. Hence by triangle inequality ($|z _1+z _2| \leq |z _1| + |z _2|$), we can write $|z| \leq |z-w| + |w|$. It is given in the question that, $|z-w| = 2$ and $|w| = \sqrt{4^2 + 3^2} = 5$.
Putting the values, we get $|z| \leq 7$. 
Using another result of triangle inequality ($\big||z _1| - |z _2| \big| \leq |z _1 + z _2|$), we can write $\big||z-w| - |w|\big| \leq |z - w + w|$.
Hence, we get $|z| \geq 3$. The minimum value is 3 and maximum value is 7.
Hence, (D) is the correct option
Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $\displaystyle |Z - \frac {4}{Z}| = 2$, then the maximum value of $\displaystyle |Z|$ is equal to

  1. $\displaystyle \sqrt 5 + 1$

  2. 2
  3. $\displaystyle 2 + \sqrt 2$
  4. $\displaystyle \sqrt 3 + 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have for any two complex numbers $\displaystyle \alpha$ and $\displaystyle \beta$
$\displaystyle ||\alpha|| \leq |\alpha - \beta|$
Now $\displaystyle ||Z|-|\frac {4}{|Z|}||\leq|Z-\frac {4}{Z}|$
$\displaystyle \Rightarrow |Z| - \frac {4}{|Z|}|\leq 2$
Set $\displaystyle |Z| = r > 0$, then $\displaystyle |r-\frac {4}{r}|\leq 2$
$\displaystyle \Rightarrow -2 \leq r - \frac {4}{r} \leq 2$
The left inequality gives
$\displaystyle r^2 + 2r - 4 \geq 0$
The corresponding roots are
$\displaystyle r = \frac {-2\pm \sqrt {20}}{2} = -1 \pm \sqrt 5$
Thus $\displaystyle r \geq \sqrt 5 - 1$ or $\displaystyle r \leq -1 - \sqrt 5$
implies that $\displaystyle r \geq \sqrt 5 - 1$ (As r > 0) ... (i)
Again consider the right inequality
$\displaystyle r - \frac {4}{r} \leq 2 \Rightarrow r^2 - 2 r - 4 \leq 0$
The corresponding roots are
$\displaystyle r = \frac {2 \pm \sqrt {20}}{2} = 1 \pm \sqrt 5$
Thus $\displaystyle 1 - \sqrt 5 \leq r \leq 1 + \sqrt 5$
But r > 0, hence $\displaystyle r \leq 1 + \sqrt 5$ .... (ii)
(i) and (ii) gives
$\displaystyle \sqrt 5 - 1 \leq r \leq \sqrt 5 + 1$
So, the greatest value is $\displaystyle \sqrt 5 + 1$.

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 $\cfrac{-D}{ya}$)