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 business maths linear programming problems structure of linear programming model linear programming problem operations research

If $a>0$, then least value of $(a^3+a^2+a+1) ^2$ is

  1. $64a^2$
  2. $16a^4$
  3. $16a^3$

  4. None of the above.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

we know that $A.M.\geq G.M.$


therefore, $\dfrac{a^3+a^2+a+1}{4}\geq \sqrt[4]{a^3*a^2*a*1}$

$\implies \dfrac{a^3+a^2+a+1}{4}\geq \sqrt[4]{a^6}$

squaring on both sides 

$\implies (\dfrac{a^3+a^2+a+1}{4})^2\geq ({a^{\dfrac{6}{4}}})^2$

$\implies ({a^3+a^2+a+1})^2\geq 16a^3$

Multiple choice economics consumption and investment functions keynesian law of consumption and propensity to consume ex ante and ex post concept of consumption function, saving function and investment function

The maximum value of multiplier is when the value of MPC is _________.

  1. Infinity, zero

  2. Infinity, one

  3. One, infinity

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

Investment multiplier refers to the number of time by which the increase in output or income exceeds the increase in investment. It is measured as the ratio between change in income and change in investment and it is denoted as 'k'.

Multiplier(k) => Change in income / change in investment = 1/ {1-MPC(c)} where c is the marginal propensity to consume. 

Therefore, the value of multiplier will be maximum when the value of MPC is either infinity or zero. 

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

If $|z^2-3|=3|z|$, then the maximum value of |z| is

  1. $1$
  2. $\displaystyle \frac {3+\sqrt {21}}{2}$
  3. $\displaystyle \frac {\sqrt {21}-3}{2}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By the law of inequality, 
$|{ z }^{ 2 }-3|\ge { |z| }^{ 2 }-3$
$ \Longrightarrow 3|z|\ge { |z| }^{ 2 }-3\ \Longrightarrow { |z| }^{ 2 }-3|z|-3\le 0\ \Longrightarrow 0\le |z|\le \displaystyle\frac { 3+\sqrt { 21 }  }{ 2 } $
Hence the maximum value of $|z|=\displaystyle\frac { 3+\sqrt { 21 }  }{ 2 } $

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

If $z$ is a complex number satisfying the equation $\left| z+i \right| +\left| z-i \right| =8$, on the complex plane then maximum value of $\left| z \right| $ is

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation |z+i| + |z-i| = 8 represents an ellipse with foci at (0, -1) and (0, 1). The sum of distances to foci is 2a = 8, so a = 4. The center is at (0,0). The maximum distance from the origin is the semi-major axis length, which is 4.

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

The minimum value of $\displaystyle \left | z-1 \right |+\left | z \right |$for complex values of z is

  1. $2$
  2. $\displaystyle \frac{1}{2}$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\left| w \right| =\left| \left( w-z \right) +z \right| $ 
Using Triangle Inequality.
$\left| w-z \right| +\left| z \right| \ge \left| \left( w-z \right) +z \right| =\left| w \right| $
$\Rightarrow \left| z \right| +\left| z-w \right| \ge \left| w \right| $
$\Rightarrow \left| z \right| +\left| z-1 \right| \ge 1$
Therefore, minimum value of $\left| z \right| +\left| z-1 \right| $ is 1
Hence, option 'D' is correct.

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

If $|z-4+3i|\le 1$ and $m$ and $n$ are the least and greatest values of $|z|$ and $k$ is the least value of $\displaystyle \frac { { x }^{ 4 }+{ x }^{ 2 }+4 }{ x } $ on the interval $(0,\infty)$, then $k$ is equal to

  1. $m$
  2. $n$
  3. $m+n$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have, 

$1\ge \left| z-\left( 4-3i \right)  \right| $
$\Rightarrow 1\ge \left| z \right| -\left| 4-3i \right| \quad ,\quad \left| 4-3i \right| -\left| z \right| $
$\Rightarrow 1\ge \left| z \right| -5\quad ,\quad 5-\left| z \right| $
$\left| z \right| \le 6,\left| z \right| \ge 4\Rightarrow 4\le \left| z \right| \le 6\Rightarrow m=4,n=6$
Let $y=\displaystyle\frac { 4+{ x }^{ 2 }+{ x }^{ 4 } }{ x } ={ x }^{ 3 }+x+\displaystyle\frac { 4 }{ x } ={ x }^{ 3 }+x+\frac { 1 }{ x } +\frac { 1 }{ x } +\frac { 1 }{ x } +\frac { 1 }{ x } $
$\because x\in \left( 0,\infty  \right) $, then ${ x }^{ 3 },x,\frac { 1 }{ x } ,\frac { 1 }{ x } ,\frac { 1 }{ x } ,\frac { 1 }{ x } $ are all positive numbers whose product is 1.
Thus their sum y will be least when 
${ x }^{ 3 }=x=\displaystyle\frac { 1 }{ x } \Rightarrow x=1$
So least value of $y=6,k=6$
So $k=n$

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

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

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

$\left| z+\dfrac { 2 }{ z }  \right| =2$


$\left| z+\dfrac { 2 }{ z }  \right| \ge \left| z \right| -\dfrac { 2 }{ \left| z \right|  } $  ....{ $\because \left| { z } _{ 1 }{ +z } _{ 2 } \right| \ge \left| { z } _{ 1 } \right| -\left| { z } _{ 2 } \right| $}

$\Rightarrow 2\ge \left| z \right| -\dfrac { 2 }{ \left| z \right|  } \ \Rightarrow { \left| z \right|  }^{ 2 }-2\left| z

\right| -2\le 0\ \Rightarrow \left| z \right| \le \sqrt { 3 } +1$

Ans: C

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

If $\displaystyle z\epsilon C \; and \; \left | z+4 \right |\leq 3$ then the greatest value of $\left | z+1 \right |$ is

  1. 5

  2. 6

  3. 4

  4. 3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\left| z+4 \right| \le 3$      ...(1)

$\left| \left( z+4 \right) -3 \right| \le \left| z+4 \right| +\left| -3 \right| \ \Rightarrow \left| z+1 \right| \le \left| z+4 \right| +3$

$\Rightarrow \left| z+1 \right| \le 6$       ....{ $\because \quad \left| z+4 \right| \le 3$}

Ans: B

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

The maximum value of |z| where z satisfies the condition $\displaystyle \left | z + \frac{2}{z} \right | = 2$ 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

Given, $\displaystyle \left | z + \frac{2}{z} \right | = 2   $

$   \Rightarrow |z| - \dfrac{2}{|z|} \leq 2      $
$ \Rightarrow |z|^2 - 2 |z| - 2 \leq 0$
$\Rightarrow |z| \leq \displaystyle \frac{2 \pm \sqrt{4 + 8}}{2} \leq 1 \pm \sqrt 3$
Hence, max. value of |z| is $1 + \sqrt 3$

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

If $|z| \leq 1$ then the minimum and maximum value of |z - 3| are

  1. 4, 2

  2. 3, 4

  3. 4, 6

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

Given |z| <= 1, the point z lies within or on the unit circle centered at the origin. The distance |z - 3| represents the distance from z to the point (3, 0). The minimum distance is 3 - 1 = 2, and the maximum distance is 3 + 1 = 4.

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 $\displaystyle \sqrt 5 + 1$.