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 concepts of seven and eight digit numbers comparison of numbers comparing numbers operations on rational numbers indian place value chart largest and smallest numbers writing and expanding numbers

The ascending order of minimum values of the function  $P:\sin ^{ -1 }{ x } -\cos ^{ -1 }{ x } $, $Q=\tan ^{ -1 }{ x } -\cot ^{ -1 }{ x } $, $R=\sec ^{ -1 }{ x } -\csc ^{ -1 }{ x } $

  1. P, Q, R

  2. P, R, Q

  3. Q, P, R

  4. Q, R, P
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For P: sin^-1(x) - cos^-1(x) ranges from -pi/2 to pi/2, minimum is -pi/2. For Q: tan^-1(x) - cot^-1(x) ranges from -pi/2 to pi/2, minimum is -pi/2. For R: sec^-1(x) - csc^-1(x) ranges from -pi/2 to pi/2, minimum is -pi/2. However, evaluating the functions at their domain boundaries reveals P < Q < R.

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

Maximum value of $z = 6 x + 11 y$ ,  subject to $2 x + y \leq 104 , x + 2 y \leq 76 , x \geq 0 , y \geq 0$ is

  1. $240$
  2. $540$
  3. $440$
  4. $410$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By evaluating the objective function at the corner points of the feasible region defined by the given linear constraints, we find the maximum value. The corner points include the origin, intersection axes points, and the simultaneous solution of the constraint boundary lines, yielding a maximum value of 440.

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

If $a,b >0$, $a+b=1$, then the least value of $(1+\dfrac 1a)(1+\dfrac 1b)$, is

  1. $3$
  2. $6$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given, $a+b=1$
we know that, $A.M.\geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

$\implies \dfrac{1}{2}\geq \sqrt{ab}$

$\implies \sqrt{ab}\leq \dfrac{1}{2}$

squaring on both sides

$\implies ab \leq \dfrac{1}{4}$  --------------(1)

Similarly

$\implies \dfrac{1+a+1+b}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{2+(a+b)}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{2+1}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{3}{2}\geq \sqrt{(1+a)(1+b)}$

squaring on both sides

$\implies \dfrac{1}{(1+a)(1+b)}\leq \dfrac{4}{9}$  ---------------(2)

multiplying (1) and (2) we get

$\implies \dfrac{ab}{(1+a)(1+b)}\leq \dfrac{1}{4}*\dfrac{4}{9}$

$\implies \dfrac{ab}{(1+a)(1+b)}\leq \dfrac{1}{9}$

$\implies \dfrac{1}{(1+a)(1+b)}\leq \dfrac{1}{9ab}$

$\implies \dfrac{(1+a)(1+b)}{ab}\geq 9$

$\implies \dfrac{(1+a)}{a}*\dfrac{(1+b)}{b}\geq 9$

$\implies (1+\dfrac{1}{a})(1+\dfrac{1}{b})\geq 9$

Therefore, the minimum value of $ (1+\dfrac{1}{a})(1+\dfrac{1}{b})$ is $ 9$

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

If $l,m,n$ be three positive roots of the equation $x^3-ax^2+bx+48=0$, then the minimum value of $\dfrac 1l +\dfrac 2m+\dfrac 3n$ is

  1. $1$
  2. $2$
  3. $\dfrac {-3}{2}$
  4. $\dfrac 52$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

we Know that, $A.M.\geq G.M.$

$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$

let $a=\dfrac{1}{l}, b=\dfrac{2}{m}, c=\dfrac{3}{n}$

Therefore,

$\dfrac{1}{3}(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq \sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$


$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$

Given, the roots of the polynomial $x^3-ax^2+bx+48=0$ are $l,m,n$
Therefore, the product of the roots $lmn=-(\dfrac{48}{1})=-48$

Substituting $lmn=-48$ in the above equation

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{6}{-48})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1}{-8})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(-\dfrac{1}{2})^3}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times(-\dfrac{1}{2})$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq (-\dfrac{3}{2})$

therefore, the minimum value is $-\dfrac{3}{2}$

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

For any positive real number $a$ and for any $n \in N$, the greatest value of 
$\dfrac {a^n}{1+a+a^2....a^{2n}}$ is

  1. $\dfrac 1{2n}$
  2. $\dfrac 1{2n+1}$
  3. $\dfrac 1{2n-1}$

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

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


Therefore, $\dfrac{1+a+a^2+...+a^{2n}}{2n+1}\geq \sqrt[(2n+1)]{1*a*a^2*...*a^{2n}}$

$\implies \dfrac{1+a+a^2+....+a^{2n}}{2n+1}\geq \sqrt[(2n+1)]{a^{(1+2+...+2n)}}$

We know that sum of first $n$ numbers is $1+2+...+n=\dfrac{n(n+1)}{2}$

Therefore $1+2+...+2n=\dfrac{2n(2n+1)}{2}=n(2n+1)$

$\implies \dfrac{1+a+...+a^{2n}}{2n+1}\geq (a^{n(2n+1)})^{\dfrac{1}{2n+1}}$

$\implies \dfrac{1+a+...+a^{2n}}{2n+1}\geq a^n$

$\implies \dfrac{a^n}{1+a+...+a^{2n}}\leq \dfrac{1}{2n+1}$

Therefore the greatest value of $\dfrac{a^n}{1+a+...+a^{2n}}$ is $\dfrac{1}{2n+1}$

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

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.