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 real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

The smaller value of n for which $x^{2} - 2x - 3$ and $x^{3} - 2x^{2} - nx - 3$ have an H.C.F. involving $x$ is

  1. 0

  2. 1

  3. 2

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

For the two polynomials to have an HCF involving x, they must share a common root. x^2 - 2x - 3 factors to (x-3)(x+1). Testing x=3 in the second polynomial: 27 - 18 - 3n - 3 = 0 => 6 - 3n = 0 => n=2.

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

What is the smallest integer n for which $\displaystyle { 25 }^{ n }>{ 5 }^{ 12 }$?

  1. $6$
  2. $7$
  3. $8$
  4. $9$

  5. 10
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${25}^{n}>{5}^{12}$

${({5}^{2})}^{n}>{5}^{12}$
${{5}^{2n}}>{5}^{12}$
Comparing the power
$2n>12$
$n>\dfrac{12}{2}$
$n>6$
Smallest integer greater than $6$ will be $7$

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

Which of the following has the greatest value?

  1. $(6^{2} \times 6)^{4}$
  2. $(36)^{5}$
  3. $(36^{2} \times 6^{3})^{2}$
  4. $(216)^{4}$
  5. $(6^{4})^{4}$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

  1. ${ ({ 6 }^{ 2 }\times 6) }^{ 4 }={ { (6 }^{ 3 }) }^{ 4 }={ 6 }^{ 12 }$
  2. ${ 36 }^{ 5 }={ ({ 6 }^{ 2 }) }^{ 5 }={ 6 }^{ 10 }$
  3. ${ ({ 36 }^{ 2 }\times { 6 }^{ 3 }) }^{ 2 }={ ({ 6 }^{ 4 }\times { 6 }^{ 3 }) }^{ 2 }={ ({ 6 }^{ 7 }) }^{ 2 }={ 6 }^{ 14 }$
  4. ${ 216 }^{ 4 }={ ({ 6 }^{ 3 }) }^{ 4 }={ 6 }^{ 12 }$
  5. ${ ({ 6 }^{ 4 }) }^{ 4 }={ 6 }^{ 16 }$
  • Therefore option $E$ has maximum value

Multiple choice maths calculations and mental strategies 4 mental additions and subtractions of decimals multiplication and division of decimals mental multiplication and division

If k is an integer, and if $0.02468 \times 10^k$ is greater than 10,000, what is the least possible value of k?

  1. 7

  2. 4

  3. 6

  4. 5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiplying 0.02468 by a positive power of 10 will shift the decimal point to the right. Simply shift the decimal point to the right until the result is greater than 10,000. Keep track of how many times you shift the decimal point. Shifting the decimal point 5 times results in 2,468. This is still less than 10,000. Shifting one more place yields 24,680, which is greater than 10,000.

Multiple choice maths calculating and mental strategies 3 finding percentage of a number how many in all? problems on percentage

Which is the greatest ?

  1. $\dfrac{50}{3}\%$
  2. $\dfrac{2}{15}$
  3. $0.17$
  4. $6\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Option A$\Rightarrow \begin{pmatrix}\dfrac{50}{3}\end{pmatrix}\times\begin{pmatrix}\dfrac{1}{100}\end{pmatrix}=\dfrac{1}{6}=0.166$
Option B $\Rightarrow \dfrac{2}{15}=0.133$
Option C $\Rightarrow 6\%=\dfrac{6}{100}=0.06$
Clearly $\Rightarrow 0.17$ is greater

Multiple choice maths part number dividing fractions division of a fractions division of a fraction

The least number among $\displaystyle \frac{4}{9}, \, \sqrt{\frac{9}{49}},$ 0.45 and $(0.8)^2$ is

  1. $\displaystyle \frac{4}{9}$
  2. $\displaystyle \sqrt{\frac{9}{49}}$

  3. 0.45
  4. $(0.8)^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
: Decimal equivalents of the given numbers.
$\displaystyle \frac{4}{9} \, = \, 0.44; \, \sqrt{\frac{9}{49}} \, = \, \frac{3}{7} \, = \, 0.43$
$\displaystyle 0.45 \, and \, (0.8)^2 \, = \, 0.64$
$\displaystyle \therefore$ Least number is 0.43
$\displaystyle = \, \sqrt{\frac{9}{49}}$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

The smallest integer n such that $\displaystyle \left(\frac{1+i}{1-i}\right)^{n}= 1$ is

  1. 16

  2. 12

  3. 8

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

$\displaystyle \left ( \frac{1 + i}{1 - i} \right )^n = 1$         ${ \because -i = \displaystyle \Rightarrow \frac{1}{i}}$
$\displaystyle \Rightarrow\left ( \frac{1 + i}{\displaystyle 1 + \frac{1}{i}} \right )^n = 1$
$i^n = 1$
so min value of $n =4$

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If $i=\sqrt{-1}$, then select from the following having the greatest value.

  1. $i^4+i^3+i^2+i$
  2. $i^8+i^6+i^4+i^2$
  3. $i^{12}+i^9+i^6+i^3$
  4. $i^{16}+i^{12}+i^8+i^4$
  5. $i^{20}+i^{15}+i^{10}+i^5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $i=\sqrt {-1}$

The value of option $A$ is $1-i-1+i = 0$
The value in option $B$ is $1-1+1-1 = 0$
The value in option $C$ is $1+i-1-i = 0$
The value in option $D$ is $1+1+1+1 = 4$
The value in option $E$ is $1-i-1+i = 0$
So, the correct answer is option $D$.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

Find the least value of $n$ for which $\left (\dfrac {1 + i}{1 - i}\right )^{n} = 1$.

  1. $4$
  2. $3$
  3. $-4$
  4. $1$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$\because \dfrac {1 + i}{1 - i} = \dfrac {1 + i}{1 - i}\times \dfrac {1 + i}{1 + i}$
$= \dfrac {(1 + i)^{2}}{1 - i^{2}} = \dfrac {1 + 2i + i^{2}}{1 - i^{2}}$
$= \dfrac {1 + 2i - 1}{1 + 1} = i$
$\therefore \left (\dfrac {1 + i}{1 - i}\right )^{n} = 1$
$\Rightarrow i^{n} = 1$
Thus, $i^{n}$ will be positive integer, if $n = 4$.

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 $\dfrac{1}{2n+1}$