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 business economics and quantitative methods measures of dispersion and skewness quartile deviation or semi-interquartile range interquartile range histograms and frequency distribution diagrams

Range =

  1. Largest value - Smallest value

  2. Largest value divided by 2

  3. Both A and B

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

Range refers to the value of variation between the largest and the smallest values in a particular data set. It can be calculated as the difference between the upper limit and lower limit of a particular set of data.

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

If a $ >0,  $ then the minimum value of sum of $  \dfrac{1}{a}, 1, a^{2}, a^{3}, \dfrac{1}{a^{4}}  $ is equal to

  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sum of $\dfrac{1}{a},1,a^2,a^3,\dfrac{1}{a^4}$ is $1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}$

$AM\ge GM$
$\implies \dfrac{1+a^2+a^3+\frac{1}{a}+\frac{1}{a^4}}{5}\ge \sqrt[5]{(1)(a^2)(a^3)(\frac{1}{a})(\frac{1}{a^4})}$
$1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}\ge 5\sqrt[5]{1}$

$1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}\ge 5$
The minimum value of $1+a^2+a^3+\dfrac{1}{a}+\dfrac{1}{a^4}$ is $5$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

What is the greatest value of the positive integer n satisfying the condition $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +  ...... + \dfrac{1}{2^{n - 1}} < 2 - \dfrac{1}{1000}$?

  1. $8$
  2. $9$
  3. $10$
  4. $11$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given : $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+......+\dfrac{1}{2^{n−1}} < 2−\dfrac{1}{1000}$

Left side forms a sum of a finite geometric series with first term $1$ and common ratio $\dfrac{1}{2}$ with $n$ terms.
 Sum $ =\dfrac{a(1-r^{ n })}{(1-r)} = \dfrac{1(1-(0.5)^{ n })}{0.5} = 2-{ 2 }^{ 1-n }$
 So, $2-{ 2 }^{ 1-n } < 2-\dfrac { 1 }{ 1000 }$  
We have,
${ 2 }^{ n-1 } < 1000$ we get the max value of $n = 10$.
Hence, C is correct.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $y=a\log\left|x\right|+bx^{2}+x$ has extreme values at $x=2$ and $x=-4/3$ then 

  1. $a=12,b=-10$
  2. $a=4,b=-3/4$
  3. $a=-6,b=1/4$
  4. $none$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The derivative y' = a/x + 2bx + 1. Setting y' = 0 at x = 2 and x = -4/3 gives a system of equations: a/2 + 4b + 1 = 0 and a/(-4/3) - 8b/3 + 1 = 0. Solving these yields a=12 and b=-10.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $x^2+y^2=25$ , then $log _5 \begin {bmatrix} Max (3x+4y) \end {bmatrix}$ is

  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$log _5(3x+4y)$

Let $s=3x+4y$
given $x^2+y^2=25$
then $S=3x+4 \sqrt{25-x^2}$
$\dfrac{ds}{dx}=3+4 \dfrac{1}{2\sqrt(25-x^2)}$
$3=\dfrac{4x}{\sqrt{25-x^2}}$
$9(25-x^2)=16x^2$
$x=\pm 3$

and $x^2+y^2=25$
$y^2=25-x^2$
$y=\pm 4$
$\dfrac{d^2s}{dx^2}<0$ ; At $x=3 \,and\, y=4$

$S=3x+4y=3(3)+4(4)=25$
$log _5 (3x+4y)=log _5(s)=log _5(25)=log _5(5^2)=2$

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

The minimum value of $\displaystyle f(x)=|x-1|+|x-2|+|x-3|$ is equal to 

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

Solution:- (B) 2

The function $f$ is linear on each of the intervals $\left( - \infty, 1 \right], \left[ 1, 2\right], \left[2, 3\right] \text{ and } \left[ 3, \infty \right)$. Since a linear function on an interval always attains its minimum at one of the endpoints of the interval, and $f \left( x \right) = +\infty \text{ as } x = \pm \infty$, the function $f$ must attain its minimum at one of $x = 1, 2, 3$. Since $f(1)=3,  f \left( 2 \right) = 2 \text{ and } f \left( 3 \right) = 3$, the function $f$ attains a minimum of 2 at $x=2$.

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

For $\dfrac { { 2 }^{ 2 }+{ 4 }^{ 2 }+{ 6 }^{ 2 }+....+{ \left( 2n \right)  }^{ 2 } }{ { 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+....+{ \left( 2n-1 \right)  }^{ 2 } }$ to exceed $1.01$, the maximum value of $n$ is

  1. 149

  2. 150

  3. 151

  4. 152
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given


$\dfrac { { 2 }^{ 2 }+{ 4 }^{ 2 }+{ 6 }^{ 2 }....+{ (2n) }^{ 2 } }{ { 1 }^{ 2 }{ +3 }^{ 2 }{ +5 }^{ 2 }{ ....+(2n-1) }^{ 2 } } =\dfrac { \sum { { (2n) }^{ 2 } }  }{ \sum { { (2n-1) }^{ 2 } }  } $

$\sum { { (2n) }^{ 2 }=\sum { 4{ n }^{ 2 } } =4\times \sum { { n }^{ 2 } } =\dfrac { 4(n)(n+1)(2n+1) }{ 6 }  } $[since $\sum { { n }^{ 2 } } =\dfrac { (n)(n+1)(2n+1) }{ 6 } $]

$\sum { { (2n-1) }^{ 2 }=\sum { 4{ n }^{ 2 }+1-4n } =4\sum { { n }^{ 2 }+\sum { 1 }  }  } -4\sum { n } =\dfrac { 4(n)(n+1)(2n+1) }{ 6 } +n-\dfrac { 4(n)(n+1) }{ 2 } $[since $\sum { { n }^{ 2 }= } \dfrac { (n)(n+1) }{ 2 } $]

Now solving numerator and denominator we get

$\dfrac { { 4n }^{ 2 }+6n+2 }{ 4{ n }^{ 2 }-1 } $ to exceed $1.01$

 $n\Rightarrow$  $\in[0,150]$

Therefore maximim value of $n$ is 150.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

What is the least value of $a$ in $ \displaystyle\frac{\sqrt 2+\sqrt 3}{\sqrt{2+3}} < a$?

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac { \sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 2+3 }  } =\dfrac { \sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 5 }  } =\dfrac { (\sqrt { 2 } +\sqrt { 3 } )\times \sqrt { 5 }  }{ 5 } =\dfrac { 7.02 }{ 5 } \\ =1.40$
$\Rightarrow 1.40<a$
So, least integer value of $a$ is $2$.
Hence, option B is correct.
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The smallest between $\sqrt{17} - \sqrt{12}$ and $\sqrt{11} - \sqrt{6}$ is _________.

  1. $\sqrt{17} - \sqrt{12}$
  2. $\sqrt{11} - \sqrt{6}$

  3. Both are equal

  4. Can't be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sqrt{17} \approx 4.123$

$\sqrt{12} \approx 3.464$
$\sqrt{11} \approx 3.316$
$\sqrt{6} \approx 2.449$

$\Rightarrow \sqrt{17} - \sqrt{12} = 0.659$
$\Rightarrow \sqrt{11} - \sqrt{6} = 0.867$

Hence, $\sqrt{17}-\sqrt{12}$ is smaller.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The smallest of $\sqrt [ 3 ]{ 4 } , \sqrt [ 4 ]{ 5 } , \sqrt [ 4 ]{ 6 } , \sqrt [ 3 ]{ 8 } $ is:

  1. $\sqrt [ 3 ]{ 8 } $
  2. $\sqrt [ 4 ]{ 5 } $
  3. $\sqrt [ 3 ]{ 4 } $
  4. $\sqrt [ 4 ]{ 6 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\sqrt[3]{4}=\sqrt[12]{44}=\sqrt[12]{256}$
$\sqrt[4]{6}=\sqrt[12]{5^3}=\sqrt[12]{125}$
$\sqrt[4]{6}=\sqrt[12]{6^{3}}=\sqrt[12]{216}$
$\sqrt[3]{8}=\sqrt[12]{8^{4}}=\sqrt[12]{64^{2}}$
As $'125'$ is smallest
$\therefore \boxed{4\sqrt{5}}$ is smallest
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The smallest of $\sqrt[3]{4},    \sqrt[4]{5},     \sqrt[4]{6},    \sqrt[3]{8}$ is:

  1. $\sqrt[3]{8}$
  2. $\sqrt[4]{5}$
  3. $\sqrt[3]{4}$
  4. $\sqrt[4]{6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(B) $\sqrt[3]{4}, \sqrt[4]{5},  \sqrt[4]{6}, \sqrt[3]{8}$

$=4^{1/3}, 5^{1/4}, 6^{1/4}, 8^{1/3}$

L.C.M of 3 & 4 $=12$

So, the given surds can be written as,

$=4^{4/12}, 5^{3/12}, 6^{3/12}, 8^{4/12}$

$=(4^{4})^{1/12}, (5^{3})^{1/12}, (6^{3})^{1/12}, (8^{4})^{1/12}$

$=(256)^{1/12}, (125)^{1/12}, (216)^{1/12}, (4096)^{1/12}$

$\therefore $ The smallest one is $\sqrt[4]{5}$