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 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 some important compounds of magnesium and calcium the s-block elements chemistry

In which of following cases is the value of x maximum?

  1. $CaSO _4,xH _2O$
  2. $BaSO _4,xH _2O$
  3. $MgSO _4,xH _2O$

  4. All have the same value of x
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$MgSO _4$ has 7 moles of water of hydration.
$CaSO _4$ has 2 moles of water hydration $BaSO _4$ has 1 mole of water of hydration.
Thus $MgSO _4$ has maximum x.

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 $\sqrt[4]{5}$

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

Which of the following numbers is the least ?
$\displaystyle (0.5)^{2},\sqrt{0.49},\sqrt[3]{0.008},0.23$

  1. $\displaystyle (0.5)^{2}$
  2. $\displaystyle \sqrt{0.49}$
  3. $\displaystyle \sqrt[3]{0.008}$

  4. 0.23
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ (0.5)^{2}=0.25$
$\sqrt{0.49}=0.7;$
$ \sqrt[3]{0.008}=\sqrt[3]{.2^3}=0.2$
$0.23$
Arranging in ascending order the numbers are $0.2< 0.23< 0.25< 0.7$
$ \therefore \sqrt[3]{0.008}=0.2$ is the least

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

The smallest of $\displaystyle \sqrt{8}+\sqrt{5},\sqrt{7}+\sqrt{6},\sqrt{10}+\sqrt{3}$ and $\displaystyle \sqrt{11}+\sqrt{2}$ is 

  1. $\displaystyle \sqrt{8}+\sqrt{5}$
  2. $\displaystyle \sqrt{7}+\sqrt{6}$
  3. $\displaystyle \sqrt{10}+\sqrt{3}$
  4. $\displaystyle \sqrt{11}+\sqrt{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \sqrt{8}+\sqrt{5}=2.83+2.24=5.07$
$\displaystyle \sqrt{7}+\sqrt{6}=2.65+2.45=5.09$
$\displaystyle \sqrt{10}+\sqrt{13}=3.16+3.61=6.77$
$\displaystyle \sqrt{11}+\sqrt{12}=3.32+1.41=4.73$
$\displaystyle \therefore $ Smallest is $\displaystyle \sqrt{11}+\sqrt{2}$

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

Which of the following is smallest?

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

Let us rewrite the given set of magnitudes $\sqrt [ 4 ]{ 5 } ,\sqrt [ 5 ]{ 4 } ,\sqrt { 4 },\sqrt {3}$ as follows:


$\sqrt [ 4 ]{ 5 } ={ \left( 5 \right)  }^{ \dfrac { 1 }{ 4 }  }\ \sqrt [ 5 ]{ 4 } ={ \left( 4 \right)  }^{ \dfrac { 1 }{ 5 }  }\ \sqrt { 4 } ={ \left( 4 \right)  }^{ \dfrac { 1 }{ 2 }  }\ \sqrt { 3 } ={ \left( 3 \right)  }^{ \dfrac { 1 }{ 2 }  }$

 
We now take the LCM of the denominators of the powers to make the denominators same, then the above magnitudes will be:

$\sqrt [ 4 ]{ 5 } ={ \left( 5 \right)  }^{ \dfrac { 1\times 5 }{ 4\times 5 }  }={ \left( 5 \right)  }^{ \dfrac { 5 }{ 20 }  }={ \left( { 5 }^{ 5 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 3125 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 3125 } \\ \sqrt [ 5 ]{ 4 } ={ \left( 4 \right)  }^{ \dfrac { 1\times 4 }{ 5\times 4 }  }={ \left( 4 \right)  }^{ \dfrac { 4 }{ 20 }  }={ \left( { 4 }^{ 4 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 256 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 256 } \\ \sqrt { 4 } ={ \left( 4 \right)  }^{ \dfrac { 1\times 10 }{ 2\times 10 }  }={ \left( 4 \right)  }^{ \dfrac { 10 }{ 20 }  }={ \left( { 4 }^{ 10 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 1048576 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 1048576 } \\ \sqrt { 3 } ={ \left( 3 \right)  }^{ \dfrac { 1\times 10 }{ 2\times 10 }  }={ \left( 3 \right)  }^{ \dfrac { 10 }{ 20 }  }={ \left( { 3 }^{ 10 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 2187 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 2187 }$     

Now, the descending order is as shown below:

$\sqrt [ 20 ]{ 1048576 } >\sqrt [ 20 ]{ 3125 } >\sqrt [ 20 ]{ 2187 } >\sqrt [ 20 ]{ 256 } \\ \Rightarrow \sqrt { 4 } >\sqrt [ 4 ]{ 5 } >\sqrt { 3 } >\sqrt [ 5 ]{ 4 }$ 

Hence, the smallest magnitude is $\sqrt [ 5 ]{ 4 }$.