Tag: maths

Questions Related to maths

State True or False, if the following expression is polynomial in one variable (State reason for your answer):
$y+\dfrac {2}{y}$
  1. True

  2. False


Correct Option: B
Explanation:

The expression $y+\dfrac2y$ contain the term $\dfrac2y$, here

the exponent of $y$ is $-1$, which is not a whole number.
Therefore, the given expression is not a polynomial in one variable.

State whether true/false:

The following expression is a polynomial in one variable:
$x^{10}+y^3+t^{50}$

  1. True

  2. False


Correct Option: B
Explanation:

Clearly, the given expression $x^{10}+y^3+t^{50}$ contains three variables $x,y\space and\space t$
Hence, the given expression is not a polynomial in one variable instead in three variables.

If $\displaystyle A=\pi \left ( R^{2}-r^{2} \right )$, then $R$ is equal to

  1. $\displaystyle \sqrt{\frac{A-\pi r^{2}}{\pi }}$

  2. $\displaystyle \sqrt{\frac{A+\pi r^{2}}{\pi }}$

  3. $\displaystyle \sqrt{\frac{r^{2}\pi -A}{\pi }}$

  4. $\displaystyle \sqrt{\frac{r^{2}\pi -A}{r}}$


Correct Option: B
Explanation:
Given, $A=\pi(R^2-r^2)$
Therefore, $A =$ $\displaystyle \pi R^{2}-\pi r^{2}$
$\Rightarrow  A+\pi r^{2}=\pi R^{2}$
$\displaystyle \Rightarrow R^{2}=\frac{A+\pi r^{2}}{\pi }$
$\displaystyle \Rightarrow $ $\displaystyle R=\sqrt{\frac{A+\pi r^{2}}{\pi }}$

The sum of the reciprocals of $\displaystyle\frac{x+3}{x^2+1}$ and $\displaystyle\frac{x^2-9}{x^2+3}$ is

  1. $\displaystyle\frac{x^3+2x^2-x}{x^2-9}$

  2. $\displaystyle\frac{x^3-2x^2+x}{x^2-9}$

  3. 1

  4. 0


Correct Option: B
Explanation:

Reciprocals will be $\frac { { { x }^{ 2 } }+1 }{ x+3 } $,$\frac { { x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
Their sum will be
$\frac { { { x }^{ 2 } }+1 }{ x+3 } +\frac { { x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
 $=\frac { \left( x-3 \right) \left( { x }^{ 2 }+1 \right) +{ x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
$=\frac { { x }^{ 3 }+x-3{ x }^{ 2 }-3+{ x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
 $=\frac { { x }^{ 3 }-2{ x }^{ 2 }+x }{ { x }^{ 2 }-9 } $

If $\displaystyle x^{2}-3x+1=0$ then the value of $\displaystyle x-\frac{1}{x}$ is

  1. $\displaystyle \sqrt{5}$

  2. $\displaystyle \sqrt{3}$

  3. $\displaystyle \sqrt{2}$

  4. $\displaystyle \sqrt{6}$


Correct Option: A
Explanation:

$x^{2}-3x+1$

$\therefore x^{2}+1=3x\Rightarrow \frac{x^{2}+1}{x}=\frac{3x}{x}\Rightarrow x+\frac{1}{x}=3$
$x^{2}+\frac{1}{x}^{2}=\left ( x+\frac{1}{x} \right )^{2}-2=(3)^{2}-=9-2=7$
We know 
$\left ( x-\frac{1}{x} \right )^{2}=x^{2}+\frac{1}{x}^{2}-2\Rightarrow 7-2=5$
$\therefore x-\frac{1}{x}=\sqrt{5}$

If $\displaystyle x-\frac{1}{x}=3$; then the value of $\displaystyle \frac{3x^{2}-3}{x^{2}+2x-1}$ is

  1. 9/5

  2. 8/5

  3. 7/5

  4. 6/5


Correct Option: A
Explanation:

Given $x+\frac{1}{x}=3$ multiply by x both sides

Then $x^{2}-1=3x$
$\Rightarrow x^{2}-3x-1=0$
So $\frac{3x^{2}-3}{x^{2}+2x-1}=\frac{3(x^{2}-1)}{x^{2}-3x-1+5x}= \frac{3\times 3x}{0+5x}= \frac{9x}{5x}=\frac{9}{5}$

If $x=2$, $y=3$, then $x^x+y^y$ is equal to

  1. $30$

  2. $37$

  3. $33$

  4. $31$


Correct Option: D
Explanation:

Given $x=2, y=3$

${ x }^{ x }+{ y }^{ y }$
$={ 2 }^{ 2 }+{ 3 }^{ 3 }=4+27=31$

If $\displaystyle x^{2}-11x+1=0 $ then the value of $\displaystyle x+\frac{1}{x}$ is

  1. 10

  2. 11

  3. 12

  4. 13


Correct Option: B
Explanation:

$\therefore x^{2}+1=3x$

$\frac{x^{2}+1}{x}= \frac{11x}{x}$
$\Rightarrow x+\frac{1}{x}=3$

If $\displaystyle x+\frac{a}{x}=b$ then the value of $\displaystyle \frac{x^{2}+bx+a}{bx^{2}-x^{3}}$ is

  1. $\displaystyle \frac{6b}{a}$

  2. $\displaystyle \frac{5b}{a}$

  3. $\displaystyle \frac{2b}{a}$

  4. $\displaystyle \frac{4b}{a}$


Correct Option: C
Explanation:

Given $x+\frac{a}{x}=b$ Multiply by x both sides

$x^{2}+a=bx$
$\Rightarrow x^{2}-bx+a=0$
So $\frac{x^{2}+bx+a}{bx^{2}-x^{3}}=\frac{x^{2}-bx+a+2bx}{-x(x^{2}-bx)}=\frac{0+2bx}{-x(-a)}=\frac{2bx}{ax}=\frac{2b}{a}$

If $x<-1$, then $x^2$

  1. $=1$

  2. $< 1$

  3. $>1$

  4. none of these


Correct Option: C
Explanation:

$x<-1$
 ${ x }^{ 2 }<{ \left( -1 \right)  }^{ 2 }\quad squaring\quad both\quad side$
 ${ x }^{ 2 }>1$