Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice reciprocal equations theory of equations maths

Solve $(1-a^2)(x+a)-2a(1-x^2)=0$

  1. $x= a,\dfrac{-(1+a^2)}{2a}$
  2. $x= 0,\dfrac{-(1+a^2)}{2a}$
  3. $x= 0,\dfrac{-(1+a^2)}{4a}$
  4. $x= 2a,\dfrac{-(1+a^2)}{4a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $(1-a^2)(x+a)-2a(1-x^2)=0$

$x+a-a^2x-a^3-2a+2ax^2=0$
$x-a-a^2x-a^3+2ax^2=0$
$2ax^2+x(1-a^2)-(a+a^3)=0$
By using formula, we have
$x=\dfrac {-(1-a^2)\pm \sqrt {(1-a^2)^2-4(2a)[-(a+a^3)]}}{2(2a)}$
$=\dfrac {-(1-a^2)\pm \sqrt {(1+3a^2)^2}}{4a}$
$=\dfrac {-(1-a^2)\pm (1+3a^2)}{4a}$
$=\dfrac {-1+a^2+1+3a^2}{4a}; \dfrac {-1+a^2-1-3a^2}{4a}$
$=\dfrac {4a^2}{4a}; \dfrac {-2-2a^2}{4a}$
$=a; \dfrac {-(1+a^2)}{2a}$

Multiple choice reciprocal equations theory of equations maths

Solve the reciprocal equation $x^4-3x^3+4x^2-3x+1=0$

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

We see that it is a reciprocal equation, so we divide it by $x^2$: 


$\Rightarrow$$x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}=0$


$\Rightarrow$$(x^2+\dfrac{1}{x^2})-3(x+\dfrac{1}{x})+4=0$


We will now substitute $x+\dfrac{1}{x}=u$

By squaring it, and solve further we get

$\Rightarrow$$x^2+\dfrac{1}{x^2}=u^2-2$

We plug back into the equation to get

$\Rightarrow$$u^2-3u+2=0$

The solutions are $u _{1}=1, u _{2}=2$
So either $x+\dfrac{1}{x}=1$ or $2$

From the first solution we get 
$\Rightarrow$$x^2-x+1=0$    which has no solution
From the second one we get
$\Rightarrow$$x^2-2x+1=0$ 
$\Rightarrow$$x=1$

Multiple choice reciprocal equations theory of equations maths

If the coefficients from one end of an equation are equal in magnitude and opposite in sign to the coefficients from the other end, then the equation is said to be 

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Consider a general equation:

$a _{n}x^n+a _{n-1}x^{n-1}+a _{n-2}x^{n-2}+....................+a _1x+a _0=0$

Now if  $a _n-i=-a _i ,$        $  i=0,1,2,3,...........n$

Then this type of equation is called as reciprocal equation of second type.

Ex- $4x^4-9x^3+9x-4=0$
Multiple choice reciprocal equations theory of equations maths

An equation of the form $2x^4-3x^3+7x^2-3x+2=0$ is called a .................

  1. Reciprocal equation

  2. Radical equation

  3. Exponential equation

  4. Quadratic equation
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As 'The coefficients from beginning to end and vice versa are the same'
therefore given equation is a reciprocal equation.

Multiple choice reciprocal equations theory of equations maths

Solve for $x$:   $\dfrac{8\sqrt{x-5}}{3x-7}=\dfrac{\sqrt{3x-7}}{x-5}$

  1. $13$
  2. $23$
  3. $14$
  4. $-13$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\cfrac{8\sqrt{x-5}}{3x-7}=\cfrac{\sqrt{3x-7}}{x-5}$

$\Rightarrow 8(x-5)^{\tfrac 32}=(3x-7)^{\tfrac 32}$
Squaring on both sides we get,
$\Rightarrow 64(x-5)^{3}=(3x-7)^{3}$
Taking cube roots on both sides, we get
$\Rightarrow 4(x-5)=(3x-7)$
$\Rightarrow 4x-20=3x-7$
$\Rightarrow x=13$
Hence, option A is correct.

Multiple choice reciprocal equations theory of equations maths

Find $x$,  $2^{x^2}:2^{2x}=8:1$

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

Given, $2^{x^2}:2^{2x}=8:1$

$\Rightarrow \dfrac {2{x^2}}{2^{2x}}=\dfrac {8}{1}$
$\Rightarrow 2^{x^2}=8.2^{2x}$
$\Rightarrow 2^{x^2}=2^3.2^{2x}$
$\Rightarrow 2^{x^2}=2^{2x+3}$
$\Rightarrow x^2=2x+3$ ....As bases are equal, powers must be equal
$\Rightarrow x^2-2x-3=0$
$\Rightarrow (x-3)(x+1)=0$
$\therefore x=3,-1$

Multiple choice reciprocal equations theory of equations maths

Solve the equation: $x^{-2}-2x^{-1}=8$

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

Given, $x^{-2}-2x^{-1}=8$

$\Rightarrow \dfrac {1}{x^2}-\dfrac {2}{x}=8$
$\Rightarrow \dfrac {1-2x}{x^2}=8$
$\Rightarrow 1-2x=8x^2$
$\Rightarrow 8x^2+2x-1=0$
$\Rightarrow (2x+1)(4x-1)$
$\Rightarrow x=\dfrac {1}{4}, \dfrac {-1}{2}$

Multiple choice reciprocal equations theory of equations maths

The number of solutions $(x, y, z)$ to the system of equations $ x + 2y + 4z = 9, 4yz + 2xz + xy = 13, xyz = 3 $ such that at least two of $ x, y, z$ are integers is

  1. $3$
  2. $5$
  3. $6$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the roots of the system equation are:
$\alpha =x,\beta =2y,\gamma =4z$
$\alpha +\beta +\gamma =x+2y+4z=9$
$\alpha \beta +\beta \gamma +\gamma \alpha =2xy+8yz+yzx$
$=2(4yz+2xz+xy)\Rightarrow 26$
$\alpha \beta \gamma =8xyz\Rightarrow 24$
Thus,our polynomial should be:
$P^{3}-9P+26P-24=0$
$(P-2)(P-3)(P-4)=0$
since our roots are :
$\alpha =x,\beta =2y$ and $\gamma =4z$
$(x,2y,4z)=(2,3,4)$ or its permutations,or 6 combination.
However,note that one case if,
$x=4,2y=3$ and $4z=2$
$(x,y,z)=(4,\dfrac{3}{2},\dfrac{1}{2})$
which two of the roots are not an integer :Excluding of this case ,we have five solutions.
Multiple choice reciprocal equations theory of equations maths

Solve the equation $\sqrt{4x^2-7x-15}-\sqrt{x^2-3x}=\sqrt{x^2-9}$

  1. $2, 3$
  2. $1, 6$
  3. $-1, 3$
  4. $1, 3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given equation is $\sqrt { 4{ x }^{ 2 }-7x-15 } =\sqrt { { x }^{ 2 }-9 } +\sqrt { { x }^{ 2 }-3x } $

$\Rightarrow \sqrt { x-3 } (\sqrt { 4x+5 } )=\sqrt { x-3 } (\sqrt { x+3 } +\sqrt { x } )$
Therefore $x=3$ is one solution and $\sqrt { 4x+5 } =\sqrt { x+3 } +\sqrt { x } $
By squaring above equation on both sides , we get $x+1=\sqrt{x(x+3)}$
Again square it on both sides , we get $x^{2}+2x+1=x^{2}+3x$
$\Rightarrow x=1$
Therefore option $D$ is correct

Multiple choice reciprocal equations theory of equations maths

The roots of $a _ { 1 } x ^ { 2 } + b _ { 1 } x + c _ { 2 } = 0$ are reciprocal of the roots of the equation $a _ { 2 } x ^ { 2 } + b _ { 2 } x + c _ { 2 } = 0$

  1. $\dfrac { a _ { 1 } } { a _ { 2 } } = \dfrac { b _ { 1 } } { b _ { 2 } } = \dfrac { c _ { 1 } } { c _ { 2 } }$
  2. $\dfrac { b _ { 1 } } { b _ { 2 } } = \dfrac { c _ { 1 } } { a _ { 2 } } = \dfrac { a _ { 1 } } { c _ { 2 } }$
  3. $\dfrac { a _ { 1 } } { a _ { 2 } } = \dfrac { b _ { 1 } } { c _ { 2 } } = \dfrac { c _ { 1 } } { b _ { 2 } }$
  4. $a _ { 1 } = \dfrac { 1 } { a _ { 2 } } , b _ { 1 } = \dfrac { 1 } { b _ { 2 } } , c _ { 1 } = \dfrac { 1 } { c _ { 2 } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given:${a} _{1}{x}^{2}+{b} _{1}x+{c} _{1}=0$     ........$(1)$
${a} _{2}{x}^{2}+{b} _{2}x+{c} _{2}=0$     ........$(2)$

Let $\alpha,\,\beta$ be the roots of ${a} _{1}{x}^{2}+{b} _{1}x+{c} _{1}=0$     ........$(1)$

$\Rightarrow\,\alpha+\beta=-\dfrac{{b} _{1}}{{a} _{1}}$ and 

$\alpha\beta=\dfrac{{c} _{1}}{{a} _{1}}$

Given:Roots of $(1)$ are reciprocal to $(2)$

$\dfrac{1}{\alpha}+\dfrac{1}{\beta}=-\dfrac{{b} _{2}}{{a} _{2}}$ and $\dfrac{1}{\alpha\beta}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{\alpha+\beta}{\alpha\beta}-\dfrac{{b} _{2}}{{a} _{2}}$ and $\dfrac{1}{\alpha\beta}=\dfrac{{c} _{2}}{{a} _{2}}$

Using $\alpha+\beta=-\dfrac{{b} _{1}}{{a} _{1}}$ and $\alpha\beta=\dfrac{{c} _{1}}{{a} _{1}}$ we have

$\Rightarrow\,\dfrac{-\dfrac{{b} _{1}}{{a} _{1}}}{\dfrac{{c} _{1}}{{a} _{1}}}=-\dfrac{{b} _{2}}{{a} _{2}}$ and $\dfrac{1}{\dfrac{{c} _{1}}{{a} _{1}}}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{-{b} _{1}}{{c} _{1}}=-\dfrac{{b} _{2}}{{a} _{2}}$ and
 
$\dfrac{{a} _{1}}{{c} _{1}}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}$ and 

$\dfrac{{a} _{1}}{{c} _{1}}=\dfrac{{c} _{2}}{{a} _{2}}$

$\Rightarrow\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}$ and 

$\dfrac{{c} _{1}}{{a} _{1}}=\dfrac{{a} _{2}}{{c} _{2}}$

$\Rightarrow\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}$ and 

$\dfrac{{c} _{1}}{{a} _{2}}=\dfrac{{a} _{1}}{{c} _{2}}$

$\therefore\,\dfrac{{b} _{1}}{{b} _{2}}=\dfrac{{c} _{1}}{{a} _{2}}=\dfrac{{a} _{1}}{{c} _{2}}$

Option$(b)$ is correct.