Tag: maths

Questions Related to maths

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

reciprocal equations theory of equations maths

  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. $x= 2a,\dfrac{-(1+a^2)}{4a}$

Show answer
Correct Option: A
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
$x=\dfrac {-(1-a^2)\pm \sqrt {(1-a^2)^2-4(2a)[-(a+a^3)]}}{2(2a)}$
$=\dfrac {-(1-a^2)\pm \sqrt {(1+3a^2)^2}}{4a}$
$=\dfrac {-(1-a^2)\pm (1+3a^2)}{4a}$
$=\dfrac {-1+a^2+1+3a^2}{4a}; \dfrac {-1+a^2-1-3a^2}{4a}$
$=\dfrac {4a^2}{4a}; \dfrac {-2-2a^2}{4a}$
$=a; \dfrac {-(1+a^2)}{2a}$

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

reciprocal equations theory of equations maths

  1. $0$

  2. $1$

  3. $3$

  4. $-1$

Show answer
Correct Option: B
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$

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 

reciprocal equations theory of equations maths

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these

Show answer
Correct Option: A
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- $4x^4-9x^3+9x-4=0$

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

reciprocal equations theory of equations maths

  1. Reciprocal equation

  2. Radical equation

  3. Exponential equation

  4. Quadratic equation

Show answer
Correct Option: A
Explanation:

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

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

reciprocal equations theory of equations maths

  1. $13$

  2. $23$

  3. $14$

  4. $-13$

Show answer
Correct Option: A
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.

Find $x$,  $2^{x^2}:2^{2x}=8:1$

reciprocal equations theory of equations maths

  1. $3,-1$

  2. $3,1$

  3. $-3,-1$

  4. $-3,1$

Show answer
Correct Option: A
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
$\Rightarrow x^2-2x-3=0$
$\Rightarrow (x-3)(x+1)=0$
$\therefore x=3,-1$

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

reciprocal equations theory of equations maths

  1. $\dfrac{3}{4}, \dfrac{-1}{2}$

  2. $\dfrac{1}{4}, \dfrac{-1}{3}$

  3. $\dfrac{1}{3}, \dfrac{-1}{2}$

  4. $\dfrac{1}{4}, \dfrac{-1}{2}$

Show answer
Correct Option: D
Explanation:

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

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

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

reciprocal equations theory of equations maths

  1. $3$

  2. $5$

  3. $6$

  4. $4$

Show answer
Correct Option: B
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.

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

reciprocal equations theory of equations maths

  1. $2, 3$

  2. $1, 6$

  3. $-1, 3$

  4. $1, 3$

Show answer
Correct Option: D
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

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

reciprocal equations theory of equations maths

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

Show answer
Correct Option: B
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.