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Finding roots by iteration - class-XII

Description: finding roots by iteration
Number of Questions: 56
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Tags: complex numbers and linear inequations maths theory of equations solving equations numerically
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What are the solutions of the equation $x^2+8x+15=0$? 

  1. $3,-5$

  2. $3,5$

  3. $5,-3$

  4. $-5,-3$


Correct Option: D
Explanation:

${ x }^{ 2 }+8x+15=0\ { x }^{ 2 }+5x+3x+15=0\ (x+5)(x+3)=0\ x=-5,-3$

If $3$ times the third term of an A.P. is equal to $5$ times the fifth term. Then its $8$ term is

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: A
Explanation:

$3(a+2d)=5(a+4d)\ 3a+6d=5a+20d\ -2a=14d\ a=-7d\ a=+7d=-7d+7d={ 0 }$

If $a, b , c \in R $ and $3b^2 - 8ac < 0$ then the
equation $ax^4 + bx^3 +cx^2 +5x - 7=0$ has

  1. (a) all real roots

  2. (b) all imaginary roots

  3. (c) exactly two real and two imaginary roots

  4. (d) none


Correct Option: A

The solution to the equation ${7}^{1+x}+{7}^{1-x}=50$ is

  1. $0$

  2. $\pm 1$

  3. $2$

  4. none of these


Correct Option: B
Explanation:

$7^{1+x}+7^{1-x}=50$

$7(7^{x}+7^{-x})=50$
$7(7^{2{x}})+7=50(7^{x})$
$\implies (7^{x}-7)(7^{x+1}-1)=0$
$\implies 7^{x}=7^{1}$ or $7^{x+1}=7^{0}\implies x=\pm 1$

Solve the equation $y^2 + 2y = 40$, correct to $1$ decimal place using trial and improvement method.

  1. $5.8$

  2. $5.4$

  3. $5.7$

  4. $5.9$


Correct Option: B

The number of solution of $2\cos^2\dfrac{\pi}{2}\sin^2x=x^2+\dfrac{1}{x^2},\;0 \le x \le \dfrac{\pi}{2}$ is 

  1. Zero

  2. One

  3. Infinite many

  4. Four


Correct Option: A
Explanation:

$2\left ( \cos^2 \dfrac{\pi }{2} \right )\left ( \sin^{2}x \right )=x^{2}+\dfrac{1}{x^{2}}$

 
$=2(0) \sin^{2}x=x^{2}+\dfrac{1}{x^{2}}$ 

$=x^{2}+\dfrac{1}{x^{2}}=0$ 

not possible for any $ X\in R$ 

$\therefore $ no. of solutions = $0 $

find the value of $f(2)$ if $f(x)=x^3+x^2+x+1$

  1. 15

  2. 12

  3. 10

  4. None of these


Correct Option: A
Explanation:

$f(x)=x^3+x^2+x+1\f(2)=2^3+2^2+2+1\f(2)=8+4+2+1=15$

The value of $p$ if $\dfrac 3p+\dfrac 4p=1$

  1. 7

  2. 3

  3. 4

  4. 1


Correct Option: A
Explanation:

$\dfrac 3p+\dfrac 4p=1\3+4=p\p=7$

Simplify $(3x-11y) -(17x+13y)$ and choose the right answer. 

  1. $7x - 12y$

  2. $14x - 54y$

  3. $-3 (5x-4y)$

  4. $-2 (7x+12y)$


Correct Option: D
Explanation:
$\left(3x-11y\right)-\left(17x+13y\right)$

$=3x-11y-17x-13y$

$=-14x-24y$

$=-2\left(7x+12y\right)$

Solve $\frac { 7 y + 4 } { y + 2 } = \frac { - 4 } { 3 }$

  1. -4$/ 5$

  2. 5$/9$

  3. -9$/2$

  4. 4$/5$


Correct Option: A
Explanation:

Given $\dfrac{7 y+4}{y+2}=\dfrac{-4}{3}$

$\implies 3(7 y+4)=-4(y+2)$
$\implies 21 y+12=-4{y}-8$
$\implies 25 y=-20$
$\implies y=\dfrac{-20}{25}$
$\implies y=-\dfrac{4}{5}$

The product of $\left( { 23 x }^{ 2 }{ y }^{ 2 }z \right)$ and $\left( -15{ x }^{ 3 }{ yz }^{ 2 } \right) $ is .........................  .

  1. $-345 { x }^{ 5 } { y }^{ 3 } { z }^{ 3 }$

  2. $345 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$

  3. $145 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$

  4. $170 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$


Correct Option: A
Explanation:
$\left(23{x}^{2}{y}^{2}z\right)\times\left(-15{x}^{3}y{z}^{2}\right)$
$=-345‬{x}^{2+3}{y}^{2+1}{z}^{1+2}$
$=-345{x}^{5}{y}^{3}{z}^{3}$

The number of integers (positive, negative or zero) solutions of
$xy-6(x+y)=0$ with x is less than or equal to y is:

  1. 5

  2. 10

  3. 12

  4. none of these


Correct Option: A

The degree of polynomial $p(x)=x^ {2}-3x-4x^ {3}-6$ is

  1. $2$

  2. $1$

  3. $3$

  4. $6$


Correct Option: C
Explanation:

Given : $p{\left( x \right)} = {x}^{2} - 3x - 4 {x}^{3} - 6$


Since the greatest exponent in $p{\left( x \right)}$ is $3$, thus the degree of $p{\left( x \right)}$ is $3$.

The method of finding solution by trying out various values for the variable is called

  1. Rrror method

  2. Trial and error method

  3. Testing method

  4. Checking method


Correct Option: B
Explanation:

The required method is called "Trial and error method"

If $a\times b=\frac {a}{b}+\frac {b}{a}-ab$, then the value of $1^*2$ is

  1. 2

  2. 1/2

  3. 2/3

  4. 1


Correct Option: B
Explanation:

$1^*2=\frac {1}{2}+\frac {2}{1}-1\times 2=\frac {1+4-4}{2}=\frac {1}{2}$

If $(8x)^2 + (6x)^2 = d^2$ and $d = 200$, then $8x \times 6x$ is equal to

  1. 18,200

  2. 18,500

  3. 18,900

  4. 19,200


Correct Option: D
Explanation:

$(8x)^2+(6x)^2=d^2$
or $64x^2+36x^2=d^2$
or $100x^2=d^2$
or $d=\sqrt {100x^2}=10x$
$10x=200$
$\therefore x=\frac {200}{10}=20$
Hence, $8x\times 6x=8\times 20\times 6\times 20=19,200$

If $2\pi rh = 2\pi r^2$ and $h = 5$, then r is equal to

  1. 5

  2. 110

  3. 15

  4. 12


Correct Option: A
Explanation:

$2\pi r h=2\pi r^2$
$2\pi \times r\times 5=2\pi r^2$ or $r=5$

Is the following quadratic polynomial reducible or irreducible?
$f(x) = -2x^2-2x-1$

  1. Reducible with one real root

  2. Reducible with two real roots

  3. Irreducible

  4. None of these


Correct Option: C
Explanation:

To check whether the given quadratic polynomial is reducible or irreducible, we need to calculate the discriminant

Calculate the discriminant for the equation, $-2x^2-2x-1=0$

$D=b^{2}-4ac=(-2)^2-4(-2)(-1)=-4<0$
Quadratic equation is irreducible if $D<0$
$\therefore$ The quadratic polynomial is irreducible.

Correct option is C

If $\dfrac {1}{x}-\dfrac {1}{y}=\dfrac {1}{z}$, then z is equal to

  1. y-x

  2. x-y

  3. $\dfrac {y-x}{xy}$

  4. $\dfrac {xy}{y-x}$


Correct Option: D
Explanation:

$\dfrac {1}{x}-\dfrac {1}{y}=\dfrac {1}{z}$ or $\dfrac {y-x}{xy}=\dfrac {1}{z}$ or $\dfrac {z}{1}=\dfrac {xy}{y-x}$

Which of the following quadratics is irreducible?

  1. $2x^2 - 5x + 3$

  2. $2x^2 - 5x - 3$

  3. $5x^2 - 2x + 3$

  4. $5x^2 - 2x - 3$


Correct Option: C
Explanation:

We can check by comparing the D of quadratic equations with their relation with 0


$2x^2-5x+3=0$
$D=(-5)^2-4(2)(3)=25-24=1>0$
Reducible

$2x^2-5x-3=0$
$D=(-5)^2-4(2)(-3)=25+24=49>0$
Reducible

$5x^2-2x+3=0$
$D=(-2)^2-4(5)(3)=4-60=-56<0$
Irreducible


$5x^2-2x-3=0$
$D=(-2)^2-4(5)(-3)=4+60=64>0$
Rreducible


Therefore correct option is C


x is ........... variable

  1. Dependent

  2. Independent

  3. None

  4. both


Correct Option: B
Explanation:

Value of x does not depend on any other variable. so it is independent variable.

When multiplicity of a polynomial exist?

  1. when a factor appears in conjugate

  2. when a factor appears more than once

  3. when a factor appears more than twice

  4. None of the above


Correct Option: B
Explanation:

"multiplicity of a polynomial exist when a factor of a polynomial repeats more than once"

For eg. In the polynomial $(x -2)^3(x-3)^2(x-1 )$ 
The root $2$ have multiplicity $3$, the root  $3$ have $2$, and the root $1$ have $1$
Hence, B is correct.

Let $R=gS-4$. When $S=8,R= 16$. When $S= 10$, R is

  1. 11

  2. 14

  3. 21

  4. 20


Correct Option: C
Explanation:

$R=gS-4, S=8, R=16$
or $16=g\times 8-4$ or $16+4=g\times 8$
or $g=\frac {20}{8}=\frac {5}{2}$
Therefore, when $S=10$
$R=\frac {5}{2}\times 10-4=21$

A three-digit number beginning from the left is abc. The number is

  1. cba

  2. a+10b+100c

  3. 100a+10b+c

  4. none of these


Correct Option: C
Explanation:

Here, c is the digit in the units' place; b standing in the tens' place represents b tens; similarly, a represents a hundreds.
The number is, therefore, equal to a hundreds + b tens + c units $= 100a + 10b + c$.

Condition for an irreducible quadratic equation is-

  1. discriminant is positive

  2. discriminant is negative

  3. discriminant is zero

  4. None of the above


Correct Option: B
Explanation:

Irreducible quadratic equation can not be reduced more i.e the quadratic equation which do not have real roots   

This means the roots are imaginary
So if roots are imaginary, then discriminant $D  <  0$
Hence, option B is correct.

The factorized form of $x^5 +x^4-x-1$ is

  1. $(x+1)^2(x+1)^2(x-1)$

  2. $(x^2+1)(x^2+1)^2(x-1)$

  3. $(x^2+1)(x+1)^2(x-1)$

  4. $(x^2+1)^2(x-1)$


Correct Option: C
Explanation:

$x^5 +x^4-x-1$
$=x^4(x + 1)-1(x + 1)$
$=(x +1)(x^4-1)$
$= (x + 1) (x^2+ 1) (x^2-1)$
$ = (x + 1) (x^2+ 1) (x + 1) (x -1)$
$ = (x^2+ 1) (x + 1)^2(x -1)$

Solve the simultaneous equations using the convergent iterations:
$5x$ + $y$ + $2z$ = $19$
$2x$ + $3y$ +$8z$ = $39$
$x$ + $4y$ -$2z$ = $-2$

  1. $x=2$, $y=2$ and $\text z =1$

  2. $x=2$, $y=1$ and $\text z =4$

  3. $x=3$, $y=2$ and $\text z =2$

  4. None of these


Correct Option: B
Explanation:

$5x+y+2z=19$

$x+4y-2z=-2$

Adding above two equations we get,
$5x+y+2z+x+4y-2z=19-2$
$\implies 6x+5y=17$

Taking another equation 
$2x+3y+8z=39$

Multiplying the given last equation with $4$
$4x+16y-8z=-8$

Adding above two equations 
$2x+3y+8z+4x+16y-8z=31$
$\implies 6x+19y=31$

Now,
$6x+19y-(6x+5y)=31-17$
$\implies 19y-5y=14$
$\implies y=1$

By substituting the value of $y$ into the above equations 
we get,
$x=2$ and $z=4$
$\therefore\ x=2,y=1$ and $z=4$.

Solve the simultaneous equations using the convergent iterations:
$12x$ + $3y$ - $5z$ = $1$
$x$ + $5y$ +$3z$ = $28$
$3x$ + $7y$ $13z$ = $76$

  1. $x=1$, $y=4$ and $\text z =3$

  2. $x=1$, $y=3$ and $\text z =4$

  3. $x=2$, $y=3$ and $\text z =5$

  4. $x=3$, $y=2$ and $\text z =5$


Correct Option: B

Find the answer to the equation $x^{3}$ - $2x$ = $25$  to one decimal place using trial and improvement method. 

  1. $3.7$

  2. $3.2$

  3. $3.6$

  4. $3.9$


Correct Option: B
Explanation:

$f(x)={ x }^{ 3 }-2x=25$

$1.x=2$
${ x }^{ 3 }-2x=4$ (too small)

$2.\quad x=3\\ { x }^{ 3 }-2x=21$
$3.x=4{ x }^{ 3 }-2x=56$
$4.x<4\quad and\quad x>3$
$x=3.1$  
${ x }^{ 3 }-2x=23.6$ (close)
$ 5.x=3.2$
$ { x }^{ 3 }-2x=26.4$
$ x=3.2$

Use the Zero Product Property to solve the equation $(7x+2) (5x-4)=0$

  1. $\dfrac{2}{7}$ , $ \dfrac{-4}{5}$

  2. $\dfrac{-2}{7}$ , $ \dfrac{4}{5}$

  3. $\dfrac{-2}{5}$ , $ \dfrac{7}{5}$

  4. $\dfrac{2}{5}$ , $ \dfrac{-7}{5}$


Correct Option: B
Explanation:

$(7x+2)(5x-4)=0\ (7x+2)=0\quad ;\quad (5x-4)=0\ x=-\dfrac { 2 }{ 7 } \quad ;\quad x=\dfrac { 4 }{ 5 } $

The solution of the equation ${\left| {x + 1} \right|^2} - \left| {x + 2} \right| - 26 = 0$ is:

  1. $ \dfrac{-1 + \sqrt{(109)}}{2}$,$ \dfrac{-3 - \sqrt{(101)}}{2}$

  2. $ - 7,\sqrt {29} $

  3. $ \pm \sqrt {29} $

  4. $ - 7,29$


Correct Option: A
Explanation:

$|x+1|^2$ will be always non negative so we can expand it and remove modulus.


So, Equation becomes $x^2+2x-25 -|x+2| = 0$ 


For $x > -2$
$x^2 + x - 27 = 0$ which gives $x=\dfrac{-1 ^+ _-\sqrt{(109)}}{2}$
But since we assumed $x>-2$, $x=\dfrac{-1 + \sqrt{109}}{2}$

Now for $x<-2$
$x^2+3x -23$ which gives x = $\dfrac{-3 ^+ _- \sqrt{101}}{2}$
But since we assumed $x<-2$, x = $\dfrac{-3-\sqrt{101}}{2}$

$36$ factorized into two factors in such a way that sum of factors is minimum, then the factors are

  1. $2, 18$

  2. $9, 4$

  3. $3, 12$

  4. None of these


Correct Option: D
Explanation:

$36 = 1 \times 36$
     $= 2 \times 18$
     $= 3 \times 12$
     $= 4 \times 9$
     $= 6 \times 6$

1 + 36 = 37,  2 + 18 = 20,  3 + 12 = 15,  4 + 9 = 13,  6 + 6 = 12

Here, $12 < 13 < 15 < 20 < 37$


$\therefore \left( {6,6} \right)$ 

None of These

The first and last term of an A.P. are $1$ and $11$. If the sum of its terms is $36$, then the number of terms will be

  1. $5$

  2. $6$

  3. $7$

  4. $8$


Correct Option: B
Explanation:

$a=1\ a+(n-1)d=11\ 1+(n-1)d=11\ (n-1)d=10$

Also, Sum $= \cfrac { [a+(a+(n-1)d)] }{ 2 } n$
$\Rightarrow 36=n\times \cfrac { 12 }{ 2 }$
$\Rightarrow n=6$

Number of real roots of equation 
(x+1) (x+2) (x+3) (x+4) -8 =0 is

  1. 0

  2. 2

  3. 4

  4. 3


Correct Option: A
Explanation:

$\begin{matrix} \left( { x+1 } \right) \left( { x+2 } \right) \left( { x+4 } \right) =8 \ { x^{ 4 } }+{ 10^{ 3 } }+35{ x^{ 2 } }+50x+16=0 \ From\, \, Oescantes\, rule\, of\, sign\, of\, \, sign\,  \ There\, will\, be\, no\, positive\, \, roots\,  \ f\left( { -x } \right) =\, \, \, { x^{ 4 } }-10{ x^{ 3 } }+35{ x^{ 2 } }-50x+60=0 \ and\, posibility\, \, of\, negative\, roots\, \, and\, 0,2\, \, or\, \, 4 \ but\, no\, \, negative\, number\, making\, this\, equation\, '0'\, \, so\, it\, has\, no\, real\, roots\,  \  \end{matrix}$

If $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \neq 0 , x = c y + b z , y = a z + c x$ and $z = b x + a y ,$ then $a ^ { 2 } + b ^ { 2 } + c ^ { 2 } + 2 a b c =$

  1. 2

  2. $a + b + c$

  3. 1

  4. $ab + bc + ca$


Correct Option: C

If $f(x) = a{x^7} + b{x^3} + cx - 5 \,\,\,\,\,a,b,c$ are real constants and $f( - 7) = 7$ then the range of $f(7) + 17\cos x$ is

  1. $\left[ { - 34,0} \right]$

  2. $\left[ {0,34} \right]$

  3. $\left[ { - 34,34} \right]$

  4. $\left[ {34,\infty } \right]$


Correct Option: A
Explanation:
$f(x)=ax^7+6x^3+cx-5$

$f(-7)=7$

$f(7)+f(-7)=-10$

$f(7)+7=-10$

$f(7)=-17$

Range of $-17+17\cos x$ is form $[-34,\ 0]$



How many distinct real solutions does the equation $((x^2 - 2)^2 - 5)^2 = 1$ have ?

  1. 5

  2. 6

  3. 8

  4. 9


Correct Option: A
Explanation:

This equation is equivalent to
$(x^2 - 2)^2 - 5 = 1$  or $(x^2 -2)^2 - 5 = -1$


The first is equivalent to $x^2 - 2 = \sqrt{6}$ or $\ x^2 - 2 = - \sqrt{6}$, 
$x=\pm (2+\sqrt 6)$ or $x^2 \neq -\sqrt 6 +2 $
with $2$ and $0$ solutions respectively (since - $\sqrt{6}$ + 2 < 0).

The latter is equivalent to $x^2 - 2 = 2$ or $x^2 - 2 = -2,$ 
$x=\pm 2$ or $x=0$
with $2$ and $1$ solution(s) respectively.

So we have $2+0+2+1 = 5$ solutions in total.

$x$ and $y$ are real numbers such that ${7^x} - 16y = 0\;{\text{and}}\;{4^x} - 49y = 0,$ then the value of $\left( {y - x} \right)$ is

  1. $\dfrac{5}{2}$

  2. $\dfrac{{19}}{5}$

  3. $\dfrac{{4115}}{{2013}}$

  4. $\dfrac{{1569}}{{784}}$


Correct Option: D
Explanation:
$7^{x}=16y$

$4^{x}=49 y$

$\Rightarrow \dfrac{7^{x}}{4^{x}} = \dfrac{16}{49}$

$\Rightarrow \left( \dfrac{7}{4} \right)^{x} = \left( \dfrac{4}{7} \right)^{2} = \left( \dfrac{7}{4} \right)^{-2}$

$\Rightarrow x=-2$

$y= \dfrac{7^{x}}{16}$

$\Rightarrow y= \dfrac{1}{49 \times 16}$

So, $y-x = \dfrac{1}{49 \times 16}+2$

$=\dfrac{1}{784}+2$

$=\dfrac{1569}{784}$

if $3^{x}-3^{x-1}=18$, then $x^{x}$ is equal to

  1. $3$

  2. $8$

  3. $27$

  4. $216$


Correct Option: C
Explanation:

$3^x-3^{x-1}=18$

$\Rightarrow 3^x-\dfrac{3^x}{3}=18$
$\Rightarrow 3^x\left(\dfrac{2}{3}\right)=18$
     $3^{x-1}.2=3^2.2$
     $x-1=2$
$\therefore x=3$
    $x^x=3^3=27.$
Hence, the answer is $27.$

If $\quad y={ log } _{ x }({ log } _{ e }x)({ log } _{ e }x)\quad then\quad \dfrac { dy }{ dx } \quad equals$ to 

  1. $\dfrac { 1 }{ x{ log } _{ x }{ log } _{ x }x } $

  2. $\dfrac { 1 }{ x{ log } _{ e }x } $

  3. 0

  4. none of these


Correct Option: B
Explanation:
Given,

$y=\log _x\left(\log _e\left(x\right)\right)\left(\log _e\left(x\right)\right)$

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\log _x\left(\log _e\left(x\right)\right)\log _e\left(x\right)\right)$

$\left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'\rightarrow f=\log _x\left(\log _e\left(x\right)\right),\:g=\log _e\left(x\right)$

$=\dfrac{d}{dx}\left(\log _x\left(\log _e\left(x\right)\right)\right)\log _e\left(x\right)+\dfrac{d}{dx}\left(\log _e\left(x\right)\right)\log _x\left(\log _e\left(x\right)\right)$

$=\dfrac{1-\log _e \left(\log \left(x\right)\right)}{x\log ^2\left(x\right)}\log _e\left(x\right)+\dfrac{1}{x}\log _x\left(\log _e\left(x\right)\right)$

$=\dfrac{1}{x\log _e \left(x\right)}$

The number of ordered pairs of integers (x,y) satisfying the equation
${ x }^{ 2 }+6x+{ y }^{ 2 }=4$ is

  1. 2

  2. 4

  3. 6

  4. 8


Correct Option: D
Explanation:
Given,

$x^2+6x+y^2=4$ 

$\Rightarrow x^2+6x+9+y^2=4+9$

$\Rightarrow x^2+3^2+2(3x)+y^=13$

$\Rightarrow (x+3)^2+y^2=13$

sum of two squares is $13$

$\therefore $ when $(x+3)^2=9,x=0,-6$ and $y^2=2,-2$ $\Rightarrow 4$ ordered pairs.

$\therefore $ when $(x+3)^2=4,x=-1,-5$ and $y^2=3,-3$ $\Rightarrow 4$ ordered pairs.

A total of $8$ pairs.

The solution set of the system of equations $\log _{ 3 }{ x } +\log _{ 3 }{ y } =2+\log _{ 3 }{ 2 } \quad and\quad \log _{ 27 }{ (x+y) } =\dfrac { 2 }{ 3 } $ is :

  1. {6,3}

  2. {3,6}

  3. {6,12}

  4. {12,6}


Correct Option: A
Explanation:
Given,

$\log _3\left(x\right)+\log _3\left(y\right)=2+\log _3\left(2\right)$

$\log _3\left(x\right)=2+\log _3\left(2\right)-\log _3\left(y\right)$

$x=3^{2+\log _3\left(2\right)-\log _3\left(y\right)}$

$=3^{\log _3\left(2\right)}\cdot \:3^{-\log _3\left(y\right)}\cdot \:3^2$

$=2\cdot \:3^{-\log _3\left(y\right)}\cdot \:3^2$

$=2y^{-1}\cdot \:3^2$

$\Rightarrow x=\dfrac{18}{y}$.......(1)

Now,

$\log _{27}\left(x+y\right)=\dfrac{2}{3}$

from (1)

$\log _{27}\left(\dfrac{18}{y}+y\right)=\dfrac{2}{3}$

$\dfrac{18}{y}+y=27^{\frac{2}{3}}$

$\Rightarrow 18+y^2=9y$

$y^2-9y+18=0$

$(y-3)(y-6)=0$

$\therefore y=3,6$

$x=\dfrac{18}{y}=\dfrac{18}{3}=6$

$x=\dfrac{18}{y}=\dfrac{18}{6}=3$

$(x,y)=\left \{ 6,3 \right \}or\left \{ 3,6 \right \}$

Number of real solutions of the equation $\sqrt { \log _{ 10 }{ (-x) }  } =\log _{ 10 }{ \sqrt { { x }^{ 2 } }  } $ is :

  1. zero

  2. exactly 1

  3. exactly 2

  4. 5


Correct Option: A
Explanation:
Given,

$\sqrt{\log _{10}(-x)}=\log _{10}(\sqrt{x^2})$

$\sqrt{\log _{10}(-x)}=\log _{10}x$

$\log _{10}\left(-x\right)=\left(\log _{10}\left(x\right)\right)^2$

$\log _{10}\left(-1\right)+\log _{10}\left(x\right)=\left(\log _{10}\left(x\right)\right)^2$

let $\log _{10}\left(x\right)=u$

$\log _{10}\left(-1\right)+u=\left(u\right)^2........\log _{10}(-1) \ is \ not \ defined$

$u=\mathrm{Undefined}$

$\:\log _{10}\left(x\right)=\mathrm{Undefined}:\quad x=10^{\mathrm{Undefined}}$

$x=10^{\mathrm{Undefined}}\space\mathrm{False}$

No solution for $\:x\in \mathbb{R}$

Number of solutions satisfying, $\sqrt { 5-{ log } _{ 2 }x } =3-{ log } _{ 2 }x$ are :

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: A
Explanation:

$\sqrt { 5-{ \log } _{ 2 }x } =3-{ \log } _{ 2 }x$                  ---- ( 1 )


Let $\log _2x=y$                     ----- ( 2 )


$\Rightarrow$  $\sqrt{5-y}=3-y$

Squaring both sides, 

$\Rightarrow$  $5-y=9-6y+y^2$

$\Rightarrow$  $y^2-5y+4=0$

$\Rightarrow$  $y^2-4y-y+4=0$

$\Rightarrow$  $y(y-4)-1(y-4)=0$

$\Rightarrow$  $(y-4)(y-1)=0$

$\Rightarrow$  $y=4$ and $y=1$

Substituting $y=1$ in ( 1 ) we get,

$\Rightarrow$  $\log _2x=1$

We know, $\log _ba=x\Rightarrow a=b^x$

$\therefore$  $x=2^1=2$

Substituting $y=4$ in ( 1 ) we get,

$\Rightarrow$  $\log _2x=4$

We know, $\log _ba=x\Rightarrow a=b^x$

$\therefore$  $x=2^4=16$

Substituting $\log _2x=1$ in ( 1 ) we get,

$\sqrt{5-1}=3-1$

$\Rightarrow$  $\sqrt{4}=2$

$\therefore$  $2=2$

Substituting $\log _2x=4$ in ( 1 ) we get,

$\sqrt{5-4}=3-4$

$\Rightarrow$  $\sqrt{1}=-1$

$\therefore$  $1=-1$

Hence, we can see only $\log _2x=1$ satisfying.

$\therefore$  $\sqrt { 5-{ \log } _{ 2 }x } =3-{ \log } _{ 2 }x$ has only $1$ solution.

Number of ordered pair(s) of (x,y) satisfying the system of equations, $\log _2 xy = 5$ and $\log _{\frac{1}{2}} \frac{x}{y} = 1$ is:

  1. one

  2. two

  3. three

  4. four


Correct Option: B
Explanation:
Given,

$\log _2\left(xy\right)=5,\:\log _{\frac{1}{2}}\left(\frac{x}{y}\right)=1$

$\log _2\left(xy\right)=5$

$xy=2^5=32$

$x=\dfrac{32}{y}$

$\log _{\frac{1}{2}}\left(\dfrac{x}{y}\right)=1$

$\log _{\frac{1}{2}}\left(\dfrac{\frac{32}{y}}{y}\right)=1$

$\dfrac{\frac{32}{y}}{y}=\left(\dfrac{1}{2}\right)^1$

$\dfrac{32}{y}=\dfrac{1}{2}y$

$y^2=64$

$\Rightarrow y=\pm 8$

$x=\dfrac{32}{\pm 8}=\pm 4$

$(4,8),(-4,-8)$ Therefore $2$ ordered pairs

If $H.C.F. \left(a,b\right) = 9$ and $a  . b = 100$, then $L.C.M.\left(a,b\right) =$

  1. $200$

  2. $100/3$

  3. $100/9$

    1. $150$

Correct Option: C
Explanation:

$H.C.F.(a,b)=9\ a.b=100\ L.C.M.(a,b)=\frac { a.b }{ H.C.F } \ =\quad \frac { 100 }{ 9 } $

Find multiplicity of the polynomial
$f(x) = (x-1)^2(2x+5)^3(x^2+1)^2(x+\pi^2)^4$

  1. $4$

  2. $2$

  3. $8$

  4. $1$


Correct Option: A
Explanation:

The roots of the given function are $1,-\frac{5}{2} ,-i,i,-\pi^{2}$

The multiplicity of$x=1$ is $2$
The multiplicity of $x=-\frac{5}{2}$ is $3$
The multiplicity of $x=i $ is $2$
The multiplicity of $x=-i$ is $2$
The multiplicity of $x=-\pi^{2}$ is $4$
Therefore the multiplicity of polynomial is $4$
Therefor option $A$ is correct

The multiplicity of the root $x=1$ for the function $f(x) = x^2(x+1)^3(x-2)^2(x-1)$ is

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: A
Explanation:

After factorising $f(x)$, $(x-1)$ is appearing only once 

Multiplicity of the root is the number of times a number is zero of a polynomial.
Here $x=1$ is the zero of $f(x)$ only once as it is appearing one.
So, the multiplicity of the root $x=1$x=1 for the function is one.
Hence, option A is correct.

List the multiplicities of the zeroes of the polynomial $P(x)=x^2-14x+49$

  1. $x=5$ is a zero of multiplicity $3$.

  2. $x=7$ is a zero of multiplicity $2$.

  3. $x=6$ is a zero of multiplicity $1$.

  4. None of these


Correct Option: B
Explanation:

Given, $x^2-14x+49$

$\Rightarrow x^2-7x-7x+49$
$\Rightarrow x(x-7)-7(x-7)$
$\Rightarrow (x-7)^2$
So, $x=7$ is a zero of multiplicity $2$.

Is the following quadratic polynomial reducible or irreducible?
$f(x) = x^2 - \sqrt2$

  1. Reducible with one real root

  2. Reducible with two real roots

  3. Irreducible

  4. None of these


Correct Option: B
Explanation:

Let $\sqrt 2=a^2\Rightarrow a^2=2^{1/2}\Rightarrow a=2^{1/4}$

So $f(x)=x^2-a^2=(x+a)(x-a)=(x+2^{1/4})(x-2^{1/4})$, using $a^2-b^2=(a+b)(a-b)$
So given quadratic is reducible to two real roots 

If $x=2+\sqrt{3}$, $xy=1$, then $\cfrac { x }{ \sqrt { 2 } +\sqrt { x }  } +\cfrac { y }{ \sqrt { 2 } +\sqrt { y }  } =$.......

  1. $\sqrt{2}$

  2. $\sqrt{3}$

  3. $1$

  4. $2$


Correct Option: A
Explanation:
Given $x=2+\sqrt{3}$ and $xy=1$
$\Rightarrow\,y=\dfrac{1}{x}=\dfrac{1}{2+\sqrt{3}}=\dfrac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$
Let $\sqrt{x}=\sqrt{a}+\sqrt{b}$
then $x=a+b+2\sqrt{ab}$ by squaring both sides
We have $x=2+\sqrt{3}=a+b+2\sqrt{ab}$
$\Rightarrow\,a+b=2,\,\sqrt{ab}=\dfrac{\sqrt{3}}{2}$ or $ab=\dfrac{3}{4}$
$\Rightarrow\,{\left(a-b\right)}^{2}={\left(a+b\right)}^{2}-4ab=4-4\times \dfrac{3}{4}=4-3=1$
$\Rightarrow\,a-b=1$
$\Rightarrow\,a+b=2,\,a-b=1$
$\Rightarrow\,2a=3$
$\Rightarrow\,a=\dfrac{3}{2}$
Put $a=\dfrac{3}{2}$ in $a+b=2$
$b=2-a=2-\dfrac{3}{2}=\dfrac{4-3}{2}=\dfrac{1}{2}$
$\therefore\,a=\dfrac{3}{2},b=\dfrac{1}{2}$
So,$\sqrt{x}=\dfrac{\sqrt{3}+1}{\sqrt{2}}$
$\sqrt{y}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{2}}{\sqrt{3}+1}\\$
$\sqrt{y}=\dfrac{\sqrt{2}}{\sqrt{3}+1}\times\dfrac{\sqrt{3}-1}{\sqrt{3}-1}\\$
$=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{3-1}=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{2}\times\dfrac{\sqrt{2}}{\sqrt{2}}\\$
$=\dfrac{2\left(\sqrt{3}-1\right)}{2\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}\\$
Now substitute in the given expression:
$\dfrac{x}{\sqrt{2}+\sqrt{x}}=\dfrac{\left(2+\sqrt{3}\right)\times\sqrt{2}}{\sqrt{2}\times \sqrt{2}+\sqrt{3}+1}\\$
$=\dfrac{\left(2+\sqrt{3}\right)\times\sqrt{2}}{2+\sqrt{3}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}\\$
$=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}\times\dfrac{3-\sqrt{3}}{3-\sqrt{3}}=\dfrac{\sqrt{2}\left(3+\sqrt{3}\right)}{6}\\$
$\dfrac{y}{\sqrt{2}-\sqrt{y}}=\dfrac{\left(2+\sqrt{3}\right)\times\sqrt{2}}{\sqrt{2}\times \sqrt{2}-\sqrt{3}+1}\\$
$=\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}=\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\\$
$=\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\times\dfrac{3+\sqrt{3}}{3+\sqrt{3}}=\dfrac{\sqrt{2}\left(3-\sqrt{3}\right)}{6}\\$
Now,$\dfrac{x}{\sqrt{2}+\sqrt{x}}+\dfrac{y}{\sqrt{2}-\sqrt{y}}=\dfrac{\sqrt{2}\left(3+\sqrt{3}\right)}{6}+\dfrac{\sqrt{2}\left(3-\sqrt{3}\right)}{6}$
$=\dfrac{\sqrt{2}\left(3+\sqrt{3}+3-\sqrt{3}\right)}{6}$
$=\dfrac{6\sqrt{2}}{6}=\sqrt{2}$

The method of finding solution by trying out various values for the variable is called

  1. Error method

  2. Trial and error method

  3. Testing method

  4. Checking method


Correct Option: B
Explanation:

$\Rightarrow$  The method of finding solution by trying out various values for the variable is called $Trial\,and\,error\,method.$

$\Rightarrow$  In this method, we often make a guess of the root of the equation.
$\Rightarrow$  We fine the values of L.H.S and R.H.S of the given equations for different values of the variable.
$\Rightarrow$  The values of the variable for which L.H.S = R.H.S is the root of the equation.

If $f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}}  + \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}}  + \sqrt {{x^2} + {{\left( {y - 1} \right)}^2}}  + \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 4} \right)}^2}} $ where $x,y \in R$, then the minimum value of $f\left( {x,y} \right)$ is

  1. $2 + \sqrt 5 $

  2. $5 + \sqrt 2 $

  3. $5 - \sqrt 2 $

  4. $\sqrt 5 - 2$


Correct Option: A

$|x - 1| + |x + 3| + |x - 5| = k$
How many values does $k$ have.

  1. only one solution

  2. two solution

  3. no solution

  4. infinite solutions


Correct Option: A

Consider the equation $(1 + a + b)^{2} = 3(1 + a^{2} + b^{2})$, where a, b are real numbers.
Then

  1. there is no solution pair (a, b)

  2. there are infinitely many solution pairs (a, b)

  3. there are exactly two solution pairs (a, b)

  4. there is exactly one solution pair (a, b)


Correct Option: D
Explanation:

$ (1+a+b)^{2} = 3(1+a^{2}+b^{2}) $

$ \Rightarrow (1+a^{2}+b^{2} + 2a + 2b + 2ab) = 3 + 3a^{2} +3b^{2} $
$ \Rightarrow a^{2}+b^{2} -ab-a-b+1 = 0 $
$ \Rightarrow (a-b)^{2} + ((a-1)(b-1)) = ((a-1)-(b-1))^{2} + ((a-1)(b-1)) = 0 $
Let $ a-1 = x $ and $ b-1 = y $

$ \Rightarrow (x-y)^{2} + (xy) = 0 $
$ \Rightarrow x^{2} - xy + y^{2} = 0 $                                 ...($1$)
Assume $ y \neq 0 $ and divide by $ y^{2} $

$ \Rightarrow \left ( \dfrac{x}{y} \right )^{2} - \dfrac{x}{y} + 1 = 0 $
Substitute $ t = \dfrac{x}{y} $
$ \Rightarrow t^{2} - t + 1 = 0 $
Discriminant $ D = 1-4 = -3 < 0 $. No solution here.

Assume $ y = 0 $ in equation ($1$). It satisfies the equation for $ x = 0 $. Hence the solution is $x=0$ and $y=0$

$ \Rightarrow a=1$ and $b=1$

The equations of the plane through the points $(1,-1,2),(-3,2,-2)$ and perpendicular to the plane $x+2y+3z+7=0$ is  

  1. $x+16y+11z-7=0$

  2. $17x+8y-11z+13=0$

  3. $x+y+z-2=0$

  4. $x-5y-3z=0$


Correct Option: B
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