Tag: finding roots by iteration

Questions Related to finding roots by iteration

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

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

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

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

  3. $ \pm \sqrt {29} $

  4. $ - 7,29$

Show answer
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

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $2, 18$

  2. $9, 4$

  3. $3, 12$

  4. None of these

Show answer
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

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $5$

  2. $6$

  3. $7$

  4. $8$

Show answer
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

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. 0

  2. 2

  3. 4

  4. 3

Show answer
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 $a ^ { 2 } + b ^ { 2 } + c ^ { 2 } + 2 a b c =$

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. 2

  2. $a + b + c$

  3. 1

  4. $ab + bc + ca$

Show answer
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

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

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

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

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

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

Show answer
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 ?

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. 5

  2. 6

  3. 8

  4. 9

Show answer
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

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

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

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

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

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

Show answer
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

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

  1. $3$

  2. $8$

  3. $27$

  4. $216$

Show answer
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 

maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

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

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

  3. 0

  4. none of these

Show answer
Correct Option: B
Explanation:
Given,

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

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

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

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

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

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