Tag: solving equations numerically

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Questions Related to solving equations numerically

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

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$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 = $\dfrac{-3-\sqrt{101}}{2}$

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

$36$ 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
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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. $ab + bc + ca$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given the system of equations, this is a classic problem where the determinant of the coefficient matrix must be zero for non-trivial solutions. Solving the system leads to the identity a^2 + b^2 + c^2 + 2abc = 1.

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

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. $\left[ {34,\infty } \right]$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $[-34,\ 0]$



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

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

  1. 5

  2. 6

  3. 8

  4. 9
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

$x$ 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. $\dfrac{{1569}}{{784}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $y-x = \dfrac{1}{49 \times 16}+2$

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

$=\dfrac{1569}{784}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

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

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

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

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

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

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