Tag: descartes rule

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Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The number of real solution of $x-\dfrac{1}{x^2-4}=2-\dfrac{1}{x^2-4}$ is 

  1. $0$
  2. $1$
  3. $2$
  4. $infinitie$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that:
$x-\dfrac{1}{x^2-4}=2-\dfrac{1}{x^2-4}$

$x^3-4x-1=2x^2-8-1$
$x^3-2x^2-4x+8=0$
$x^2(x-2)-4(x-2)=0$
$(x^2-4)(x-2)=0$
$(x-2)(x+2)(x-2)=0$
$(x-2)(x+2)=0$
$x=-2,+2$

Hence, 
There are two real solutions for the given expression.
Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Let $f(x)=1+2x+3x^2+.....+(n+1)x^n,$ where n is even. Then the number of real roots of the equation $f(x)=0$ is 

  1. $0$
  2. $1$
  3. $n$
  4. $None$ $of$ $these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

f(x) = 1 + 2x + 3x^2 + ... + (n+1)x^n. This is the derivative of 1 + x + x^2 + ... + x^(n+1). Since n is even, the sum is (x^(n+2) - 1) / (x - 1). The derivative of this for x not equal to 1 is positive for all x > 0. For x < 0, the terms alternate, and analysis shows no real roots.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The general solution of the equation 
$tan \, x + tan \, 2x + \sqrt{3} \, tan \, x \, tan \, 2x = \sqrt{3}$ is 

  1. $x = \dfrac{n \pi}{3} + \dfrac{\pi}{9}, \, n \in z$
  2. $x = m \pi + \dfrac{\pi}{9}, \, n \in z$
  3. $x = \dfrac{(n + 1) \pi}{3}, n \in z$
  4. $x = \dfrac{n \pi}{3} + \dfrac{\pi}{3}, \, n \in z$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\tan x+\tan 2x+\sqrt{3}\tan x\, \tan 2x=\sqrt{3}$

$\tan x+\tan 2x=\sqrt{3}-\sqrt{3}\tan x\, \tan 2x$

$\cfrac{\tan x+\tan 2x}{1-\tan x\, \tan 2x}=\sqrt{3}$

$\tan (x+2x)=\sqrt{3}$

$\tan 3x=\sqrt{3}$

$\tan 3x=\tan \cfrac{\pi }{3}$

$\Rightarrow 3x=n\pi+ \cfrac{\pi }{3}$   ( where $n$ is an integer )

$x=\cfrac{n\pi }{3}+\cfrac{\pi }{9}$
Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $1+\surd {3}i/2$ is a root of equation $x^{4}-x^{3}+x1=0$ then its real roots are 

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

Given that 1 + i√3/2 is a root, its complex conjugate 1 - i√3/2 is also a root (coefficients are real). The sum of all roots is 1 (from x³ coefficient with opposite sign). If the remaining two roots are real and equal to r, then: 2(1) + 2r = 1, giving r = ±1. Testing shows x = 1 and x = -1 satisfy the equation. Therefore, the real roots are 1 and -1.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The equation
$\left| {\begin{array}{{20}{c}}  {{{\left( {1 + x} \right)}^2}}&{{{\left( {1 - x} \right)}^2}}&{ - \left( {2 + {x^2}} \right)} \   {2x + 1}&{3x}&{1 - 5x} \   {x + 1}&{2x}&{2 - 3x} \end{array}} \right| + \left| {\begin{array}{{20}{c}}  {{{\left( {1 + x} \right)}^2}}&{2x + 1}&{x + 1} \   {{{\left( {1 - x} \right)}^2}}&{3x}&{2x} \   {1 - 2x}&{3x - 2}&{2x - 3} \end{array}} \right| = 0$

  1. has no real solution
  2. fas $4$ real solutions

  3. has two real and two non-real solutions

  4. has infinite number of solutions, real or non-real
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the two determinants is zero. By evaluating the determinants or checking for properties, one finds that the expression simplifies to a form that has no real solutions for x.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If a,b,c and d are the real roots of the equation : $x^{4}+p _{1}x^{1}+p _{2}x^{2}+p _{3}x+p _{4}=0$ and $(1+a^{2})(1+b^{2})(1+c^{2})(1+d^{2})=k(1-p _{2}+p _{4})^{2}+(p _{3}-p _{1})^{2}$ then the value f k is:

  1. -1

  2. 1

  3. 2

  4. -2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the relationship between roots and coefficients, and the identity that (1+a²)(1+b²)(1+c²)(1+d²) can be expressed in terms of the polynomial's coefficients, we can derive that k = 1. This involves substituting the roots into the given expression and using Vieta's formulas to relate it to the coefficients p₁, p₂, p₃, p₄.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $\alpha $ and $\beta $ are the roots of ${ x }^{ 2 }+px+q=0$ and ${ \alpha  }^{ 4 } , { \beta  }^{ 4 }$ are the roots of ${ x }^{ 2 }-rx+s=0$, then the equation ${ x }^{ 2 }-4qx+2{ q }^{ 2 }-r=0$ has always two real roots.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If alpha, beta are roots of x^2 + px + q = 0, then alpha^4, beta^4 are roots of x^2 - rx + s = 0. The discriminant of the final equation x^2 - 4qx + 2q^2 - r = 0 is D = (4q)^2 - 4(2q^2 - r) = 16q^2 - 8q^2 + 4r = 8q^2 + 4r. Since r = alpha^4 + beta^4, which is always positive, D > 0, so the roots are real.