Tag: sign of quadratic expression

Questions Related to sign of quadratic expression

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. $infinitie$


Correct Option: C
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.

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. $None$ $of$ $these$


Correct Option: A

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. $x = \dfrac{n \pi}{3} + \dfrac{\pi}{3}, \, n \in z$


Correct Option: A
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 )

$x=\cfrac{n\pi }{3}+\cfrac{\pi }{9}$

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. $1,2$


Correct Option: C

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


Correct Option: A

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


Correct Option: B

Number of solutions of the equation $\cos 6x+\tan^2x+\cos 6x\tan^2x=1$ in the interval $[0, 2\pi]$ is?

  1. $4$

  2. $5$

  3. $6$

  4. $7$


Correct Option: C
Explanation:

$\left( \cos6x+1 \right) { \tan }^{ 2 }x$

so 6 solutions
30,150,210,330,0,180

At how many maximum points will a cubic equation cut the $x$ axis?

  1. $0$

  2. $2$

  3. $1$

  4. $3$


Correct Option: D
Explanation:

A cubic Equation can have at max 3 distinct roots, so it will cut the $x$ axis at $3$ points.

Find the point of intersection of $x+y=-1\x-y=15$

  1. (7, -8)

  2. (-7, 8)

  3. (7, 8)

  4. (-7, -8)


Correct Option: A
Explanation:

$x+y=-1\x=-1-y\x-y=15\-1-y-y=15\-2y=15+1\y=\dfrac{16}{-2}\y=-8\x-8=-1\x=-1+8\x=7\(x.y)=(7,-8)$

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


Correct Option: A