Quadratic Equation and Inequalities — Page 8 | JEE Mathematics

Q71

The number of real roots of the equation

\sqrt{x^{2}-4 x+3}+\sqrt{x^{2}-9}=\sqrt{4 x^{2}-14 x+6}

, is :

A 0

B 1

C 3

D 2

Correct Answer

Option B

Solution

$\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}$ $=\sqrt{4\left(x-\dfrac{12}{4}\right)\left(x-\dfrac{2}{4}\right)}$ $\Rightarrow \sqrt{\mathrm{x}-3}=0 \Rightarrow \mathrm{x}=3$ which is in domain or $\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{x}+3}=\sqrt{4 \mathrm{x}-2}$ $2 \sqrt{(x-1)(x+3)}=2 x-4$ $x^{2}+2 x-3=x^{2}-4 x+4$ $6 \mathrm{x}=7$ $\mathrm{x}=7 / 6$ (rejected)

Q72

The number of real solutions of the equation

3\left( {{x^2} + {1 \over {{x^2}}}} \right) - 2\left( {x + {1 \over x}} \right) + 5 = 0

, is

A 3

B 4

C 0

D 2

Correct Answer

Option C

Solution

$3\left(x^{2}+\dfrac{1}{x^{2}}\right)-2\left(x+\dfrac{1}{x}\right)+5=0$ $3\left[\left(x+\dfrac{1}{x}\right)^{2}-2\right]-2\left(x+\dfrac{1}{x}\right)+5=0$ Put $x+\dfrac{1}{x}=t \Rightarrow t \in(-\infty,-2] \cup[2, \infty)$

\begin{aligned} & 3 t^{2}-2 t-1=0 \\\\ & 3 t^{2}-3 t+t-1=0 \\\\ & \Rightarrow 3 t(t-1)+1(t-1)=0 \Rightarrow t=1,=-\frac{1}{3} \\\\ & \Rightarrow \quad t=1,-\frac{1}{3} \\\\ & \because t \in(-\infty,-2] \cup[2, \infty) \end{aligned}

No real value of $t \Rightarrow$ no real value of $x$ .

Q73

The equation

{x^2} - 4x + [x] + 3 = x[x]

, where

[x]

denotes the greatest integer function, has :

A exactly two solutions in (

-\infty,\infty

)

B no solution

C a unique solution in (

-\infty,\infty

)

D a unique solution in (

-\infty,1

)

Correct Answer

Option C

Solution

{x^2} - 4x + [x] + 3 = x[x]

\Rightarrow {x^2} - 4x + [x] + 3 - x[x] = 0

\Rightarrow (x - 1)(x - 3) - [x](x - 1) = 0

\Rightarrow (x - 1)(x - [x] - 3) = 0

$\therefore$

x = 1

or

x - [x] - 3 = 0

\Rightarrow \{ x\} - 3 = 0

[As

\{ x\} = x - [x]

]

\Rightarrow \{ x\} = 3

But we know,

0

\therefore

\{ x\} \ne 3

\therefore

x

has only one solution in (

-\infty,\infty

) which is

x = 1$$.

Q74

The number of real roots of the equation

x|x|-5|x+2|+6=0

, is :

A 4

B 3

C 5

D 6

Correct Answer

Option B

Solution

\begin{aligned} & x|x|-5|x+2|+6=0 \\\\ & \text{ Case-I : } \\\\ & \text{ When } xCase-II :

\begin{aligned} & \text { When } -2 \leq x Case-III :

\begin{aligned} & \text{ When } x \geq 0 \text{ then } \\\\ & x^2-5(x+2)+6=0 \\\\ \Rightarrow & x^2-5 x-4=0 \\\\ & x=\frac{5 \pm \sqrt{25+16}}{2} \\\\ & =\frac{5 \pm \sqrt{41}}{2} \\\\ & x=\frac{5 - \sqrt{41}}{2} \text{ is accepted } \\\\ \therefore & 3 \text{ real roots are possible. } \end{aligned}

Q75

Let

\alpha, \beta

be the roots of the equation

x^{2}-\sqrt{2} x+2=0

. Then

\alpha^{14}+\beta^{14}

is equal to

A

-64

B

-64 \sqrt{2}

C

-128 \sqrt{2}

D

-128

Correct Answer

Option D

Solution

Find the roots of the quadratic equation: The quadratic formula is

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

. For the quadratic equation

x^2 - \sqrt{2}x + 2 = 0

, we have

a = 1, b = -\sqrt{2}, c = 2

. Plugging these into the quadratic formula gives:

x = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2 - 4(1)(2)}}{2(1)} = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm \sqrt{6}i}{2}.

So, we have two roots, $\alpha$ and $\beta$ , which are:

\alpha = \frac{\sqrt{2} + \sqrt{6}i}{2},

\beta = \frac{\sqrt{2} - \sqrt{6}i}{2}.

Express the roots in exponential form: We can express complex numbers in the form

re^{i\theta}

. For $\alpha$ and $\beta$ , we find the magnitude

r = \sqrt{2}

and the arguments

\theta = \frac{\pi}{3}, -\frac{\pi}{3}

respectively. So, we have:

\alpha = \sqrt{2}e^{i\frac{\pi}{3}},

\beta = \sqrt{2}e^{-i\frac{\pi}{3}}.

Calculate the 14th power of the roots: To find

\alpha^{14}

and

\beta^{14}

, we use the property of exponents which says that

(a^m)^n = a^{mn}

. So, we have:

\alpha^{14} = (\sqrt{2}e^{i\frac{\pi}{3}})^{14} = 2^7e^{i\frac{14\pi}{3}} = 128e^{i\frac{2\pi}{3}},

\beta^{14} = (\sqrt{2}e^{-i\frac{\pi}{3}})^{14} = 2^7e^{-i\frac{14\pi}{3}} = 128e^{-i\frac{2\pi}{3}}.

Add the 14th powers of the roots: We want to find the real part of

\alpha^{14} + \beta^{14}

. To do this, we use the property that

e^{ix} = \cos(x) + i\sin(x)

. We have:

\alpha^{14} + \beta^{14} = 128e^{i\frac{2\pi}{3}} + 128e^{-i\frac{2\pi}{3}} = 128(2)\cos\left(\frac{2\pi}{3}\right) = -128.

Q76

The set of all

a \in \mathbb{R}

for which the equation

x|x-1|+|x+2|+a=0

has exactly one real root, is :

A

(-\infty, \infty)

B

(-6, \infty)

C

(-\infty,-3)

D

(-6,-3)

Correct Answer

Option A

Solution

x|x-1|+|x+2|+a=0

Case I : If $x<-2$ then

-x^2+x-x-2+a=0

a=x^2+2

$y=x^2+2$ is decreasing $\forall x \in(-\infty,-2)$ Case II : If $-2 \leq x<1$ then

\begin{aligned} & -x^2+x+x+2+a=0 \\\\ & a=x^2-2 x-2 \end{aligned}

$y=x^2-2 x-2$ is decreasing $\forall x \in[-2,1)$ . Case III: If $x \geq 1$ then

\begin{aligned} & x^2-x+x+2+a=0 \\\\ & a=-\left(x^2+2\right) \end{aligned}

$y=-\left(x^2+2\right)$ is decreasing $\forall x \in[1, \infty)$ $\therefore$ Exactly one real root $\forall x \in R$

Q77

Let

\alpha, \beta

be the roots of the quadratic equation

x^{2}+\sqrt{6} x+3=0

. Then

\dfrac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}

is equal to :

A 72

B 9

C 729

D 81

Correct Answer

Option D

Solution

Given quadratic equation:

x^{2}+\sqrt{6} x+3=0

We can find the roots using the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Here,

a = 1

,

b = \sqrt{6}

, and

c = 3

. Substituting these values into the quadratic formula, we have:

x = \frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4(1)(3)}}{2(1)}

Simplifying further :

x = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2}

x = \frac{-\sqrt{6} \pm \sqrt{-6}}{2}

Since the discriminant is negative, the roots are complex numbers. We can express them using the imaginary unit

i

:

x = \frac{-\sqrt{6} \pm \sqrt{6}i}{2}

$\Rightarrow$

x = \frac{-1}{2}\sqrt{6} \pm \frac{1}{2}\sqrt{6}i

$\therefore$

\alpha, \beta=\sqrt{3} \mathrm{e}^{ \pm \frac{3 \pi \mathrm{i}}{4}}

.

The required expression can be rewritten in terms of the argument of the exponential form of the roots, which simplifies the calculation:

\begin{aligned} & =\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned}

Here, the exponential power of

\sqrt{3}

in the numerator is larger by 8 compared to the denominator, so we can divide the numerator and denominator by

(\sqrt{3})^{8} = 81

to simplify:

\begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned}

Since the cosine function has a period of

2\pi

, we can reduce the arguments of the cosine function in the numerator and denominator. We have

69\pi/4 = \pi/4 + 17\pi = \pi/4

,

42\pi/4 = 2\pi/4 + 10\pi = \pi/2

,

45\pi/4 = \pi/4 + 11\pi = \pi/4

, and

30\pi/4 = 2\pi/4 + 7\pi = \pi/2

. Therefore, the required expression simplifies to :

\begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{\pi}{2}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{\pi}{2}\right)} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}+(\sqrt{3})^{6}\cdot 0}{(\sqrt{3})^{7}\sqrt{2}+(\sqrt{3})^{2}\cdot 0} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}}{(\sqrt{3})^{7}\sqrt{2}} \\\\ & = (\sqrt{3})^{15-7} \\\\ & = (\sqrt{3})^{8} \\\\ & = 81. \end{aligned}

Q78

Let

\alpha, \beta, \gamma

be the three roots of the equation

x^{3}+b x+c=0

. If

\beta \gamma=1=-\alpha

, then

b^{3}+2 c^{3}-3 \alpha^{3}-6 \beta^{3}-8 \gamma^{3}

is equal to :

A 21

B 19

C

\dfrac{169}{8}

D

\dfrac{155}{8}

Correct Answer

Option B

Solution

Given cubic equation is :

x^3+b x+c=0

$\because \alpha, \beta, \gamma$ are the roots of above equation. And $\beta \gamma=1=-\alpha$

\begin{aligned} & \text{ So, product of roots }=-c \\\\ & \Rightarrow \alpha \beta \gamma=-c \\\\ & \Rightarrow(-1)(1)=-c \\\\ & \Rightarrow c=1 \end{aligned}

Since, $\alpha=-1$ is the root. So,

\begin{aligned} & \Rightarrow-1-b+c=0 \\\\ & \Rightarrow c-b=1 \\\\ & \Rightarrow 1-b=1 \Rightarrow b=0 \end{aligned}

The given equation becomes $x^3+1=0$ So, roots are $-1,-\omega,-\omega^2$

\begin{aligned} & \therefore b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3 \\\\ & =0+2-3(-1)^3-6(-\omega)^3-8\left(-\omega^2\right)^3 \\\\ & =2+3+6 \omega^3+8 \omega^6 \\\\ & =5+6+8=19 \end{aligned}

Concept : For a cubic equation, $a x^3+b x^2+c x+d=0$ Sum of roots $=\dfrac{-b}{a}$ Product of roots taken two at a time $=\dfrac{c}{a}$ Product of roots $=\dfrac{-d}{a}$

Q79

Let

A = \{ x \in R:[x + 3] + [x + 4] \le 3\} ,

$$B = \left\{ {x \in R:{3^x}{{\left( {\sum\limits_{r = 1}^\infty {{3 \over {{{10}^r}}}} } \right)}^{x - 3}}

A

B \subset C,A \ne B

B

A \subset B,A \ne B

C

A = B

D

A \cap B = \phi

Correct Answer

Option C

Solution

We have,

\begin{aligned} & A=\{x \in R:[x+3]+[x+4] \leq 3\} \\\\ & \text{ Here, }[x+3]+[x+4] \leq 3 \\\\ & \Rightarrow [x]+3+[x]+4 \leq 3 \\\\ & (\because[x+n]=[x]+n, n \in I) \\\\ & \Rightarrow 2[x]+4 \leq 0 \Rightarrow[x] \leq-2 \\\\ & \Rightarrow x \in(-\infty,-1) \\\\ & A \equiv(-\infty,-1) ...........(i) \end{aligned}

Also,

B=\left\{x \in R: 3^x\left(\sum\limits_{r=1}^{\infty} \frac{3}{10^r}\right)^{x-3}3 \Rightarrow x<-1 \\\\ \Rightarrow & x \in(-\infty,-1) \\\\ \Rightarrow & B \equiv(-\infty,-1) ...........(ii) \end{array}

From equations (i) and (ii), we get

A=B

Q80

The sum of all the roots of the equation

\left|x^{2}-8 x+15\right|-2 x+7=0

is :

A

11+\sqrt{3}

B

9+\sqrt{3}

C

9-\sqrt{3}

D

11-\sqrt{3}

Correct Answer

Option B

Solution

\begin{aligned} & \text{ We have, }\left|x^2-8 x+15\right|-2 x+7=0 \\\\ & \Rightarrow |(x-3)(x-5)|-2 x+7=0 \end{aligned}

Now, when $x \leq 3$ or $x \geq 5$ , then

\begin{array}{ccrl} & x^2-8 x+15-2 x+7 =0 \\\\ &\Rightarrow x^2-10 x+22=0 \\\\ &\Rightarrow x^2-10 x+25-3=0 \\\\ &\Rightarrow (x-5)^2-3=0 \\\\ &\Rightarrow (x-5)= \pm \sqrt{3} \\\\ &\Rightarrow x=5+\sqrt{3}, 5-\sqrt{3} \end{array}

Since, $x \leq 3$ or $x \geq 5$

\therefore x=5+\sqrt{3}

When, $3 < x < 5$ , then

\begin{array}{rlrl} & -\left(x^2-8 x+15\right)-2 x+7 =0 \\\\ & \Rightarrow -x^2+8 x-15-2 x+7 =0 \\\\ & \Rightarrow -x^2+6 x-8 =0 \\\\ & \Rightarrow x^2-6 x+8 =0 \\\\ & \Rightarrow (x-4)(x-2) =0 \Rightarrow x=2,4 \end{array}

Since, $3 < x < 5$

\therefore x=4

\therefore \text{ Sum of roots }=(5+\sqrt{3})+4=9+\sqrt{3}