Quadratic Equation and Inequalities — Page 13 | JEE Mathematics

Q121

All the pairs (x, y) that satisfy the inequality

{2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1

also satisfy the equation

A sin x = |sin y|

B sin x = 2sin y

C 2 sin x = sin y

D 2 |sin x | = 3 sin y

Correct Answer

Option A

Solution

{2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }} \le {2^{2{{\sin }^2}y}}

$\Rightarrow$

\sqrt {{{\sin }^2}x - 2\sin x + 5} \le 2{\sin ^2}y

\Rightarrow \sqrt {{{\left( {\sin x - 1} \right)}^2} + 4} \le 2{\sin ^2}y

it is true when sinx = 1, |siny| = 1 so sinx = |siny|

Q122

Let

\lambda \ne 0

be in R. If

\alpha

and

\beta

are the roots of the equation, x2 - x + 2

\lambda

= 0 and

\alpha

and

\gamma

are the roots of the equation,

3{x^2} - 10x + 27\lambda = 0

, then

{{\beta \gamma } \over \lambda }

is equal to:

A 36

B 9

C 27

D 18

Correct Answer

Option D

Solution

$\alpha$ and $\beta$ are the roots of the equation x2 - x + 2 $\lambda$ = 0 .....(1) $\therefore$

\alpha + \beta = 1,\,\alpha \beta = 2\lambda

$\alpha$ and $\gamma$ are the roots of the equation,

3{x^2} - 10x + 27\lambda = 0

......(2) $\therefore$

\alpha + \gamma

=

{{10} \over 3}

,

\alpha \gamma

=

{{27\lambda } \over 3}

= 9 $\lambda$ Multiplying equation (1) by 3 and subtracting form equation (2) we get -7x + 21 $\lambda$ = 0 $\Rightarrow$ x = 3 $\lambda$ $\therefore$ $\alpha$ = 3 $\lambda$ As $\alpha$ is root of equation (1) so $\alpha$ 2 - $\alpha$ + 2 $\lambda$ = 0 $\Rightarrow$ 9 $\lambda$ 2 - 3 $\lambda$ + 2 $\lambda$ = 0 $\Rightarrow$ $\lambda$ =

{1 \over 9}

$\Rightarrow$ $\alpha$ = 3 $\times$

{1 \over 9}

=

{1 \over 3}

Also

\alpha \beta = 2\lambda

=

{2 \over 9}

$\Rightarrow$ $\beta$ =

{2 \over 3}

Also

\alpha \gamma

= 9 $\lambda$ = 9 $\times$

{1 \over 9}

= 1 $\Rightarrow$ $\gamma$ = 3 $\therefore$

{{\beta \gamma } \over \lambda }

=

{{{2 \over 3}.3} \over {{1 \over 9}}}

= 18

Q123

If

\alpha

and

\beta

are the roots of the quadratic equation, x2 + x sin

\theta

- 2 sin

\theta

= 0,

\theta \in \left( {0,{\pi \over 2}} \right)

, then

{{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}

is equal to :

A

{{{2^{12}}} \over {{{\left( {\sin \theta - 8} \right)}^6}}}

B

{{{2^6}} \over {{{\left( {\sin \theta + 4} \right)}^{12}}}}

C

{{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}

D

{{{2^{12}}} \over {{{\left( {\sin \theta - 4} \right)}^{12}}}}

Correct Answer

Option C

Solution

Given

\alpha + \beta = - \sin \theta

and

\alpha \beta = - 2\sin \theta

{{\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){\alpha ^{12}}{\beta ^{12}}} \over {\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}}

\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } = \sqrt {{{\sin }^2}\theta + 8\sin \theta }

Hence required quantity

{{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {2\sin \theta } \right)}^{12}}} \over {{{\sin }^{12}}\theta {{\left( {\sin \theta + 8} \right)}^{12}}}} = {{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}

Q124

All the values of

m

for which both roots of the equation

{x^2} - 2mx + {m^2} - 1 = 0

are greater than

- 2

but less then 4, lie in the interval

A

- 2 < m < 0

B

m > 3

C

- 1 < m < 3

D

1 < m < 4

Correct Answer

Option C

Solution

Equation

{x^2} - 2mx + {m^2} - 1 = 0

{\left( {x - m} \right)^2} - 1 = 0

or

\left( {x - m + 1} \right)\left( {x - m - 1} \right) = 0

x = m - 1,m + 1

m - 1 > - 2

and

m + 1 < 4

\Rightarrow m > - 1

and

m<3

or

\,\,\, - 1 < m < 3

Q125

If

\alpha

and

\beta

are the roots of the equation 2x(2x + 1) = 1, then

\beta

is equal to :

A

- 2\alpha \left( {\alpha + 1} \right)

B

2\alpha \left( {\alpha + 1} \right)

C

2{\alpha ^2}

D

2\alpha \left( {\alpha - 1} \right)

Correct Answer

Option A

Solution

$\alpha$ and $\beta$ are the roots of the equation 4x2 + 2x – 1 = 0. $\therefore$ $\alpha$ + $\beta$ =

- {1 \over 2}

$\Rightarrow$ -1 = 2 $\alpha$ + 2 $\beta$ and 4 $\alpha$ 2 + 2 $\alpha$ - 1 = 0 $\Rightarrow$ 4 $\alpha$ 2 + 2 $\alpha$ + 2 $\alpha$ + 2 $\beta$ = 0 $\Rightarrow$ $\beta$ =

- 2\alpha \left( {\alpha + 1} \right)

Q126

The equation

{e^{\sin x}} - {e^{ - \sin x}} - 4 = 0

has:

A infinite number of real roots

B no real roots

C exactly one real root

D exactly four real roots

Correct Answer

Option B

Solution

Given equation is

{e^{\sin x}} - {e^{ - \sin x}} - 4 = 0

Put

{e^{{\mathop{\rm sinx}\nolimits} \,}} = t

in the given equation, we get

{t^2} - 4t - 1 = 0

\Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}

\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}

\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm 2\sqrt 5 } \over 2}

\,\,\,\,\,\,\,\,\,\,\, = 2 \pm \sqrt 5

\Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5

\,\,\,\,\,

(as

t = {e^{\sin x}}

)

\Rightarrow {e^{\sin x}} = 2 - \sqrt 5

and

\,\,\,\,\,\,\,\,\,\,\,\,\,

{e^{\sin x}} = 2 + \sqrt 5

\Rightarrow {e^{\sin x}} = 2 - \sqrt 5 < 0

and

\,\,\,\,\,\,\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1

So, rejected Hence given equation has no solution. $\therefore$ The equation has no real roots.

Q127

The number of all possible positive integral values of

\alpha

for which the roots of the quadratic equation, 6x2

-

11x +

\alpha

= 0 are rational numbers is :

A 3

B 2

C 4

D 5

Correct Answer

Option A

Solution

For rational D must be perfect square D = 121 $-$ 24 $\alpha$ for 121 $-$ 24 $\alpha$ to be perfect square a must be 3, 4, 5 So, ans $\alpha$ = 3

Q128

The sum of the solutions of the equation

\left| {\sqrt x - 2} \right| + \sqrt x \left( {\sqrt x - 4} \right) + 2 = 0

(x > 0) is equal to:

A 9

B 12

C 4

D 10

Correct Answer

Option D

Solution

Case 1 : When

\sqrt x \ge 2

then

\left| {\sqrt x - 2} \right| = \sqrt x - 2

$\therefore$ The given equation becomes,

\left( {\sqrt x - 2} \right)

+

\sqrt x \left( {\sqrt x - 4} \right) + 2

= 0 $\Rightarrow$

\left( {\sqrt x - 2} \right)

+

x - 4\sqrt x

+ 2 = 0 $\Rightarrow$

x - 3\sqrt x

= 0 $\Rightarrow$

\sqrt x \left( {\sqrt x - 3} \right)

= 0 $\therefore$

\sqrt x

= 0 or 3

\sqrt x

= 0 is not possible as

\sqrt x \ge 2

. So,

\sqrt x

= 3 or

x

= 9 Case 2 : When

\sqrt x < 2

then

\left| {\sqrt x - 2} \right| =

- \left( {\sqrt x - 2} \right)

=

2 - \sqrt x

$\therefore$ The given equation becomes,

\left( {2 - \sqrt x } \right)

+

\sqrt x \left( {\sqrt x - 4} \right) + 2

= 0 $\Rightarrow$

{2 - \sqrt x }

+

x - 4\sqrt x

+ 2 = 0 $\Rightarrow$

x - 5\sqrt x + 4

= 0 $\Rightarrow$

x - 4\sqrt x - \sqrt x + 4

= 0 $\Rightarrow$

\sqrt x \left( {\sqrt x - 4} \right)

-\left( {\sqrt x - 4} \right)

= 0 $\Rightarrow$

\left( {\sqrt x - 4} \right)

\left( {\sqrt x - 1} \right)

= 0 $\therefore$

\sqrt x

= 4 or 1

\sqrt x

= 4 is not possible as

\sqrt x < 2

. $\therefore$

\sqrt x

= 1 or

x

= 1 So, Sum of all solutions = 9 + 1 = 10

Q129

If

\alpha, \beta

are the roots of the equation

x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\dfrac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\dfrac{2}{3}}}-1\right)=0

, then the equation, whose roots are

\alpha+\dfrac{1}{\beta}

and

\beta+\dfrac{1}{\alpha}

, is :

A

3 x^{2}-20 x-12=0

B

3 x^{2}-10 x-4=0

C

3 x^{2}-10 x+2=0

D

3 x^{2}-20 x+16=0

Correct Answer

Option B

Solution

{3^{\sqrt {{{\log }_3}5} }} - {5^{\sqrt {{{\log }_5}3} }} = {3^{\sqrt {{{\log }_3}5} }} - {\left( {{3^{{{\log }_3}5}}} \right)^{\sqrt {{{\log }_5}3} }}

{3^{{{\left( {{{\log }_3}5} \right)}^{{1 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = 0

Note : In the given equation 'x' is missing. So

\alpha + \beta + {1 \over \alpha } + {1 \over \beta } = (\alpha + \beta ) + {{\alpha + \beta } \over {\alpha \beta }}

= 5 - {5 \over 3} = {{10} \over 3}

\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = 2 + \alpha \beta + {1 \over {\alpha \beta }} = 2 - 3 - {1 \over 3} = {{ - 4} \over 3}

So Equation must be option (B).

Q130

Let

\mathrm{S}

be the set of positive integral values of

a

for which $$\frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32}

A 0

B

\infty

C 3

D 1

Correct Answer

Option A

Solution

$x^2-8 x+32>0 \forall x \in R$ as discriminant of this quadratic is $64-4 \times 32<0$

\Rightarrow a x^2+2(a+1) x+9 a+4<0 \forall x \in R

$\Rightarrow$ Only possible when $a<0$ and $D<0$ $\Rightarrow$ Since $S$ is set of positive values of $a \Rightarrow S$ is a null set

\Rightarrow n(S)=0