Quadratic Equation and Inequalities — Page 14 | JEE Mathematics

Q131

The number of real roots of the equation, e4x + e3x – 4e2x + ex + 1 = 0 is :

A 1

B 2

C 3

D 4

Correct Answer

Option A

Solution

e4x + e3x – 4e2x + ex + 1 = 0 Dividing by e2x, we get e2x + ex - 4 +

{1 \over {{e^x}}}

+

{1 \over {{e^{2x}}}}

= 0 $\Rightarrow$

\left( {{e^{2x}} + {1 \over {{e^{2x}}}}} \right) + \left( {{e^x} + {1 \over {{e^x}}}} \right)

- 4 = 0 $\Rightarrow$

{\left( {{e^x} + {1 \over {{e^x}}}} \right)^2} - 2

+

\left( {{e^x} + {1 \over {{e^x}}}} \right)

- 4 = 0 Let

{{e^x} + {1 \over {{e^x}}} = z}

(z2 – 2) + (z) – 4 = 0 $\Rightarrow$ z2 + z – 6 = 0 $\Rightarrow$ z = –3, 2 $\therefore$

{{e^x} + {1 \over {{e^x}}} = 2}

$\Rightarrow$ (ex – 1)2 = 0 $\Rightarrow$ x = 0. $\therefore$ Number of real roots = 1

Q132

Let a, b

\in

R, a

\ne

0 be such that the equation, ax2 – 2bx + 5 = 0 has a repeated root

\alpha

, which is also a root of the equation, x2 – 2bx – 10 = 0. If

\beta

is the other root of this equation, then

\alpha

2 +

\beta

2 is equal to :

A 28

B 24

C 26

D 25

Correct Answer

Option D

Solution

Roots of equation ax2 – 2bx + 5 = 0 are $\alpha$ , $\alpha$ . $\therefore$ $\alpha$ + $\alpha$ =

{{2b} \over a}

$\Rightarrow$ 2 $\alpha$ =

{{2b} \over a}

$\Rightarrow$ $\alpha$ =

{{b} \over a}

....(1) and $\alpha$ 2 =

{5 \over a}

....(2) From (1) and (2), we get

{{{b^2}} \over {{a^2}}} = {5 \over a}

$\Rightarrow$ b2 = 5a $\alpha$ , $\beta$ are the roots of equation x2 – 2bx – 10 = 0 $\therefore$ $\alpha$ + $\beta$ = 2b and $\alpha$ $\beta$ = -10 As $\alpha$ is a root of the equation x2 – 2bx – 10 = 0. $\therefore$ $\alpha$ 2 - 2b $\alpha$ - 10 = 0 $\Rightarrow$

{{{b^2}} \over {{a^2}}} - {{2{b^2}} \over a} - 10 = 0

$\Rightarrow$

{{5a} \over {{a^2}}} - {{10a} \over a} - 10 = 0

$\Rightarrow$

{5 \over a} - 10 - 10 = 0

$\Rightarrow$

a

=

{1 \over 4}

$\Rightarrow$

{\alpha ^2}

= 20 $\alpha$ $\beta$ = -10 $\Rightarrow$

{\beta ^2}

= 5 $\therefore$

{\alpha ^2} + {\beta ^2}

= 20 + 5 = 25

Q133

The sum of all the real roots of the equation

({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0

is

A

{\log _e}3

B

- {\log _e}3

C

{\log _e}6

D

- {\log _e}6

Correct Answer

Option B

Solution

({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0

Let

{e^x} = t

$\therefore$

({t^2} - 4)(6{t^2} - 5t + 1) = 0

\Rightarrow ({t^2} - 4)(2t - 1)(3t - 1) = 0

$\therefore$ t = 2, $-$ 2,

{1 \over 2}

,

{1 \over 3}

$\therefore$

{e^x} = 2 \Rightarrow x = \ln 2

{e^x} = - 2

(not possible)

{e^x} = {1 \over 2} \Rightarrow x = - \ln 2

{e^x} = {1 \over 3} \Rightarrow x = - \ln 3

$\therefore$ Sum of all real roots

= \ln 2 - \ln 2 - \ln 3

= - \ln 3

Q134

If the equations x2 + bx−1 = 0 and x2 + x + b = 0 have a common root different from −1, then

\left| b \right|

is equal to :

A

\sqrt 2

B 2

C 3

D

\sqrt 3

Correct Answer

Option D

Solution

Given, x2 + bx $-$ 1 = 0 . . . . .(1) and x2 + x + b = 0 . . . . .

(2) Performing (1) $-$ (2) we get, bx $-$ 1 $-$ x $-$ b = 0 $\Rightarrow$ x(b $-$ 1) = b + 1 $\Rightarrow$ x =

{{b + 1} \over {b - 1}}

putting value of x in equation (2),

{\left( {{{b + 1} \over {b - 1}}} \right)^2} + \left( {{{b + 1} \over {b - 1}}} \right) + b = 0

$\Rightarrow$ (b + 1)2 + (b + 1) (b $-$ 1) + b (b $-$ 1)2 = 0 $\Rightarrow$ b2 + 2b + 1 + b2 $-$ 1 + b (b2 $-$ 2b + 1) = 0 $\Rightarrow$ 2b3 + 2b + b3 $-$ 2b2 + b = 0 $\Rightarrow$ b3 + 3b = 0 $\Rightarrow$ b(b2 + 3) = 0 b2 = $-$ 3, b = 0 $\therefore$ b =

\pm \sqrt 3 i

$\Rightarrow$

\left| b \right|

=

\sqrt 3

Q135

If tanA and tanB are the roots of the quadratic equation, 3x2

-

10x

-

25 = 0, then the value of 3 sin2(A + B)

-

10 sin(A + B).cos(A + B)

-

25 cos2(A + B) is :

A

-

10

B 10

C

-

25

D 25

Correct Answer

Option C

Solution

As tan A and tan B are the roots of 3x2 $-$ 10x $-$ 25 = 0, So, tan(A + B) =

{{\tan A + \tan B} \over {1 - \tan A\tan B}}

=

{{{{10} \over 3}} \over {1 + {{25} \over 3}}}

=

{{10/3} \over {28/3}}

=

{5 \over {14}}

Now, cos2 (A + B) = $-$ 1 + 2 cos2 (A + B) =

{{1 - {{\tan }^2}(A + B)} \over {1 + {{\tan }^2}(A + B)}}\,

$\Rightarrow$ cos2(A + B) =

{{196} \over {221}}

\therefore\,\,\,

3sin2(A + B) $-$ 10sin(A + B)cos(A + B) $-$ 25 cos2(A + B) = cos2(A + B) [ 3tan2(A + B) $-$ 10tan(A + B) $-$ 25] =

{{75 - 700 - 4900} \over {196}} \times {{196} \over {221}}

= $-$

{{5525} \over {196}}

$\times$

{{196} \over {221}}

= $-$ 25

Q136

If m is chosen in the quadratic equation (m2 + 1) x2 – 3x + (m2 + 1)2 = 0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-

A

4\sqrt 3

B

8\sqrt 3

C

8\sqrt 5

D

10\sqrt 5

Correct Answer

Option C

Solution

Given quadratic equation (m2 + 1) x2 – 3x + (m2 + 1)2 = 0 Let roots of the equation $\alpha$ and $\beta$ .

$\therefore$ Sum of roots = $\alpha$ + $\beta$ =

{3 \over {{m^2} + 1}}

Product of roots = $\alpha$ $\beta$ = m2 + 1

{3 \over {{m^2} + 1}}

is maximum when m = 0 Hence equation becomes x2 – 3x + 1 = 0

\alpha + \beta = 3

,

\alpha \beta = 1

\left| {\alpha - \beta } \right|

=

\sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta }

=

\sqrt {{{\left( 3 \right)}^2} - 4.1}

=

\sqrt 5

\left| {{\alpha ^3} - {\beta ^3}} \right| = \left| {(\alpha - \beta )({\alpha ^2} + {\beta ^2} + \alpha \beta )} \right|

=

\sqrt 5 \left| {{{\left( {\alpha + \beta } \right)}^2} - \alpha \beta } \right|

= \sqrt 5 (9 - 1) = 8\sqrt 5

Q137

Let

\alpha

and

\beta

be the roots of the quadratic equation x2 sin

\theta

– x(sin

\theta

cos

\theta

+ 1) + cos

\theta

= 0 (0 <

\theta

< 45o), and

\alpha

<

\beta

. Then

\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{{{\left( { - 1} \right)}^n}} \over {{\beta ^n}}}} \right)}

is equal to :

A

{1 \over {1 + \cos \theta }} + {1 \over {1 - \sin \theta }}

B

{1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}

C

{1 \over {1 - \cos \theta }} - {1 \over {1 + \sin \theta }}

D

{1 \over {1 + \cos \theta }} - {1 \over {1 - \sin \theta }}

Correct Answer

Option B

Solution

D = (1 + sin $\theta$ cos $\theta$ )2 $-$ 4sin $\theta$ cos $\theta$ = (1 $-$ sin $\theta$ cos $\theta$ )2 $\Rightarrow$ roots are $\beta$ = cosec $\theta$ and $\alpha$ = cos $\theta$

\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{\left( { - {1 \over \beta }} \right)}^n}} \right)} = \sum\limits_{n = 0}^\infty {{{\left( {\cos \theta } \right)}^n}} + \sum\limits_{n = 0}^n {{{\left( { - \sin \theta } \right)}^n}}

= {1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}

Q138

If

\alpha

,

\beta

and

\gamma

are three consecutive terms of a non-constant G.P. such that the equations

\alpha

x 2 + 2

\beta

x +

\gamma

= 0 and x2 + x – 1 = 0 have a common root, then

\alpha

(

\beta

+

\gamma

) is equal to :

A

\alpha

\gamma

B 0

C

\beta

\gamma

D

\alpha

\beta

Correct Answer

Option C

Solution

Let the common ratio of G.P is r Then the equation of

\alpha {x^2} + 2\beta x + \gamma = 0

$\Rightarrow$

\alpha {x^2} + 2\alpha rx + \alpha {r^2} = 0

$\Rightarrow$ x2 + 2rx + r2 = 0 ........(i) Equation (i) and x2 + x - 1 = 0 ......... (ii) has a common root.

Now (i) - (ii) $\Rightarrow$ (2r-1)x + (r2+1) = 0 $\Rightarrow$ x =

{{ - \left( {{r^2} + 1} \right)} \over {2r - 1}}

....... (iii) Now putting (iii) in equation (ii) $\Rightarrow$ (r2+1)2 - (r2+1) (2r - 1) - (2r - 1)2 = 0 $\Rightarrow$ r4 - 2r3 - r2 + 2r + 1 = 0 .......... (iv) dividing equation (iv) by r2 $\Rightarrow$ r2 - 2r - 1 +

{2 \over r} + {1 \over {{r^2}}} = 0

$\Rightarrow$

{\left( {r - {1 \over r}} \right)^2} - 2\left( {r - {1 \over r}} \right) + 1 = 0

$\Rightarrow$

{\left( {r - {1 \over r} - r} \right)^2} = 0

$\Rightarrow$

{{{r^2} - 1} \over r} = 1

$\Rightarrow$ r2 = 1 + r Now $\alpha$ ( $\beta$ + $\gamma$ ) $\Rightarrow$ $\alpha$ ( $\alpha$ r + $\alpha$ r2) $\Rightarrow$

\alpha ^2 r (1 + r)

[ $\because$ r2 = 1 + r] $\Rightarrow$

\alpha ^2 r . r^2

$\Rightarrow$

( \alpha r )\,( \alpha r^2 )

$\Rightarrow$ $\beta$ $\gamma$

Q139

If both the roots of the quadratic equation x2

-

mx + 4 = 0 are real and distinct and they lie in the interval [1, 5], then m lies in the interval :

A (

-

5,

-

4)

B (4, 5)

C (5, 6)

D (3, 4)

Correct Answer

Option B

Solution

x2 $-$ mx + 4 = 0 Case-I : D > 0 m2 $-$ 16 > 0 $\Rightarrow$ m

\in

( $-$ $\infty$ , $-$ 4) $\cup$ (4, $\infty$ ) Case-II :

\Rightarrow \,\,1 < {{ - b} \over {2a}} < 5

\Rightarrow \,\,1 < {m \over 2} < 5 \Rightarrow \,m \in \left( {2,10} \right)

Case-III : f(1) > 0 and f(5) > 0 1 $-$ m + 4 > 0 and 25 $-$ 5m + 4 > 0 m < 5 and m <

{{29} \over 5}

Case-IV : Let one root is x = 1 1 $-$ m + 4 = 0 m = 5 Now equation x2 $-$ 5x + 4 = 0 (x $-$ 1) (x $-$ 4) = 0 x = 1 i.e. m = 5 is also included hence m

\in

(4, 5] So given option is (4, 5)

Q140

Let f(x) be a quadratic polynomial such that f(–1) + f(2) = 0. If one of the roots of f(x) = 0 is 3, then its other root lies in :

A (–3, –1)

B (1, 3)

C (–1, 0)

D (0, 1)

Correct Answer

Option C

Solution

Let the other root is $\alpha$ . $\therefore$ f(x) = a(x – 3) (x – $\alpha$ ) f(2) = a( $\alpha$ – 2) f(–1) = 4a(1 + $\alpha$ ) Given f(–1) + f(2) = 0 $\Rightarrow$ a( $\alpha$ – 2 + 4 + 4 $\alpha$ ) = 0 $\Rightarrow$ 5 $\alpha$ = -2 As a $\ne$ 0 $\Rightarrow$ $\alpha$ =

-\frac{2}{5}

= - 0.4 $\therefore$ $\alpha$

\in

(–1, 0)