Quadratic Equation and Inequalities — Page 4 | JEE Mathematics

Q31

The number of real roots of the equation 5 + |2x – 1| = 2x (2x – 2) is

A 2

B 1

C 3

D 4

Correct Answer

Option B

Solution

When 2x $\ge$ 1 5 + 2x –1 = 2x (2x – 2) Let 2x = t $\Rightarrow$ 5 + t – 1 = t (t – 2) $\Rightarrow$ t = 4, – 1(rejected) $\Rightarrow$ 2x = 4 $\Rightarrow$ x = 2 Now when 2x < 1 5 + 1 – 2x = 2x (2x – 2) Let 2x = t $\Rightarrow$ 5 + 1 – t = t (t – 2) $\Rightarrow$ 0 = t2 – t – 6 $\Rightarrow$ 0 = (t – 3) (t – 2) $\Rightarrow$ t = 3, – 2 2x = 3, 2x = – 2 (rejected) $\therefore$ Only one real roots.

Q32

Let p, q

\in

R. If 2 -

\sqrt 3

is a root of the quadratic equation, x2 + px + q = 0, then :

A p2 – 4q – 12 = 0

B q2 – 4p – 16 = 0

C q2 + 4p + 14 = 0

D p2 – 4q + 12 = 0

Correct Answer

Option A

Solution

If a quadratic equation with rational coefficient has one irrational root then other root will be the conjugate of the irrational root.

Here x2 + px + q = 0 has one root 2 -

\sqrt 3

. $\therefore$ Other root will be 2 +

\sqrt 3

.

Sum of the roots = -p = 4 and product of the roots = q = 1 You can see p2 – 4q – 12 = 0 satisfy value of p = -4 and q = 1.

Q33

Let p and q be two positive numbers such that p + q = 2 and p4+q4 = 272. Then p and q are roots of the equation :

A x2 – 2x + 8 = 0

B x2 - 2x + 136=0

C x2 – 2x + 16 = 0

D x2 – 2x + 2 = 0

Correct Answer

Option C

Solution

{p^2} + {q^2} = {(p + q)^2} - 2pq

= 4 - 2pq

Now,

{\left( {{p^2} + {q^2}} \right)^2} = {p^4} + {q^4} + 2{p^2}{q^2}

\Rightarrow {\left( {4 - 2pq} \right)^2} = 272 + 2{p^2}{q^2}

\Rightarrow 16 + 4{p^2}{q^2} - 16pq = 272 + 2{p^2}{q^2}

\Rightarrow 2{p^2}{q^2} - 16pq - 256 = 0

\Rightarrow {p^2}{q^2} - 8pq - 128 = 0

\Rightarrow (pq - 16)(pq + 8) = 0

\Rightarrow pq = 16, - 8

Here, pq = - 8 is not possible as p and q are positive. $\therefore$ pq = 16 Now, the equation whose roots are p and q is

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

Q34

The set of all real values of

\lambda

for which the quadratic equations, (

\lambda

2 + 1)x2 – 4

\lambda

x + 2 = 0 always have exactly one root in the interval (0, 1) is :

A (–3, –1)

B (2, 4]

C (0, 2)

D (1, 3]

Correct Answer

Option D

Solution

Given quadratic equation,

({\lambda ^2} + 1){x^2} - 4\lambda x + 2 = 0

Here coefficient of x2 is ( $\lambda$ 2 + 1) which is always positive.

So quadratic equation is upward parabola.

So, either f(0) < 0 and f(1) > 1 or f(0) > 0 and f(1) < 0 $\therefore$ In both those cases, f(0) f(1) $\le$ 0 $\Rightarrow$ 2( $\lambda$ 2 $-$ 4 $\lambda$ + 3) $\le$ 0 $\Rightarrow$ $\lambda$

\in

[1, 3] At $\lambda$ = 1 : Quadratic equation becomes 2x2 $-$ 4x + 2 = 0 $\Rightarrow$ (x $-$ 1)2 = 0 $\Rightarrow$ x = 1, 1 As both roots can't lie between (0, 1) So, $\lambda$ = 1 can't be possible.

At $\lambda$ = 3 : 10x2 $-$ 12x + 2 = 0 $\Rightarrow$ 5x2 $-$ 6x + 1 = 0 $\Rightarrow$ (5x $-$ 1) (x $-$ 1) = 0 $\Rightarrow$ x = 1,

{1 \over 5}

In the interval (0, 1) exactly one root

{1 \over 5}

present. $\therefore$ $\lambda$

\in

(1, 3]

Q35

The product of the roots of the equation 9x2 - 18|x| + 5 = 0 is :

A

{{5} \over {9}}

B

{{5} \over {27}}

C

{{25} \over {81}}

D

{{25} \over {9}}

Correct Answer

Option C

Solution

9{x^2} - 18\left| x \right| + 5 = 0

$\Rightarrow$

9{x^2} - 15\left| x \right| - 3\left| x \right| + 5 = 0

( $\because$ x2 =

{\left| x \right|^2}

) $\Rightarrow$

3\left| x \right|(3\left| x \right| - 5) - (3\left| x \right| - 5) = 0

$\Rightarrow$

\left| x \right| = {1 \over 3},\,{5 \over 3}

$\Rightarrow$

x = \pm {1 \over 3}, \pm \,{5 \over 3}

$\therefore$ Product of roots =

\left( \frac{1}{3} \right) \left( -\frac{1}{3} \right) \left( \frac{5}{3} \right) \left( -\frac{5}{3} \right)

=

{{25} \over {81}}

Q36

Let

\alpha

and

\beta

be the roots of the equation 5x2 + 6x – 2 = 0. If Sn =

\alpha

n +

\beta

n, n = 1, 2, 3...., then :

A 5S6 + 6S5 = 2S4

B 5S6 + 6S5 + 2S4 = 0

C 6S6 + 5S5 + 2S4 = 0

D 6S6 + 5S5 = 2S4

Correct Answer

Option A

Solution

$\alpha$ and $\beta$ be the roots of the equation 5x2 + 6x – 2 = 0. $\Rightarrow$ 5 $\alpha$ 2 + 6 $\alpha$ - 2 = 0 $\Rightarrow$ 5 $\alpha$ n + 2 + 6 $\alpha$ n + 2 - 2 $\alpha$ n = 0 ......(

1) (By multiplying $\alpha$ n) Similarly 5 $\beta$ n + 2 + 6 $\beta$ n + 2 - 2 $\beta$ n = 0 ......(

2) By adding (1) & (2) 5Sn+2 + 6Sn+1 – 2Sn = 0 For n = 4 5S6 + 6S5 = 2S4

Q37

If

\alpha

and

\beta

are the roots of the equation, 7x2 – 3x – 2 = 0, then the value of

{\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}

is equal to :

A

{1 \over {24}}

B

{{27} \over {32}}

C

{{27} \over {16}}

D

{3 \over 8}

Correct Answer

Option C

Solution

Given, 7x2 – 3x – 2 = 0 $\therefore$ $\alpha$ + $\beta$ =

{3 \over 7}

$\alpha$ $\beta$ = -

{2 \over 7}

{\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}

=

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

=

{{{3 \over 7} + {2 \over 7}\left( {{3 \over 7}} \right)} \over {1 - {{\left( {\alpha + \beta } \right)}^2} + 2\alpha \beta + {{\left( { - {2 \over 7}} \right)}^2}}}

=

{{{3 \over 7} + {2 \over 7}\left( {{3 \over 7}} \right)} \over {1 - {{\left( {{3 \over 7}} \right)}^2} + 2\left( { - {2 \over 7}} \right) + {{\left( { - {2 \over 7}} \right)}^2}}}

=

{{27} \over {16}}

Q38

Let S be the set of all real roots of the equation, 3x(3x – 1) + 2 = |3x – 1| + |3x – 2|. Then S :

A contains exactly two elements.

B is an empty set.

C is a singleton.

D contains at least four elements.

Correct Answer

Option C

Solution

Let 3x = t ; t

>

0 t(t – 1) + 2 = |t – 1| + |t – 2| t2 – t + 2 = |t – 1| + |t – 2| Case-I : t

<

1 t2 – t + 2 = 1 – t + 2 – t $\Rightarrow$ t2 + 2 = 3 – t $\Rightarrow$ t2 + t – 1 = 0 $\Rightarrow$ t =

{{ - 1 \pm \sqrt 5 } \over 2}

$\Rightarrow$ t =

{{\sqrt 5 - 1} \over 2}

[ As t

>

0] Case-II : 1 $\le$ t

<

2 $\Rightarrow$ t2 – t + 2 = t – 1 + 2 – t $\Rightarrow$ t2 – t + 1 = 0 D

<

0 so no real solution.

Case-III : t $\ge$ 2 $\Rightarrow$ t2 – t + 2 = t – 1 + t – 2 $\Rightarrow$ t2 – 3t - 5 = 0 $\Rightarrow$ D

<

0 so no real solution.

Q39

Let

\alpha = {{ - 1 + i\sqrt 3 } \over 2}

. If

a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}}

and

b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}}

, then a and b are the roots of the quadratic equation :

A x2 + 101x + 100 = 0

B x2 + 102x + 101 = 0

C x2 – 102x + 101 = 0

D x2 – 101x + 100 = 0

Correct Answer

Option C

Solution

\alpha = {{ - 1 + i\sqrt 3 } \over 2}

= $\omega$

a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}}

=

\left( {1 + \omega } \right)\sum\limits_{k = 0}^{100} {{\omega ^{2k}}}

=

\left( {1 + \omega } \right){{1\left( {1 - {{\left( {{\omega ^2}} \right)}^{101}}} \right)} \over {1 - {\omega ^2}}}

=

{{1 - {\omega ^{202}}} \over {1 - \omega }}

=

{{1 - \omega } \over {1 - \omega }}

= 1

b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}}

=

\sum\limits_{k = 0}^{100} {{\omega ^{3k}}}

= 101 [as

{\omega ^3}

= 1] $\therefore$ roots are 101 and 1 Then equation is = x2 – 102x + 101 = 0

Q40

Let

\alpha

and

\beta

be the roots of the equation x2 - x - 1 = 0. If pk =

{\left( \alpha \right)^k} + {\left( \beta \right)^k}

, k

\ge

1, then which one of the following statements is not true?

A (p1 + p2 + p3 + p4 + p5) = 26

B p5 = 11

C p3 = p5 – p4

D p5 = p2 · p3

Correct Answer

Option D

Solution

x2 - x - 1 = 0 $\therefore$ $\alpha$ 2 - $\alpha$ - 1 = 0 $\Rightarrow$ $\alpha$ 2 = $\alpha$ + 1 $\therefore$ $\alpha$ 3 = $\alpha$ 2 + $\alpha$ = $\alpha$ + 1 + $\alpha$ = 2 $\alpha$ + 1 Now $\alpha$ 4 = 2 $\alpha$ 2 + $\alpha$ = 2( $\alpha$ + 1) + $\alpha$ = 3 $\alpha$ + 2 Now $\alpha$ 5 = 3 $\alpha$ 2 + 2 $\alpha$ = 3( $\alpha$ + 1) + 2 $\alpha$ = 5 $\alpha$ + 3 Given pk =

{\left( \alpha \right)^k} + {\left( \beta \right)^k}

$\therefore$ p5 =

{\left( \alpha \right)^5} + {\left( \beta \right)^5}

= 5 $\alpha$ + 3 + 5 $\beta$ + 3 = 5( $\alpha$ + $\beta$ ) + 6 = 5(1) + 6 [As $\alpha$ + $\beta$ = 1] = 11 Now p2 · p3 = ( $\alpha$ 2 + $\beta$ 2).( $\alpha$ 3 + $\beta$ 3) = ( $\alpha$ + 1 + $\beta$ + 1)(2 $\alpha$ + 1 + 2 $\beta$ + 1) = ( $\alpha$ + $\beta$ + 2)(2( $\alpha$ + $\beta$ ) + 2) = (1 + 2)(2 + 2) = 12 $\therefore$ p5 $\ne$ p2 ·p3 So option (D) is wrong.