Quadratic Equation and Inequalities — Page 12 | JEE Mathematics

Q111

If one real root of the quadratic equation 81x2 + kx + 256 = 0 is cube of the other root, then a value of k is

A

-

81

B

-

300

C 100

D 144

Correct Answer

Option B

Solution

81x2 + kx + 256 = 0 ; x = $\alpha$ , $\alpha$ 3 $\Rightarrow$ $\alpha$ 4 =

{{256} \over {81}}

$\Rightarrow$ $\alpha$ = $\pm$

{{4} \over {3}}

Now $-$

{k \over {81}}

= $\alpha$ + $\alpha$ 3 = $\pm$

{{100} \over {27}}

$\Rightarrow$ k = $\pm$ 300

Q112

Let [t] denote the greatest integer

\le

t. Then the equation in x, [x]2 + 2[x+2] - 7 = 0 has :

A no integral solution.

B exactly two solutions.

C exactly four integral solutions.

D infinitely many solutions.

Correct Answer

Option D

Solution

{[x]^2} + 2[x + 2] - 7 = 0

$\Rightarrow$

{[x]^2} + 2[x] + 4 - 7 = 0

Using the property [x + n] = [x] + n ; n

\in

I $\Rightarrow$

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

let [x] = y

{y^2} + 3y - y - 3 = 0

$\Rightarrow$

(y - 1)(y + 3) = 0

[x] = 1\,or\,[x] = - 3

$\therefore$

x \in \left[ {1,2} \right)\,\& \, \in \left[ { - 3, - 2} \right)

Q113

Let

\alpha

and

\beta

be the roots of equation

p{x^2} + qx + r = 0,

p \ne 0.

If

p,\,q,\,r

in A.P. and

{1 \over \alpha } + {1 \over \beta } = 4,

then the value of

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

is :

A

{{\sqrt {34} } \over 9}

B

{{2\sqrt 13 } \over 9}

C

{{\sqrt {61} } \over 9}

D

{{2\sqrt 17 } \over 9}

Correct Answer

Option B

Solution

Let

p,q,r

are in

AP

\Rightarrow 2q = p + r\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Given

{1 \over \alpha } + {1 \over \beta } = 4 \Rightarrow {{\alpha + \beta } \over {\alpha \beta }} = 4

We have

\alpha + \beta = - q/p

and

\alpha \beta = {r \over p}

\Rightarrow {{ - {q \over p}} \over {{r \over p}}} = 4 \Rightarrow q = - 4r\,\,\,\,\,\,\,\,...\left( {ii} \right)

From

(i),

we have

2\left( { - 4r} \right) = p + r \Rightarrow p = - 9r

q = - 4r

Now

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

= \sqrt {{{\left( {{{ - q} \over p}} \right)}^2} - {{4r} \over p}}

= {{\sqrt {{q^2} - 4pr} } \over {\left| p \right|}}

= {{\sqrt {16{r^2} + 36{r^2}} } \over {\left| { - 9r} \right|}}

= {{2\sqrt {13} } \over 9}

Q114

Let

\alpha

and

\beta

be the roots of x2 - 3x + p=0 and

\gamma

and

\delta

be the roots of x2 - 6x + q = 0. If

\alpha, \beta, \gamma, \delta

form a geometric progression.Then ratio (2q + p) : (2q - p) is:

A 9 : 7

B 5 : 3

C 3 : 1

D 33 :31

Correct Answer

Option A

Solution

$\alpha$ and $\beta$ are the roots of x2 $-$ 3x + p = 0 $\therefore$ $\alpha$ + $\beta$ = 3 and

\alpha \beta

= p $\gamma$ and $\delta$ are the roots of x2 $-$ 6x + q = 0 $\therefore$ $\gamma$ + $\delta$ = 6 and

\gamma \delta

= q Given that, $\alpha$ , $\beta$ , $\gamma$ , $\delta$ are in G.P.

$\therefore$ Let $\alpha$ = a, $\beta$ = ar, $\gamma$ = ar2 and $\delta$ = ar3 As $\alpha$ + $\beta$ = 3 $\Rightarrow$ a + ar = 3 $\Rightarrow$ a(1 + r) = 3 ...(

1) Also $\gamma$ + $\delta$ = 6 $\Rightarrow$ ar2 + ar3 = 6 $\Rightarrow$ ar2 (1 + r) = 6 ...

(2) Dividing (2) by (1), r2 = 2

\Rightarrow r = \sqrt 2

$\therefore$

a = {3 \over {1 + \sqrt 2 }}

So,

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

,

\beta = {{3\sqrt 2 } \over {1 + \sqrt 2 }}

,

\gamma = {{3 \times 2} \over {1 + \sqrt 2 }}

,

\delta = {{3(2\sqrt {2)} } \over {1 + \sqrt 2 }}

$\therefore$

p = \alpha \beta = {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}

and

q = \gamma \delta = {{36\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}

Now,

{{2q + p} \over {2q - p}}

= {{{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} + {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}} \over {{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} - {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}}}

= {{72\sqrt 2 + 9\sqrt 2 } \over {72\sqrt 2 - 9\sqrt 2 }}

= {{81} \over {63}}

= {9 \over 7}

Q115

If an angle A of a

\Delta

ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are :

A secA, cotA

B sinA, secA

C secA, tanA

D tanA, cosA

Correct Answer

Option C

Solution

Here, 9x2 + 27x + 20 = 0

\therefore\,\,\,

x =

{{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}

$\Rightarrow$

\,\,\,

x =

{{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}

$\Rightarrow$

\,\,\,

x = $-$

{4 \over 3}

, $-$

{5 \over 3}

Given, cosA = $-$

{3 \over 5}

\therefore\,\,\,

sec A =

{1 \over {\cos A}}

= $-$

{5 \over 3}

Here, A is an obtuse angle.

\therefore\,\,\,

tan A = $-$

\sqrt {{{\sec }^2}A - 1} = - {4 \over 3}.

Hence, roots of the equation are sec A and tan A.

Q116

STATEMENT - 1 : For every natural number

n \ge 2,

{1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + ........ + {1 \over {\sqrt n }} > \sqrt n .

STATEMENT - 2 : For every natural number

n \ge 2,

,

\sqrt {n\left( {n + 1} \right)} < n + 1.

$

A Statement - 1 is false, Statement - 2 is true

B Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for statement - 1

C Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1

D Statement - 1 is true, Statement - 2 is false

Correct Answer

Option B

Solution

Statements

2

is

\sqrt {n\left( {n + 1} \right)} < n + 1,n \ge 2

\Rightarrow \sqrt n < \sqrt {n + 1} ,n \ge 2

which is true

\Rightarrow \sqrt 2 < \sqrt 3 < \sqrt 4 < \sqrt 5 < - - - - - - \sqrt n

Now

\sqrt 2 < \sqrt n \Rightarrow {1 \over {\sqrt 2 }} > {1 \over {\sqrt n }}

\sqrt 3 < \sqrt n \Rightarrow {1 \over {\sqrt 3 }} > {1 \over {\sqrt n }};

\sqrt n \le \sqrt n \Rightarrow {1 \over {\sqrt n }} \ge {1 \over {\sqrt n }}

Also

{1 \over {\sqrt 1 }} > {1 \over {\sqrt n }}

$\therefore$ Adding all, we get

{1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + ....... + {1 \over n} > {n \over {\sqrt n }} = \sqrt n

Hence both the statements are correct and statement

2

is a correct explanation of statement

-1.

Q117

The sum of all the real values of x satisfying the equation 2(x

-

1)(x2 + 5x

-

50) = 1 is :

A 16

B 14

C

-

4

D

-

5

Correct Answer

Option C

Solution

We know, 2x = 1 only when x = 0.

Similarly, 2(x $-$ 1)(x2 + 5x $-$ 50) = 1 when (x $-$ 1)(x2 + 5x $-$ 50) = 0 $\Rightarrow$ (x - 1)(x + 10)(x - 5) = 0 $\therefore$ x = 1, -10, 5 Sum of real values of x = 1 + (-10) + 5 = -4

Q118

If f(x) is a quadratic expression such that f (1) + f (2) = 0, and

-

1 is a root of f (x) = 0, then the other root of f(x) = 0 is :

A

-

{5 \over 8}

B

-

{8 \over 5}

C

{5 \over 8}

D

{8 \over 5}

Correct Answer

Option D

Solution

Let $\alpha$ and $\beta$ = - 1 are the roots of the polynomial, then we get f(x) = x2 + (1 - $\alpha$ )x - $\alpha$ $\therefore$ f(1) = 2 - 2 $\alpha$ and f(2) = 6 - 3 $\alpha$ Also given, f (1) + f (2) = 0 $\therefore$ 2 - 2 $\alpha$ + 6 - 3 $\alpha$ = 0 $\Rightarrow$ $\alpha$ =

{8 \over 5}

Q119

If

\alpha

and

\beta

be two roots of the equation x2 – 64x + 256 = 0. Then the value of

{\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}

is :

A 1

B 3

C 2

D 4

Correct Answer

Option C

Solution

x2 – 64x + 256 = 0 $\alpha$ + $\beta$ = 64, $\alpha$ $\beta$ = 256

{\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}

=

{{{\alpha ^{{3 \over 8}}}} \over {{\beta ^{{5 \over 8}}}}} + {{{\beta ^{{3 \over 8}}}} \over {{\alpha ^{{5 \over 8}}}}}

=

{{\alpha + \beta } \over {{{\left( {\alpha \beta } \right)}^{{5 \over 8}}}}}

=

{{64} \over {{{\left( {256} \right)}^{{5 \over 8}}}}}

= 2

Q120

If

\alpha

and

\beta

are the roots of the equation x2 + px + 2 = 0 and

{1 \over \alpha }

and

{1 \over \beta }

are the roots of the equation 2x2 + 2qx + 1 = 0, then

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

is equal to :

A

{9 \over 4}\left( {9 - {q^2}} \right)

B

{9 \over 4}\left( {9 + {q^2}} \right)

C

{9 \over 4}\left( {9 - {p^2}} \right)

D

{9 \over 4}\left( {9 + {p^2}} \right)

Correct Answer

Option C

Solution

$\alpha$ and $\beta$ are the roots of the equation x2 + px + 2 = 0 $\therefore$

\alpha + \beta = - p,\,\alpha \beta = 2

{1 \over \alpha }

and

{1 \over \beta }

are the roots of the equation 2x2 + 2qx + 1 = 0 $\therefore$

{1 \over \alpha } + {1 \over \beta } = - q,\,{1 \over {\alpha \beta }} = {1 \over 2}

$\Rightarrow$

{{\alpha + \beta } \over {\alpha \beta }} = - q \Rightarrow {{ - p} \over 2} = - q

\Rightarrow p = 2q

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

= 2 + {1 \over 2} + 2 = {9 \over 2}

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

= 2 + {1 \over 2} - \left[ {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right]

= {5 \over 2} - \left[ {{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \over {\alpha \beta }}} \right]

= {5 \over 2} - \left[ {{{{p^2} - 4} \over 2}} \right]

= {{9 - {p^2}} \over 2}

\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \left( {{{9 - {p^2}} \over 2}} \right)\left( {{9 \over 2}} \right)

= {9 \over 4}(9 - {p^2})