Quadratic Equation and Inequalities — Page 11 | JEE Mathematics

Q101

Let the equation

x(x+2)(12-k)=2

have equal roots. Then the distance of the point

\left(k, \dfrac{k}{2}\right)

from the line

3 x+4 y+5=0

is

A 15

B 12

C

5 \sqrt{3}

D

15 \sqrt{5}

Correct Answer

Option A

Solution

Given the equation $x(x+2)(12-k) = 2$ , we want it to have equal roots.

To achieve this, we need to manipulate it into a quadratic form in terms of $x$ : $x^2 + 2x - \dfrac{2}{12-k} = 0$ For this quadratic equation to have equal (repeated) roots, the discriminant $D$ must be zero.

The discriminant $D$ for the equation $ax^2 + bx + c = 0$ is given by: $D = b^2 - 4ac$ Substituting $a = 1$ , $b = 2$ , and $c = -\dfrac{2}{12-k}$ into the discriminant formula, we get: $4 - 4\left(-\dfrac{2}{12-k}\right) = 0$ Simplifying the expression: $1 + \dfrac{2}{12-k} = 0$ Solving for $k$ : $\dfrac{2}{12-k} = -1 \quad \Rightarrow \quad 2 = -(12 - k) \quad \Rightarrow \quad 2 = -12 + k \quad \Rightarrow \quad k = 14$ Now, consider the point $(k, \dfrac{k}{2})$ , which becomes $(14, 7)$ .

We need to find its distance from the line $3x + 4y + 5 = 0$ .

The formula for the distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is: $d = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ Substituting $(x_1, y_1) = (14, 7)$ and the line coefficients $A = 3$ , $B = 4$ , $C = 5$ : $d = \dfrac{|3(14) + 4(7) + 5|}{\sqrt{3^2 + 4^2}}$ Calculating the numerator: $3 \times 14 + 4 \times 7 + 5 = 42 + 28 + 5 = 75$ And the denominator: $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ Therefore, the distance is: $d = \dfrac{75}{5} = 15$ Thus, the distance is 15.

Q102

The number of distinct real roots of the equation x7

-

7x

-

2 = 0 is

A 5

B 7

C 1

D 3

Correct Answer

Option D

Solution

Given equation

{x^7} - 7x - 2 = 0

Let

f(x) = {x^7} - 7x - 2

f'(x) = 7{x^6} - 7 = 7({x^6} - 1)

and

f'(x) = 0 \Rightarrow x = \, + \,1

and

f( - 1) = - 1 + 7 - 2 = 5 > 0

f(1) = 1 - 7 - 2 = - 8 < 0

So, roughly sketch of f(x) will be So, number of real roots of f(x) = 0 and 3

Q103

The value of

\lambda

such that sum of the squares of the roots of the quadratic equation, x2 + (3 –

\lambda

)x + 2 =

\lambda

has the least value is -

A 1

B 2

C

{{15} \over 8}

D

{4 \over 9}

Correct Answer

Option B

Solution

$\alpha$ + $\beta$ = $\lambda$ $-$ 3 $\alpha$ $\beta$ = 2 $-$ $\lambda$ $\alpha$ 2 + $\beta$ 2 = ( $\alpha$ + $\beta$ )2 $-$ 2 $\alpha$ $\beta$ = ( $\lambda$ $-$ 3)2 $-$ 2

\left( {2 - \lambda } \right)

= $\lambda$ 2 + 9 $-$ 6 $\lambda$ $-$ 4 + 2 $\lambda$ = $\lambda$ 2 $-$ 4 $\lambda$ + 5 = ( $\lambda$ $-$ 2)2 + 1 $\therefore$ $\lambda$ = 2

Q104

Difference between the corresponding roots of

{x^2} + ax + b = 0

and

{x^2} + bx + a = 0

is same and

a \ne b,

then

A

a + b + 4 = 0

B

a + b - 4 = 0

C

a - b - 4 = 0

D

a - b + 4 = 0

Correct Answer

Option A

Solution

Let

\alpha ,\beta

and

\gamma ,\delta

be the roots of the equations

{x^2} + ax + b = 0

and

{x^2} + bx + a = 0

respectively. $\therefore$

\alpha + \beta = - a,\alpha \beta = b

and

\gamma + \delta = - b,\gamma \delta = a.

Given

\left| {\alpha - \beta } \right| = \left| {\gamma - \delta } \right|

\Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( {\gamma - \delta } \right)^2}

\Rightarrow {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta = {\left( {\gamma + \delta } \right)^2} - 4\gamma \delta

\Rightarrow {a^2} - 4b = {b^2} - 4a

\Rightarrow \left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0

\Rightarrow a + b + 4 = 0

( as

a \ne b

)

Q105

If

\alpha \ne \beta

but

{\alpha ^2} = 5\alpha - 3

and

{\beta ^2} = 5\beta - 3

then the equation having

\alpha /\beta

and

\beta /\alpha \,\,

as its roots is

A

3{x^2} - 19x + 3 = 0

B

3{x^2} + 19x - 3 = 0

C

3{x^2} - 19x - 3 = 0

D

{x^2} - 5x + 3 = 0

Correct Answer

Option A

Solution

We have

{\alpha ^2} = 5\alpha - 3

and

{\beta ^2} = 5\beta - 3;

\Rightarrow \alpha \,\,\& \,\,\beta

are roots of equation,

{x^2} = 5x - 3

or

{x^2} - 5x + 3 = 0

$\therefore$

\alpha + \beta = 5

and

\alpha \beta = 3

Thus, the equation having

{\alpha \over \beta }\,\,\& \,\,{\beta \over \alpha }

as its roots is

{x^2} - x\left( {{\alpha \over \beta } + {\beta \over \alpha }} \right) + {{\alpha \beta } \over {\alpha \beta }} = 0

\Rightarrow {x^2} - x\left( {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right) + 1 = 0

or

3{x^2} - 19x + 3 = 0

Q106

If

a,\,b,\,c

are distinct

+ ve

real numbers and

{a^2} + {b^2} + {c^2} = 1

then

ab + bc + ca

is

A less than 1

B equal to 1

C greater than 1

D any real no.

Correct Answer

Option A

Solution

As

\,\,\,\,\,{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} > 0

\Rightarrow 2\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) > 0

\Rightarrow 2 > 2\left( {ab + bc + ca} \right)

\Rightarrow ab + bc + ca < 1

Q107

If the sum of the roots of the quadratic equation

a{x^2} + bx + c = 0

is equal to the sum of the squares of their reciprocals, then

{a \over c},\,{b \over a}

and

{c \over b}

are in

A Arithmetic - Geometric Progression

B Arithmetic Progression

C Geometric Progression

D Harmonic Progression

Correct Answer

Option D

Solution

a{x^2} + bx + c = 0,

\alpha + \beta = {{ - b} \over a},\alpha \beta = {c \over a}

As for given condition,

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

\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}

= {{{{{b^2}} \over {{a^2}}} - {{2c} \over a}} \over {{{{c^2}} \over {{a^2}}}}}

On simplification

2{a^2}c = a{b^2} + b{c^2}

\Rightarrow {{2a} \over b} = {c \over a} + {b \over c}

\Rightarrow {c \over a},{a \over b},{b \over c}

are in

A.P.

$\therefore$

{a \over c},{b \over a},\,\,\& \,\,

are in

H.P.

Q108

The number of integral values of m for which the equation (1 + m2 )x2 – 2(1 + 3m)x + (1 + 8m) = 0 has no real root is :

A 2

B infinitely many

C 1

D 3

Correct Answer

Option B

Solution

(1 + m2 )x2 – 2(1 + 3m)x + (1 + 8m) = 0 Given equation has no real solution, $\therefore$ Discriminant (D) < 0 $\Rightarrow$ 4(1 + 3m)2 - 4(1 + m2)(1 + 8m) < 0 $\Rightarrow$ 4[9m2 + 6m + 1 - 8m - 1 - 8m3 - m2] < 0 $\Rightarrow$ -8m3 + 8m2 - 2m < 0 $\Rightarrow$ -2m(4m2 - 4m + 1) < 0 $\Rightarrow$ m(2m - 1)2 > 0 $\therefore$ m > 0 and m $\ne$

{{1 \over 2}}

So we can say number of integral values of m are infinitely many.

Q109

The value of

a

for which the sum of the squares of the roots of the equation

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

assume the least value is

A

1

B

0

C

3

D

2

Correct Answer

Option A

Solution

Given quadratic equation,

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

Let $\alpha$ and $\beta$ are the roots of the equation. $\therefore$ $\alpha$ + $\beta$ =

a - 2

and $\alpha$ $\beta$ =

- a - 1

Now

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

$\Rightarrow$

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

$\Rightarrow$

{\alpha ^2} + {\beta ^2} = {a^2} - 2a + 6

$\Rightarrow$

{\alpha ^2} + {\beta ^2} = {\left( {a - 1} \right)^2} + 5

$\Rightarrow$ The value of

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

will be minimum, when

{a - 1}

= 0 $\Rightarrow$

{a = 1}

Q110

Let

\alpha

,

\beta

be two roots of the equation x2 + (20)1/4x + (5)1/2 = 0. Then

\alpha

8 +

\beta

8 is equal to

A 10

B 100

C 50

D 160

Correct Answer

Option C

Solution

x2 + (20)1/4x + (5)1/2 = 0 $\Rightarrow$ x2 +

\sqrt 5

= - (20)1/4x Squaring both sides, we get

{\left( {{x^2} + \sqrt 5 } \right)^2} = \sqrt {20} {x^2}

$\Rightarrow$ x4 = $-$ 5 $\Rightarrow$ x8 = 25 $\Rightarrow$ $\alpha$ 8 + $\beta$ 8 = 50