Quadratic Equation and Inequalities — Page 5 | JEE Mathematics

Q41

Let

\alpha

and

\beta

be two real roots of the equation (k + 1)tan2x -

\sqrt 2

.

\lambda

tanx = (1 - k), where k(

\ne

- 1) and

\lambda

are real numbers. if tan2 (

\alpha

+

\beta

) = 50, then a value of

\lambda

is:

A 5

\sqrt 2

B 10

C 5

D 10

\sqrt 2

Correct Answer

Option B

Solution

Let tan $\alpha$ and tan $\beta$ are the roots of (k + 1)tan2x -

\sqrt 2

. $\lambda$ tanx - (1 - k) = 0 $\therefore$ tan $\alpha$ + tan $\beta$ =

{{\sqrt 2 \lambda } \over {k + 1}}

and an $\alpha$ .tan $\beta$ =

{{k - 1} \over {k + 1}}

Now tan( $\alpha$ + $\beta$ ) =

{{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }}

=

{{{{\lambda \sqrt 2 } \over {k + 1}}} \over {1 - {{k - 1} \over {k + 1}}}}

=

{{{\lambda \sqrt 2 } \over 2}}

=

{{\lambda \over {\sqrt 2 }}}

Given

{{{{\lambda ^2}} \over 2}}

= 50 $\Rightarrow$ $\lambda$ = 10

Q42

The integer 'k', for which the inequality x2

-

2(3k

-

1)x + 8k2

-

7 > 0 is valid for every x in R, is :

A 4

B 2

C 3

D 0

Correct Answer

Option C

Solution

{x^2} - 2(3k - 1)x + 8{k^2} - 7 > 0

Now, D < 0

\Rightarrow 4{(3k - 1)^2} - 4 \times 1 \times (8{k^2} - 7) < 0

\Rightarrow 9{k^2} - 6k + 1 - 8{k^2} + 7 < 0

\Rightarrow {k^2} - 6k + 8 < 0

\Rightarrow (k - 4)(k - 2) < 0

2 < k < 4 then k = 3

Q43

Let

\alpha

and

\beta

be the roots of x2

-

6x

-

2 = 0. If an =

\alpha

n

-

\beta

n for n

\ge

1, then the value of

{{{a_{10}} - 2{a_8}} \over {3{a_9}}}

is :

A 3

B 2

C 4

D 1

Correct Answer

Option B

Solution

Given, $\alpha$ and $\beta$ be the roots of

{x^2} - 6x - 2 = 0

\begin{array}{ll}{\alpha + \beta = 6} \\ {\alpha \beta = - 2} \end{array}

and

{\alpha ^2} - 6\alpha - 2 = 0 \Rightarrow {\alpha ^2} - 2 = 6\alpha

{\beta ^2} - 6\beta - 2 = 0 \Rightarrow {\beta ^2} - 2 = 6\beta

{{{a_{10}} - 2{a_8}} \over {3{a_9}}} = {{\left( {{\alpha ^{10}} - {\beta ^{10}}} \right) - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}

= {{\left( {{\alpha ^{10}} - 2{\alpha ^8}} \right) - \left( {{\beta ^{10}} - 2{\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}

Now,

= {{{\alpha ^8}\left( {{\alpha ^2} - 2} \right) - {\beta ^8}\left( {{\beta ^2} - 2} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}

= {{{\alpha ^8}(6\alpha ) - {\beta ^8}(6\beta )} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}} = {{6\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}} = {6 \over 3} = 2

Q44

The value of

3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}

is equal to

A 1.5 +

\sqrt 3

B 2 +

\sqrt 3

C 3 + 2

\sqrt 3

D 4 +

\sqrt 3

Correct Answer

Option A

Solution

Let

x = 3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}

So,

x = 3 + {1 \over {4 + {1 \over x}}} = 3 + {1 \over {{{4x + 1} \over x}}}

\Rightarrow (x - 3) = {x \over {(4x + 1)}}

\Rightarrow (4x + 1)(x - 3) = x

\Rightarrow 4{x^2} - 12x + x - 3 = x

\Rightarrow 4{x^2} - 12x - 3 = 0

x = {{12 \pm \sqrt {{{(12)}^2} + 12 \times 4} } \over {2 \times 4}} = {{12 \pm \sqrt {12(16)} } \over 8}

= {{12 \pm 4 \times 2\sqrt 3 } \over 8} = {{3 \pm 2\sqrt 3 } \over 2}

x = {3 \over 2} \pm \sqrt 3 = 1.5 \pm \sqrt 3

. But only positive value is accepted So, x =

1.5 + \sqrt 3

Q45

If

\alpha

and

\beta

are the distinct roots of the equation

{x^2} + {(3)^{1/4}}x + {3^{1/2}} = 0

, then the value of

{\alpha ^{96}}({\alpha ^{12}} - 1) + {\beta ^{96}}({\beta ^{12}} - 1)

is equal to :

A 56

\times

325

B 56

\times

324

C 52

\times

324

D 28

\times

325

Correct Answer

Option C

Solution

As,

({\alpha ^2} + \sqrt 3 ) = - {(3)^{1/4}}.\alpha

\Rightarrow ({\alpha ^4} + 2\sqrt 3 {\alpha ^2} + 3) = \sqrt 3 {\alpha ^2}

(On squaring) $\therefore$

({\alpha ^4} + 3) = ( - )\sqrt 3 {\alpha ^2}

\Rightarrow {\alpha ^8} + 6{\alpha ^4} + 9 = 3{\alpha ^4}

(Again squaring) $\therefore$

{\alpha ^8} + 3{\alpha ^4} + 9 = 0

\Rightarrow {\alpha ^8} = - 9 - 3{\alpha ^4}

(Multiply by $\alpha$ 4) So,

{\alpha ^{12}} = - 9{\alpha ^4} - 3{\alpha ^8}

$\therefore$

{\alpha ^{12}} = - 9{\alpha ^4} - 3( - 9 - 3{\alpha ^4})

\Rightarrow {\alpha ^{12}} = - 9{\alpha ^4} + 27 + 9{\alpha ^4}

Hence,

{\alpha ^{12}} = {(27)^2}

\Rightarrow {({\alpha ^{12}})^8} = {(27)^8}

\Rightarrow {\alpha ^{96}} = {(3)^{24}}

Similarly

{\beta ^{96}} = {(3)^{24}}

$\therefore$

{\alpha ^{96}}({\alpha ^{12}} - 1) + {\beta ^{96}}({\beta ^{12}} - 1) = {(3)^{24}} \times 52

$\Rightarrow$ Option (3) is correct.

Q46

Let [x] denote the greatest integer less than or equal to x. Then, the values of x

\in

R satisfying the equation

{[{e^x}]^2} + [{e^x} + 1] - 3 = 0

lie in the interval :

A

\left[ {0,{1 \over e}} \right)

B [loge2, loge3)

C [1, e)

D [0, loge2)

Correct Answer

Option D

Solution

{[{e^x}]^2} + [{e^x} + 1] - 3 = 0

\Rightarrow {[{e^x}]^2} + [{e^x}] + 1 - 3 = 0

Let

[{e^x}] = t

\Rightarrow {t^2} + t - 2 = 0

\Rightarrow t = - 2,1

[{e^x}] = - 2

(Not possible) or

[{e^x}] = 1

$\therefore$

1 \le {e^x} < 2

\Rightarrow \ln (1) \le x < \ln (2)

\Rightarrow 0 \le x < \ln (2)

\Rightarrow x \in [0,\ln 2)

Q47

The number of real roots of the equation

{e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0

is :

A 2

B 4

C 6

D 1

Correct Answer

Option A

Solution

{e^{6x}} - {e^{4x}} - 2{e^{3x}} - 12{e^{2x}} + {e^x} + 1 = 0

\Rightarrow {\left( {{e^{3x}} - 1} \right)^2} - {e^x}\left( {{e^{3x}} - 1} \right) = 12{e^{2x}}

{\left( {{e^{3x}} - 1} \right)^2}\left( {{e^x} - {e^{ - x}} - {e^{ - 2x}}} \right) = 12

\Rightarrow \underbrace {{e^x} - {e^{ - x}} - {e^{ - 2x}}}_{increa{\mathop{\rm sing}\nolimits} \,(let\,f(x))} = {{12} \over {\underbrace {{e^{3x}} - 1}_{decrea{\mathop{\rm sing}\nolimits} \,(let\,g(x))}}}

$\Rightarrow$ No. of real roots = 2

Q48

Let

\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\}

and

\beta = \mathop {\min }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\}

. If

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

is a quadratic equation whose roots are

\alpha

1/5 and

\beta

1/5, then the value of c

-

b is equal to :

A 42

B 47

C 43

D 50

Correct Answer

Option A

Solution

\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\}

= \max \{ {2^{6\sin 3x}}{.2^{8\cos 3x}}\}

= max\{ {2^{6\sin 3x + 8\cos 3x}}\}

and

\beta = \min \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} = \min \{ {2^{6\sin 3x + 8\cos 3x}}\}

Now range of

6sin3x + 8cos3x

= \left[ { - \sqrt {{6^2} + {8^2}} , + \sqrt {{6^2} + {8^2}} } \right] = [ - 10,10]

$\alpha$ = 210 & $\beta$ = 2 $-$ 10 So, $\alpha$ 1/5 = 22 = 4 $\Rightarrow$ $\beta$ 1/5 = 2 $-$ 2 = 1/4 quadratic 8x2 + bx + c = 0

- {b \over 8} = {{17} \over 4}

$\Rightarrow$ b = -34

{c \over 8} = 1

$\Rightarrow$ c = 8 $\therefore$ c – b = 8 + 34 = 42

Q49

If [x] be the greatest integer less than or equal to x, then

\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]}

is equal to :

A 0

B 4

C

-

2

D 2

Correct Answer

Option B

Solution

\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]}

= [4] + [-4.5] + [5] + [-5.5] + [6] +..... + [-49.5] + [50] = 4 - 5 + 5 - 6 + 6 ......-50 + 50 = 4

Q50

The number of real solutions of the equation, x2

-

|x|

-

12 = 0 is :

A 2

B 3

C 1

D 4

Correct Answer

Option A

Solution

|x|2 $-$ |x| $-$ 12 = 0 $\Rightarrow$ (|x| + 3)(|x| $-$ 4) = 0 $\Rightarrow$ |x| = 4 $\Rightarrow$ x = $\pm$ 4