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Quadratic Equation and Inequalities

JEE Mathematics · 144 questions · Page 3 of 15 · Click an option or "Show Solution" to reveal answer

Q21
If α\alpha and β\beta are the roots of the equation x2−x+1=0,{x^2} - x + 1 = 0, then α2009+β2009={\alpha ^{2009}} + {\beta ^{2009}} =
A  −1\, - 1
B  1\, 1
C  2\, 2
D  −2\, - 2
Correct Answer
Option B
Solution
x2−x+1=0{x^2} - x + 1 = 0
⇒x=1±1−42\Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}
x=1±3i2x = {{1 \pm \sqrt 3 i} \over 2}
α=12+i32=−ω2\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}
β=12−i32=−ω\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega
α2009+β2009=(−ω2)2009+(−ω)2009{\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}
=−ω2−ω=1= - {\omega ^2} - \omega = 1
Q22
If the equations x2+2x+3=0{x^2} + 2x + 3 = 0 and ax2+bx+c=0,a{x^2} + bx + c = 0, a, b, c ∈ R,a,\,b,\,c\, \in \,R, have a common root, then a :b :c a\,:b\,:c\, is
A 1:2:31:2:3
B 3:2:13:2:1
C 1:3:21:3:2
D 3:1:23:1:2
Correct Answer
Option A
Solution

Given equations are

            x2+2x+3=0     ...(i)\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + 2x + 3 = 0\,\,\,\,\,...\left( i \right)
            ax2+bx+c=0   ...(ii)\,\,\,\,\,\,\,\,\,\,\,\,a{x^2} + bx + c = 0\,\,\,...\left( {ii} \right)

Roots of equation

(i)(i)

are imaginary roots. According to the question

(ii)(ii)

will also have both roots same as

(i).(i).

Thus

a1=b2=c3=λ(say){a \over 1} = {b \over 2} = {c \over 3} = \lambda \left( {say} \right)
⇒a=λ,b=2λ,c=3λ\Rightarrow a = \lambda ,b = 2\lambda ,c = 3\lambda

Hence, required ratio is

1:2:31:2:3
Q23
The quadratic equations x2−6x+a=0{x^2} - 6x + a = 0 and x2−cx+6=0{x^2} - cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is
A 1
B 4
C 3
D 2
Correct Answer
Option D
Solution

Let the roots of equation

x2−6x+a=0{x^2} - 6x + a = 0

be α\alpha and

44

β\beta and that of the equation

x2−cx+6=0{x^2} - cx + 6 = 0

be α\alpha and

3β.3\beta .

Then

                       α+4β=6;       4αβ=a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a

and

             α+3β=c;       3αβ=6\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 3\beta = c;\,\,\,\,\,\,\,3\alpha \beta = 6
⇒               a=8\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 8

∴\therefore The equation becomes

x2−6x+8=0{x^2} - 6x + 8 = 0
⇒(x−2)(x−4)=0\Rightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0

⇒\Rightarrow roots are

22

and

44
⇒α=2,β=1\Rightarrow \alpha = 2,\beta = 1

∴\therefore Common root is

2.2.
Q24
Let α\alpha and β\beta be the roots of equation x2−6x−2=0{x^2} - 6x - 2 = 0. If an=αn−βn,{a_n} = {\alpha ^n} - {\beta ^n}, for n≥1,n \ge 1, then the value of a10−2a82a9{{{a_{10}} - 2{a_8}} \over {2{a_9}}} is equal to :
A 33
B −3 - 3
C 66
D −6 - 6
Correct Answer
Option A
Solution

Given equation, x2 - 6x - 2 = 0 Roots are α\alpha and β\beta. So,

α+β=6\alpha + \beta = 6

and

αβ=−2\alpha \beta = - 2

In the question given,

an=αn−βn{a_n} = {\alpha ^n} - {\beta ^n}

∴\therefore

a8=α8−β8{a_8} = {\alpha ^8} - {\beta ^8}

and

a9=α9−β9{a_9} = {\alpha ^9} - {\beta ^9}

and

a10=α10−β10{a_{{10}}} = {\alpha ^{{10}}} - {\beta ^{10}}

Now, the given equation

a10−2a82a9{{{a_{10}} - 2{a_8}} \over {2{a_9}}}

=

α10−β10−2(α8−β8)2(α9−β9){{{\alpha ^{10}} - {\beta ^{10}} - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

=

α10−β10+αβ(α8−β8)2(α9−β9){{{\alpha ^{10}} - {\beta ^{10}} + \alpha \beta \left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

(as

αβ=−2\alpha \beta = - 2

) =

α10−β10+α9β−αβ92(α9−β9){{{\alpha ^{10}} - {\beta ^{10}} + {\alpha ^9}\beta - \alpha {\beta ^9}} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

=

α9(α+β)−β9(α+β)2(α9−β9){{{\alpha ^9}\left( {\alpha + \beta } \right) - {\beta ^9}\left( {\alpha + \beta } \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

=

(α+β)(α9−β9)2(α9−β9){{\left( {\alpha + \beta } \right)\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

=

(α+β)2{{\left( {\alpha + \beta } \right)} \over 2}

=

62{6 \over 2}

(as

α+β{ {\alpha + \beta } }

= 6) = 3

Q25
If for a positive integer n, the quadratic equation x(x+1)+(x+1)(x+2)x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right)+....+(x+n−1‾)(x+n) + .... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right)=10n = 10n has two consecutive integral solutions, then n is equal to :
A 9
B 10
C 11
D 12
Correct Answer
Option C
Solution
∑r=1n(x+r−1)(x+r)=10n\sum\limits_{r = 1}^n {\left( {x + r - 1} \right)\left( {x + r} \right)} = 10n

⇒\Rightarrow

∑r=1n(x2+xr+(r−1)x+r2−r)=10n\sum\limits_{r = 1}^n {\left( {{x^2} + xr + \left( {r - 1} \right)x + {r^2} - r} \right)} = 10n

⇒\Rightarrow

∑r=1n(x2+(2r−1)x+r(r−1))=10n\sum\limits_{r = 1}^n {\left( {{x^2} + \left( {2r - 1} \right)x + r\left( {r - 1} \right)} \right)} = 10n

⇒\Rightarrow

nx2+{1+3+5+....+(2n−1)}xn{x^2} + \left\{ {1 + 3 + 5 + .... + \left( {2n - 1} \right)} \right\}x
+{1.2+2.3+...+(n−1)n}+ \left\{ {1.2 + 2.3 + ... + \left( {n - 1} \right)n} \right\}

= 10n ⇒\Rightarrow

nx2+n2x+n(n2−1)3=10nn{x^2} + {n^2}x + {{n\left( {{n^2} - 1} \right)} \over 3} = 10n

⇒\Rightarrow

x2+nx+(n2−31)3=0{x^2} + nx + {{\left( {{n^2} - 31} \right)} \over 3} = 0

Let α\alpha and α\alpha + 1 be its two solutions ∴\therefore α\alpha + (α\alpha + 1) = -n ⇒\Rightarrow α\alpha =

−n−12{{ - n - 1} \over 2}

....(1) Also α\alpha(α\alpha + 1) =

(n2−31)3{{\left( {{n^2} - 31} \right)} \over 3}

......(2) Putting value of (1) in (2), we get

−(n+12)(1−n2)=(n2−31)3- \left( {{{n + 1} \over 2}} \right)\left( {{{1 - n} \over 2}} \right) = {{\left( {{n^2} - 31} \right)} \over 3}

⇒\Rightarrow n2 = 121 ⇒\Rightarrow n = 11

Q26
The sum of all real values of xx satisfying the equation (x2−5x+5)x2+4x−60 =1{\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}}\, = 1 is :
A 66
B 55
C 33
D −4-4
Correct Answer
Option C
Solution

Given equation,

(x2−5x+5)x2+4x−60=1{\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}} = 1

Case 1 : When x2 - 5x + 5 = 1 and x2 + 4x - 60 is any real no then this equation satisfy.

Note : When we put any real number as a power of 1 the value stays always 1 (1 any real no = 1). x2 - 5x + 5 = 1 (x - 1)(x - 4) = 0 ∴\therefore x = 1, 4 Case 2 : When x2 - 5x + 5 is a real no and x2 + 4x - 60 = 0 then the given equation satisfy.

As we know if power of any real no is zero then it will become 1((any real number)0 = 1).

For, x2 + 4x - 60 = 0 (x - 6)(x + 10) = 0 ∴\therefore x = 6, -10 Case 3 : When x2 - 5x + 5 = -1 and x2 + 4x - 60 is even this equation satisfy.

As we know (-1)even = 1.

For, x2 - 5x + 5 = -1 (x - 2)(x - 3) = 0 ∴\therefore x = 2, 3 But x can't be 3 because when x = 3 the value of x2 + 4x - 60 becomes 32 + 4.3 - 60 = - 39 which is an odd number, then (-1)-39 = -1.

So for x = 3 equation does not satisfy.

∴\therefore The sum of all the real values = 1 + 4 + 6 + (-10) + 2 = 3

Q27
Let p, q and r be real numbers (p ≠ \ne q, r ≠ \ne 0), such that the roots of the equation 1x+p+1x+q=1r{1 \over {x + p}} + {1 \over {x + q}} = {1 \over r} are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :
A p2+q22{{{p^2} + {q^2}} \over 2}
B p2 + q2
C 2(p2 + q2)
D p2 + q2 + r2
Correct Answer
Option B
Solution

Given,

1x+p+1x+q=1r{1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}

⇒\Rightarrow

   \,\,\,
x+p+x+q(x+p)(x+q)=1r{{x + p + x + q} \over {\left( {x + p} \right)\left( {x + q} \right)}} = {1 \over r}

⇒\Rightarrow

   \,\,\,

(2x + p + q) r = x2 + px + qx + pq ⇒\Rightarrow

   \,\,\,

x2 + (p + q −- 2r) x + pq −- pr −- qr = 0 Let α\alpha and β\beta are the roots,

∴   \therefore\,\,\,

α\alpha + β\beta = −- (p + q −- 2r) and α\alpha β\beta = pq −- pr −- qr Given that, α\alpha = −- β\beta ⇒\Rightarrow α\alpha + β\beta = 0

∴   \therefore\,\,\,

−- (p + q −- 2r) = 0 Now, α\alpha2 + β\beta2 = (α\alpha + β\beta)2 −- 2α\alpha β\beta = (−- (p + q −- 2r))2 −- 2 (pq −- pr −- qr) = p2 +q2 + 4r2 + 2pq −- 4pr −- 4qr −- 2pq + 2pr + 2qr = p2 + q2 + 4r2 −- 2pr −- 2qr = p2 + q2 −- 2r (p + q −- 2r) = p2 + q2 −- 2r (0) = p2 + q2

Q28
If λ\lambda ∈ \in R is such that the sum of the cubes of the roots of the equation, x2 + (2 −- λ\lambda ) x + (10 −- λ\lambda ) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :
A 424\sqrt 2
B 252\sqrt 5
C 272\sqrt 7
D 20
Correct Answer
Option B
Solution

Let α\alpha, β\beta are the roots of the equation, ∴\therefore α\alpha + β\beta = λ\lambda −- 2 and α\alphaβ\beta = 10 −- λ\lambda

α3+β3{\alpha ^3} + {\beta ^3}

= (α\alpha + β\beta)3 −- 3α\alphaβ\beta (α\alpha + β\beta) = (λ\lambda −- 2)3 −- 3(10 −- λ\lambda)(λ\lambda −- 2) =

λ3\lambda ^3

−- 3

λ2\lambda ^2

−- 24λ\lambda + 52 Let

f(λf(\lambda

) =

λ3\lambda ^3

−- 3

λ2\lambda ^2

−- 24λ\lambda + 52 ∴\therefore

df(λ)dλ{{df(\lambda )} \over {d\lambda }}

= 3

λ2\lambda ^2

−- 6λ\lambda −- 24 ∴\therefore at maximum of minimum

df(λ)dλ{{df(\lambda )} \over {d\lambda }}

= 0 ∴\therefore

λ2\lambda ^2

−- 2λ\lambda −- 8 = 0 ⇒\Rightarrow (λ\lambda + 2) (λ\lambda −- 4) = 0 ⇒\Rightarrow λ\lambda = −-2, 4

d2f(λ)dλ2{{{d^2}f(\lambda )} \over {d{\lambda ^2}}}

= 2λ\lambda −- 2 When λ\lambda = −-2

d2f(λ)dλ2{{{d^2}f(\lambda )} \over {d{\lambda ^2}}}

= −- 6 < 0 ∴\therefore at λ\lambda = −-2, f(λ\lambda) has maximum value. When λ\lambda = 4

d2f(λ)dλ2{{{d^2}f(\lambda )} \over {d{\lambda ^2}}}

= 6 > 0 ∴\therefore at λ\lambda = 4, f(λ\lambda) has minimum value.

∴\therefore When λ\lambda = 4 equation is, x2 −- 2x + 6 = 0 ∴\therefore (α\alpha −- β\beta)2 = (α\alpha + β\beta)2 −- 4

αβ\alpha \beta

⇒\Rightarrow x2 −- 4 ×\times 6 = −- 20 ⇒\Rightarrow (α\alpha −- β\beta) =

25i2\sqrt 5 i

⇒\Rightarrow

∣α−β∣\left| {\alpha - \beta } \right|

=

252\sqrt 5

(ans)

Q29
If λ\lambda be the ratio of the roots of the quadratic equation in x, 3m2x2 + m(m – 4)x + 2 = 0, then the least value of m for which λ+1λ=1,\lambda + {1 \over \lambda } = 1, is
A −2+2 - 2 + \sqrt 2
B 4−-32\sqrt 2
C 2 −- 3\sqrt 3
D 4 −- 23\sqrt 3
Correct Answer
Option B
Solution

3m2x2 + m(m −- 4) x + 2 = 0

λ+1λ=1,αβ+βα=1,α2+β2=αβ\lambda + {1 \over \lambda } = 1,{\alpha \over \beta } + {\beta \over \alpha } = 1,{\alpha ^2} + {\beta ^2} = \alpha \beta

(α\alpha + β\beta)2 = 3α\alphaβ\beta

(−m(m−4)3m2)2=3(2)3m2,(m−4)29m2=63m{\left( { - {{m\left( {m - 4} \right)} \over {3{m^2}}}} \right)^2} = {{3\left( 2 \right)} \over {3{m^2}}},{{{{\left( {m - 4} \right)}^2}} \over {9{m^2}}} = {6 \over {3m}}
(m−4)2=18,m=4±18,  4±32{\left( {m - 4} \right)^2} = 18,m = 4 \pm \sqrt {18,} \,\,4 \pm 3\sqrt 2
Q30
Let S = { xx ∈ \in R : xx ≥ \ge 0 and 2∣x−3∣+x(x−6)+6=02\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0}. Then S
A contains exactly four elements
B is an empty set
C contains exactly one element
D contains exactly two elements
Correct Answer
Option D
Solution

Given,

2∣x−3∣+x(x−6)+6=02\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0

Case 1 : When

x−3≥0,\sqrt x - 3 \ge 0,

then equation becomes

2(x−3)+x(x−6)+6=02\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0
⇒    2x−6+x−6x+6=0\Rightarrow \,\,\,\,2\sqrt x - 6 + x - 6\sqrt x + 6 = 0
⇒    x −4x=0\Rightarrow \,\,\,\,x\, - 4\sqrt x = 0
⇒    x(x−4)=0\Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 4} \right) = 0
∴   \therefore\,\,\,
x=0,4\sqrt x = 0,4

but as

x−3≥0\sqrt x - 3 \ge 0

or

x≥3\sqrt x \ge 3

then

x≠0\sqrt x \ne 0
∴   \therefore\,\,\,
x=4\sqrt x = 4

value is possible. Case 2 : When

x−3<0\sqrt x - 3 < 0

or

x<3.\sqrt x < 3.

There equation becomes

−2(x−3)+x(x−6)+6=0- 2\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0
⇒     −2x+6+x−6x+6=0\Rightarrow \,\,\,\,\, - 2\sqrt x + 6 + x - 6\sqrt x + 6 = 0
⇒    x−8x+12=0\Rightarrow \,\,\,\,x - 8\sqrt x + 12 = 0
⇒    x−6x−2x+12=0\Rightarrow \,\,\,\,x - 6\sqrt x - 2\sqrt x + 12 = 0
⇒    x(x−6)−2(x−6)=0\Rightarrow \,\,\,\,\sqrt x \left( {\sqrt x - 6} \right) - 2\left( {\sqrt x - 6} \right) = 0
⇒    (x−2)(x−6)=0\Rightarrow \,\,\,\,\left( {\sqrt x - 2} \right)\left( {\sqrt x - 6} \right) = 0
∴   \therefore\,\,\,
x=2,6\sqrt x = 2,6

as

x<3\sqrt x < 3

so

x≠6\sqrt x \ne 6
∴   \therefore\,\,\,
x=2\sqrt x = 2

is possible. So, total possible value of

x=2,4\sqrt x = 2,4

or for x possible values are 4, 16.

∴   \therefore\,\,\,

Set S contains exactly two elements 4 and 16.

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