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Quadratic Equation and Inequalities

JEE Mathematics · 144 questions · Page 6 of 15 · Click an option or "Show Solution" to reveal answer

Q51
The set of all values of K > −-1, for which the equation (3x2+4x+3)2−(k+1)(3x2+4x+3)(3x2+4x+2)+k(3x2+4x+2)2=0{(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0 has real roots, is :
A (1,52]\left( {1,{5 \over 2}} \right]
B [2, 3)
C [−12,1)\left[ { - {1 \over 2},1} \right)
D (12,32]−{1}\left( {{1 \over 2},{3 \over 2}} \right] - \{ 1\}
Correct Answer
Option A
Solution
(3x2+4x+3)2−(k+1)(3x2+4x+3)(3x2+4x+2)+k(3x2+4x+2)2=0{(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0

Let

3x2+4x+3=a3{x^2} + 4x + 3 = a

and

3x2+4x+2=b⇒b=a−13{x^2} + 4x + 2 = b \Rightarrow b = a - 1

Given equation becomes

⇒a2−(k+1)ab+kb2=0\Rightarrow {a^2} - (k + 1)ab + k{b^2} = 0
⇒a(a−kb)−b(a−kb)=0\Rightarrow a(a - kb) - b(a - kb) = 0
⇒(a−kb)(a−b)=0⇒a=kb\Rightarrow (a - kb)(a - b) = 0 \Rightarrow a = kb

or a = b (reject) ∵\because a = kb

⇒3x2+4x+3=k(3x2+4x+2)\Rightarrow 3{x^2} + 4x + 3 = k(3{x^2} + 4x + 2)
⇒3(k−1)x2+4(k−1)x+(2k−3)=0\Rightarrow 3(k - 1){x^2} + 4(k - 1)x + (2k - 3) = 0

for real roots D ≥\ge 0

⇒16(k−1)2−4(3(k−1))(2k−3)≥0\Rightarrow 16{(k - 1)^2} - 4(3(k - 1))(2k - 3) \ge 0
⇒4(k−1){4(k−1)−3(2k−3)}≥0\Rightarrow 4(k - 1)\{ 4(k - 1) - 3(2k - 3)\} \ge 0
⇒4(k−1){−2k+5}≥0\Rightarrow 4(k - 1)\{ - 2k + 5\} \ge 0
⇒−4(k−1){2k−5}≥0\Rightarrow - 4(k - 1)\{ 2k - 5\} \ge 0
⇒(k−1)(2k−5)≤0\Rightarrow (k - 1)(2k - 5) \le 0

∴\therefore

k∈[1,52]k \in \left[ {1,{5 \over 2}} \right]

∴\therefore

k≠1k \ne 1

∴\therefore

k∈(1,52]k \in \left( {1,{5 \over 2}} \right]
Q52
cosec18∘^\circ is a root of the equation :
A x2 + 2x −- 4 = 0
B 4x2 + 2x −- 1 = 0
C x2 −- 2x + 4 = 0
D x2 −- 2x −- 4 = 0
Correct Answer
Option D
Solution
cos⁡ec18∘=1sin⁡18∘=45−1=5+1\cos ec18^\circ = {1 \over {\sin 18^\circ }} = {4 \over {\sqrt 5 - 1}} = \sqrt 5 + 1

Let

cos⁡ec18∘=x=5+1\cos ec18^\circ = x = \sqrt 5 + 1
⇒x−1=5\Rightarrow x - 1 = \sqrt 5

Squaring both sides, we get

x2−2x+1=5{x^2} - 2x + 1 = 5
⇒x2−2x−4=0\Rightarrow {x^2} - 2x - 4 = 0
Q53
The numbers of pairs (a, b) of real numbers, such that whenever α\alpha is a root of the equation x2 + ax + b = 0, α\alpha2 −- 2 is also a root of this equation, is :
A 6
B 2
C 4
D 8
Correct Answer
Option A
Solution

Consider the equation x2 + ax + b = 0 If has two roots (not necessarily real α\alpha & β\beta) Either α\alpha = β\beta or α\alpha ≠\ne β\beta Case (1) If α\alpha = β\beta, then it is repeated root.

Given that α\alpha2 −- 2 is also a root So, α\alpha = α\alpha2 −- 2 ⇒\Rightarrow (α\alpha + 1)(α\alpha −- 2) = 0 ⇒\Rightarrow α\alpha = −-1 or α\alpha = 2 When α\alpha = −-1 then (a, b) = (2, 1) α\alpha = 2 then (a, b) = (−-4, 4) Case (2) If α\alpha ≠\ne β\beta Then (I) α\alpha = α\alpha2 −- 2 and β\beta = β\beta2 −- 2 Hence, (a, b) = (−-(α\alpha + β\beta), α\alphaβ\beta) (−-1, −-2) (II) α\alpha = β\beta2 −- 2 and β\beta = α\alpha2 −- 2 Then α\alpha −- β\beta = β\beta2 −- α\alpha2 = (β\beta −- α\alpha) (β\beta + α\alpha) Since α\alpha ≠\ne β\beta we get α\alpha + β\beta = β\beta2 + α\alpha2 −- 4 α\alpha + β\beta = (α\alpha + β\beta)2 −- 2α\alphaβ\beta −- 4 Thus −-1 = 1 −-2 α\alphaβ\beta −- 4 which implies α\alphaβ\beta = −-1 Therefore (a, b) = (−-(α\alpha + β\beta), α\alphaβ\beta) = (1, −-1) (III) α\alpha = α\alpha2 −- 2 = β\beta2 −- 2 and α\alpha ≠\ne β\beta ⇒\Rightarrow α\alpha = −- β\beta Thus α\alpha = 2, β\beta = −-2 α\alpha = −-1, β\beta = 1 Therefore (a, b) = (0, −-4) & (0, 1) (IV) β\beta = α\alpha2 −- 2 = β\beta2 −- 2 and α\alpha ≠\ne β\beta is same as (III) Therefore we get 6 pairs of (a, b) Which are (2, 1), (−-4, 4), (−-1, −-2), (1, −-1), (0, −-4) Option (a)

Q54
Let α\alpha be a root of the equation 1 + x2 + x4 = 0. Then, the value of α\alpha1011 + α\alpha2022 −- α\alpha3033 is equal to :
A 1
B α\alpha
C 1 + α\alpha
D 1 + 2α\alpha
Correct Answer
Option A
Solution

Given, α\alpha is a root of the equation 1 + x2 + x4 = 0 ∴\therefore α\alpha will satisfy the equation.

∴\therefore 1 + α\alpha2 + α\alpha4 = 0

α2=−1±1−42{\alpha ^2} = {{ - 1 \pm \sqrt {1 - 4} } \over 2}
=−1±3i2= {{ - 1 \pm \sqrt 3 i} \over 2}

∴\therefore

α2=ω ar ω2{\alpha ^2} = \omega \,ar\,{\omega ^2}

Now,

α1011+α2022−α3033{\alpha ^{1011}} + {\alpha ^{2022}} - {\alpha ^{3033}}
=α . (α2)505+(α2)1011−α . (α2)1516= \alpha \,.\,{({\alpha ^2})^{505}} + {({\alpha ^2})^{1011}} - \alpha \,.\,{({\alpha ^2})^{1516}}
=α(ω)505+(ω)1011−α . (ω)1516= \alpha {(\omega )^{505}} + {(\omega )^{1011}} - \alpha \,.\,{(\omega )^{1516}}
=α . (ω3)168 . ω+(ω3)337−α . (ω3)505 . ω= \alpha \,.\,{({\omega ^3})^{168}}\,.\,\omega + {({\omega ^3})^{337}} - \alpha \,.\,{({\omega ^3})^{505}}\,.\,\omega
=α ω+1−α ω= \alpha \,\omega + 1 - \alpha \,\omega
=1= 1
Q55
Let f(x) be a quadratic polynomial such that f(−-2) + f(3) = 0. If one of the roots of f(x) = 0 is −-1, then the sum of the roots of f(x) = 0 is equal to :
A 113{{11} \over 3}
B 73{{7} \over 3}
C 133{{13} \over 3}
D 143{{14} \over 3}
Correct Answer
Option A
Solution

∵\because x = −-1 be the roots of f(x) = 0 ∴\therefore Let

f(x)=A(x+1)(x−1)f(x) = A(x + 1)(x - 1)

...... (i) Now,

f(−2)+f(3)=0f( - 2) + f(3) = 0
⇒A[−1(−2−b)+4(3−b)]=0\Rightarrow A[ - 1( - 2 - b) + 4(3 - b)] = 0
b=143b = {{14} \over 3}

∴\therefore Second root of f(x) = 0 will be

143{{14} \over 3}

∴\therefore Sum of roots

=143−1=113= {{14} \over 3} - 1 = {{11} \over 3}
Q56
The number of distinct real roots of x4 −- 4x + 1 = 0 is :
A 4
B 2
C 1
D 0
Correct Answer
Option B
Solution
f(x)=x4−4x+1=0f(x) = {x^4} - 4x + 1 = 0
f′(x)=4x3−4f'(x) = 4{x^3} - 4
=4(x−1)(x2+1+x)= 4(x - 1)({x^2} + 1 + x)

⇒\Rightarrow Two solution

Q57
Let $$A = \{ x \in R:|x + 1|
A A−B=(−1,1)A - B = ( - 1,1)
B B−A=R−(−3,1)B - A = R - ( - 3,1)
C A∩B=(−3,−1]A \cap B = ( - 3, - 1]
D A∪B=R−[1,3)A \cup B = R - [1,3)
Correct Answer
Option B
Solution

A = (−-3, 1) and B = (−- ∞\infty, −-1] ∪\cup [3, ∞\infty) So, A −- B = (−-1, 1) B −- A = (−- ∞\infty, −-3] ∪\cup [3, ∞\infty) = R −- (−-3, 3) A ∩\cap B = (−-3, −-1] and A ∪\cup B = (−- ∞\infty, 1) ∪\cup [3, ∞\infty) = R −- [1, 3)

Q58
Let a, b ∈\in R be such that the equation ax2−2bx+15=0a{x^2} - 2bx + 15 = 0 has a repeated root α\alpha. If α\alpha and β\beta are the roots of the equation x2−2bx+21=0{x^2} - 2bx + 21 = 0, then α2+β2{\alpha ^2} + {\beta ^2} is equal to :
A 37
B 58
C 68
D 92
Correct Answer
Option B
Solution
ax2−2bx+15=0a{x^2} - 2bx + 15 = 0

has repeated root so

b2=15a{b^2} = 15a

and

α=15b\alpha = {{15} \over b}

∵\because α\alpha is a root of

x2−2bx+21=0{x^2} - 2bx + 21 = 0

So

225b2=9⇒b2=25{{225} \over {{b^2}}} = 9 \Rightarrow {b^2} = 25

Now

α2+β2=(α+β)2−2αβ=4b2−42=100−42=58{\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta = 4{b^2} - 42 = 100 - 42 = 58
Q59
If the sum of the squares of the reciprocals of the roots α\alpha and β\beta of the equation 3x2 + λ\lambdax −- 1 = 0 is 15, then 6(α\alpha3 + β\beta3)2 is equal to :
A 18
B 24
C 36
D 96
Correct Answer
Option B
Solution
3x2+λx−1=03{x^2} + \lambda x - 1 = 0

Given, two roots are α\alpha and β\beta. ∴\therefore Sum of roots

=α+β=−λ3= \alpha + \beta = {-\lambda \over 3}

And product of roots

=αβ=−13= \alpha \beta = {-1 \over 3}

Given that, Sum of square of reciprocal of roots α\alpha and β\beta is 15. ∴\therefore

1α2+1β2=15{1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} = 15
⇒α2+β2α2β2=15\Rightarrow {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} = 15
⇒(α+β)2−2αβ(αβ)2=15\Rightarrow {{{{(\alpha + \beta )}^2} - 2\alpha \beta } \over {{{(\alpha \beta )}^2}}} = 15
⇒λ29+2×1319=15\Rightarrow {{{{{\lambda ^2}} \over 9} + 2 \times {1 \over 3}} \over {{1 \over 9}}} = 15
⇒λ2+6919=15\Rightarrow {{{{{\lambda ^2} + 6} \over 9}} \over {{1 \over 9}}} = 15
⇒λ2+6=15\Rightarrow {\lambda ^2} + 6 = 15
⇒λ2=9\Rightarrow {\lambda ^2} = 9

Now,

6(α3+β3)26{({\alpha ^3} + {\beta ^3})^2}
=6{(α+β)(α2+β2−αβ)}2= 6{\{ (\alpha + \beta )({\alpha ^2} + {\beta ^2} - \alpha \beta )\} ^2}
=6(α+β)2[(α+β)2−2αβ−αβ]2= 6{(\alpha + \beta )^2}{\left[ {{{(\alpha + \beta )}^2} - 2\alpha \beta - \alpha \beta } \right]^2}
=6(−λ3)2[(−λ3)2−3 . −13]2= 6{\left( {{-\lambda \over 3}} \right)^2}{\left[ {{{\left( {{-\lambda \over 3}} \right)}^2} - 3\,.\,{-1 \over 3}} \right]^2}
=6×λ29×[λ29+1]= 6 \times {{{\lambda ^2}} \over 9} \times \left[ {{{{\lambda ^2}} \over 9} + 1} \right]
=6×99×[99+1]2= 6 \times {{9} \over 9} \times {\left[ {{{9} \over 9} + 1} \right]^2}
=6×(2)2= 6 \times {\left( {{2}} \right)^2}
=6×4=24= {{6 \times 4}} = 24
Q60
Let S1={x∈R−{1,2}:(x+2)(x2+3x+5)−2+3x−x2≥0}{S_1} = \left\{ {x \in R - \{ 1,2\} :{{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0} \right\} and S2={x∈R:32x−3x+1−3x+2+27≤0}{S_2} = \left\{ {x \in R:{3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0} \right\}. Then, S1∪S2{S_1} \cup {S_2} is equal to :
A (−∞,−2]∪(1,2)( - \infty , - 2] \cup (1,2)
B (−∞,−2]∪[1,2]( - \infty , - 2] \cup [1,2]
C (−2,1]∪[2,∞)( - 2,1] \cup [2,\infty )
D (−∞,2]( - \infty ,2]
Correct Answer
Option B
Solution

Given,

(x+2)(x2+3x+5)−2+3x−x2≥0{{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0
x2+3x+5{x^2} + 3x + 5

is a quadratic equation

a=1>0a = 1 > 0

and

D=(−3)2−4 . 1 . 5=−11<0D = {( - 3)^2} - 4\,.\,1\,.\,5 = - 11 < 0

∴\therefore

x2+3x+5>0{x^2} + 3x + 5 > 0

(always) So, we can ignore this quadratic term

(x+2)−2+3x−x2≥0{{(x + 2)} \over { - 2 + 3x - {x^2}}} \ge 0
⇒x+2−(x2−3x+2)≥0\Rightarrow {{x + 2} \over { - ({x^2} - 3x + 2)}} \ge 0
⇒x+2x2−3x+2≤0\Rightarrow {{x + 2} \over {{x^2} - 3x + 2}} \le 0
⇒x+2x2−2x−x+2≤0\Rightarrow {{x + 2} \over {{x^2} - 2x - x + 2}} \le 0
⇒x+2(x−1)(x−2)≤0\Rightarrow {{x + 2} \over {(x - 1)(x - 2)}} \le 0

∴\therefore

x∈(−α,−2]∪(1,2)x \in ( - \alpha , - 2] \cup (1,2)

∴\therefore

S1=(−α,−2]∪(1,2){S_1} = ( - \alpha , - 2] \cup (1,2)

Now,

32x−3x+1−3x+2+27≤0{3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0
⇒(3x)2−3 . 3x−32 . 3x+27≤0\Rightarrow {({3^x})^2} - 3\,.\,{3^x} - {3^2}\,.\,{3^x} + 27 \le 0

Let

3x=t{3^x} = t
⇒t2−3 . t−32 . t+27≤0\Rightarrow {t^2} - 3\,.\,t - {3^2}\,.\,t + 27 \le 0
⇒t(t−3)−9(t−3)≤0\Rightarrow t(t - 3) - 9(t - 3) \le 0
⇒(t−3)(t−9)≤0\Rightarrow (t - 3)(t - 9) \le 0

∴\therefore

3≤t≤93 \le t \le 9
⇒31≤3x≤32\Rightarrow {3^1} \le {3^x} \le {3^2}
⇒1≤x≤2\Rightarrow 1 \le x \le 2

∴\therefore

x∈[1,2]x \in [1,2]

∴\therefore

S2=[1,2]{S_2} = [1,2]

∴\therefore

S1∪S2=(−α,2]∪(1,2)∪[1,2]{S_1} \cup {S_2} = ( - \alpha ,2] \cup (1,2) \cup [1,2]
=(−α,2]∪[1,2]= ( - \alpha ,2] \cup [1,2]
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