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Quadratic Equation and Inequalities

JEE Mathematics · 144 questions · Page 7 of 15 · Click an option or "Show Solution" to reveal answer

Q61
Let S be the set of all integral values of α\alpha for which the sum of squares of two real roots of the quadratic equation 3x2+(α−6)x+(α+3)=03{x^2} + (\alpha - 6)x + (\alpha + 3) = 0 is minimum. Then S :
A is an empty set
B is a singleton
C contains exactly two elements
D contains more than two elements
Correct Answer
Option A
Solution

Given quadratic equation,

3x2+(α−6)x+(α+3)=03{x^2} + (\alpha - 6)x + (\alpha + 3) = 0

Let, a and b are the roots of the equation, ∴\therefore

a+b=−α−63a + b = - {{\alpha - 6} \over 3}

and

ab=α+33ab = {{\alpha + 3} \over 3}

For real roots,

D≥0D \ge 0
⇒(α−6)2−4 . 3 . (α+9)≥0\Rightarrow {(\alpha - 6)^2} - 4\,.\,3\,.\,(\alpha + 9) \ge 0
⇒α2−12α+36−12α−36≥0\Rightarrow {\alpha ^2} - 12\alpha + 36 - 12\alpha - 36 \ge 0
⇒α2−24α≥0\Rightarrow {\alpha ^2} - 24\alpha \ge 0
⇒α(α−24)≥0\Rightarrow \alpha (\alpha - 24) \ge 0

∴\therefore

α>24\alpha > 24

or

α<0\alpha < 0

∴\therefore Real roots of the equation possible for

α>24\alpha > 24

or

α<0\alpha < 0

. Now, sum of square of roots

=a2+b2= {a^2} + {b^2}
=(a+b)2−2ab= {(a + b)^2} - 2ab
=(α−6)29−2 . (α+3)3= {{{{(\alpha - 6)}^2}} \over 9} - 2\,.\,{{(\alpha + 3)} \over 3}
=α2−12α+36−6α−189= {{{\alpha ^2} - 12\alpha + 36 - 6\alpha - 18} \over 9}
=α2−18α+189=f(x)= {{{\alpha ^2} - 18\alpha + 18} \over 9} = f(x)

∴\therefore Sum of square of roots are minimum when

a2+b2={a^2} + {b^2} =

minimum. ∴\therefore

f(α)min⁡=α2−18α+189f{(\alpha )_{\min }} = {{{\alpha ^2} - 18\alpha + 18} \over 9}

Value of quadratic equation

α2−18α+18{\alpha ^2} - 18\alpha + 18

is minimum at

α=−b2a=−(−18)2 . 1=9\alpha = - {b \over {2a}} = - {{( - 18)} \over {2\,.\,1}} = 9

But for real roots α\alpha should be less than 0 or greater than 24. So, there is no value of α\alpha in the range

α>24∪α<0\alpha > 24 \cup \alpha < 0

where sum of squares of two real roots is minimum. ∴\therefore S is an empty set.

Q62
If α,β,γ,δ\alpha, \beta, \gamma, \delta are the roots of the equation x4+x3+x2+x+1=0x^{4}+x^{3}+x^{2}+x+1=0, then α2021+β2021+γ2021+δ2021\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021} is equal to :
A −-4
B −-1
C 1
D 4
Correct Answer
Option B
Solution

When,

x5=1{x^5} = 1

then

x5−1=0{x^5} - 1 = 0
⇒(x−1)(x4+x3+x2+x+1)=0\Rightarrow (x - 1)({x^4} + {x^3} + {x^2} + x + 1) = 0

Given,

x4+x3+x2+x+1=0{x^4} + {x^3} + {x^2} + x + 1 = 0

has roots α\alpha, β\beta, γ\gamma and 8. ∴\therefore Roots of

x5−1=0{x^5} - 1 = 0

are 1, α\alpha, β\beta, γ\gamma and 8.

We know, Sum of pth power of nth roots of unity = 0.

(If p is not multiple of n) or n (If p is multiple of n) ∴\therefore Here, Sum of pth power of nth roots of unity

=1p+αp+βp+γp+8p={0;If p is not multiple of 55;If p is multiple of 5= {1^p} + {\alpha ^p} + {\beta ^p} + {\gamma ^p} + {8^p} = \left\{ \begin{array}{lll}0 & ; & {\mathrm{If\,p\,is\,not\,multiple\,of\,5}} \\ 5 & ; & {\mathrm{If\,p\,is\,multiple\,of\,5}} \end{array} \right.

Here,

p=2021p = 2021

, which is not multiple of 5. ∴\therefore

12021+α2021+β2021+γ2021+82021=0{1^{2021}} + {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {8^{2021}} = 0
⇒α2021+β2021+γ2021+82021=−1\Rightarrow {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {8^{2021}} = - 1
Q63
The minimum value of the sum of the squares of the roots of x2+(3−a)x+1=2ax^{2}+(3-a) x+1=2 a is:
A 4
B 5
C 6
D 8
Correct Answer
Option C
Solution
α+β=a−3, αβ=1−2a\alpha + \beta = a - 3,\,\alpha \beta = 1 - 2a
⇒α2+β2=(a−3)2−2(1−2a)\Rightarrow {\alpha ^2} + {\beta ^2} = {(a - 3)^2} - 2(1 - 2a)
=a2−6a+9−2+4a= {a^2} - 6a + 9 - 2 + 4a
=a2−2a+7= {a^2} - 2a + 7
=(a−1)2+6= {(a - 1)^2} + 6

So,

α2+β2≥6{\alpha ^2} + {\beta ^2} \ge 6
Q64
Let λ≠0\lambda \ne 0 be a real number. Let α,β\alpha,\beta be the roots of the equation 14x2−31x+3λ=014{x^2} - 31x + 3\lambda = 0 and α,γ\alpha,\gamma be the roots of the equation 35x2−53x+4λ=035{x^2} - 53x + 4\lambda = 0. Then 3αβ{{3\alpha } \over \beta } and 4αγ{{4\alpha } \over \gamma } are the roots of the equation
A 7x2−245x+250=07{x^2} - 245x + 250 = 0
B 49x2−245x+250=049{x^2} - 245x + 250 = 0
C 49x2+245x+250=049{x^2} + 245x + 250 = 0
D 7x2+245x−250=07{x^2} + 245x - 250 = 0
Correct Answer
Option B
Solution

14x2−31x+3λ=014 x^{2}-31 x+3 \lambda=0

α+β=3114….(1) and αβ=3λ14...(2)35x2−53x+4λ=0α+γ=5335…(3) and αγ=4λ35…(4)(2)(4)⇒βγ=3×354×14=158⇒β=158γ\begin{aligned} & \alpha+\beta=\frac{31}{14} \ldots .(1) \text{ and } \alpha \beta=\frac{3 \lambda}{14}\quad...(2) \\\\ & 35 x^{2}-53 x+4 \lambda=0 \\\\ & \alpha+\gamma=\frac{53}{35} \ldots(3) \text{ and } \alpha \gamma=\frac{4 \lambda}{35} \quad\ldots(4) \\\\ & \frac{(2)}{(4)} \Rightarrow \frac{\beta}{\gamma}=\frac{3 \times 35}{4 \times 14}=\frac{15}{8} \Rightarrow \beta=\frac{15}{8} \gamma \end{aligned}

(1) −(3)⇒β−γ=3114−5335=155−10670=710-(3) \Rightarrow \beta-\gamma=\dfrac{31}{14}-\dfrac{53}{35}=\dfrac{155-106}{70}=\dfrac{7}{10} 158γ−γ=710⇒γ=45\dfrac{15}{8} \gamma-\gamma=\dfrac{7}{10} \Rightarrow \gamma=\dfrac{4}{5} ⇒β=158×45=32\Rightarrow \beta=\dfrac{15}{8} \times \dfrac{4}{5}=\dfrac{3}{2} ⇒α=3114−β=3114−32=57\Rightarrow \alpha=\dfrac{31}{14}-\beta=\dfrac{31}{14}-\dfrac{3}{2}=\dfrac{5}{7} ⇒λ=143αβ=143×57×32=5\Rightarrow \lambda=\dfrac{14}{3} \alpha \beta=\dfrac{14}{3} \times \dfrac{5}{7} \times \dfrac{3}{2}=5 so, sum of roots 3αβ+4αγ=(3αγ+4αββγ)\dfrac{3 \alpha}{\beta}+\dfrac{4 \alpha}{\gamma}=\left(\dfrac{3 \alpha \gamma+4 \alpha \beta}{\beta \gamma}\right)

=(3×4λ35+4×3λ14)βγ=12λ(14+35)14×35βγ=49×12×5490×32×45=5\begin{aligned} & =\frac{\left(3 \times \frac{4 \lambda}{35}+4 \times \frac{3 \lambda}{14}\right)}{\beta \gamma}=\frac{12 \lambda(14+35)}{14 \times 35 \beta \gamma} \\\\ & =\frac{49 \times 12 \times 5}{490 \times \frac{3}{2} \times \frac{4}{5}}=5 \end{aligned}

Product of roots

=3αβ×4αγ=12α2βγ=12×254932×45=25049=\frac{3 \alpha}{\beta} \times \frac{4 \alpha}{\gamma}=\frac{12 \alpha^{2}}{\beta \gamma}=\frac{12 \times \frac{25}{49}}{\frac{3}{2} \times \frac{4}{5}}=\frac{250}{49}

So, required equation is x2−5x+25049=0x^{2}-5 x+\dfrac{250}{49}=0 ⇒49x2−245x+250=0\Rightarrow 49 \mathrm{x}^{2}-245 \mathrm{x}+250=0

Q65
 Let S={x∈[−6,3]−{−2,2}:∣x+3∣−1∣x∣−2≥0} and  \text{ Let } S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\} \text{ and } T={x∈Z:x2−7∣x∣+9≤0}. T=\left\{x \in \mathbb{Z}: x^{2}-7|x|+9 \leq 0\right\} \text{. } Then the number of elements in S∩T\mathrm{S} \cap \mathrm{T} is :
A 7
B 5
C 4
D 3
Correct Answer
Option D
Solution
∣x2∣−7∣x∣+9≤0|{x^2}| - 7|x| + 9 \le 0
⇒∣x∣∈[7−132,7+132]\Rightarrow |x| \in \left[ {{{7 - \sqrt {13} } \over 2},{{7 + \sqrt {13} } \over 2}} \right]

As

x∈Zx \in Z

So, x can be

± 2,± 3,± 4,± 5\pm \,2, \pm \,3, \pm \,4, \pm \,5

Out of these values of x,

x=3,−4,−5x = 3, - 4, - 5

satisfy S as well

n(S∩T)=3n(S \cap T) = 3
Q66
Let α\alpha, β\beta be the roots of the equation x2−2x+6=0x^{2}-\sqrt{2} x+\sqrt{6}=0 and 1α2+1,1β2+1\dfrac{1}{\alpha^{2}}+1, \dfrac{1}{\beta^{2}}+1 be the roots of the equation x2+ax+b=0x^{2}+a x+b=0. Then the roots of the equation x2−(a+b−2)x+(a+b+2)=0x^{2}-(a+b-2) x+(a+b+2)=0 are :
A non-real complex numbers
B real and both negative
C real and both positive
D real and exactly one of them is positive
Correct Answer
Option B
Solution
α+β=2\alpha + \beta = \sqrt 2

,

αβ=6\alpha \beta = \sqrt 6
1α2+1+1β2+1=2+α2+β26{1 \over {{\alpha ^2}}} + 1 + {1 \over {{\beta ^2}}} + 1 = 2 + {{{\alpha ^2} + {\beta ^2}} \over 6}
=2+2−266=−a= 2 + {{2 - 2\sqrt 6 } \over 6} = - a
(1α2+1)(1β2+1)=1+1α2+1β2+1α2β2\left( {{1 \over {{\alpha ^2}}} + 1} \right)\left( {{1 \over {{\beta ^2}}} + 1} \right) = 1 + {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} + {1 \over {{\alpha ^2}{\beta ^2}}}
=76+2−266=b= {7 \over 6} + {{2 - 2\sqrt 6 } \over 6} = b
⇒a+b=−56\Rightarrow a + b = {{ - 5} \over 6}

So, equation is

x2+17x6+76=0{x^2} + {{17x} \over 6} + {7 \over 6} = 0

OR

6x2+17x+7=06{x^2} + 17x + 7 = 0

Both roots of equation are −-ve and distinct

Q67
If 1(20−a)(40−a)+1(40−a)(60−a)+…+1(180−a)(200−a)=1256\dfrac{1}{(20-a)(40-a)}+\dfrac{1}{(40-a)(60-a)}+\ldots+\dfrac{1}{(180-a)(200-a)}=\dfrac{1}{256}, then the maximum value of a\mathrm{a} is :
A 198
B 202
C 212
D 218
Correct Answer
Option C
Solution
120(120−a−140−a+140−a−160−a + .... + 1180−a−1200−a)=1256{1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {40 - a}} + {1 \over {40 - a}} - {1 \over {60 - a}}\, + \,....\, + \,{1 \over {180 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}
⇒120(120−a−1200−a)=1256\Rightarrow {1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}
⇒120(180(20−a)(200−a))=1256\Rightarrow {1 \over {20}}\left( {{{180} \over {(20 - a)(200 - a)}}} \right) = {1 \over {256}}
⇒(20−a)(200−a)=9.256\Rightarrow (20 - a)(200 - a) = 9.256

OR

a2−220a+1696=0{a^2} - 220a + 1696 = 0
⇒a=212, 8\Rightarrow a = 212,\,8
Q68
The number of integral values of k, for which one root of the equation 2x2−8x+k=02x^2-8x+k=0 lies in the interval (1, 2) and its other root lies in the interval (2, 3), is :
A 2
B 0
C 1
D 3
Correct Answer
Option C
Solution
f(1)>0⇒k>6f(2)<0⇒k<8f(3)>0⇒k>6k∈(6,8)\begin{aligned} & f(1)>0 \Rightarrow k>6 \\\\ & f(2)<0 \Rightarrow k<8 \\\\ & f(3)>0 \Rightarrow k>6 \\\\ & k \in(6,8) \end{aligned}

Only 1 integral value of kk is 7.

Q69
The equation e4x+8e3x+13e2x−8ex+1=0,x∈R\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R} has :
A two solutions and both are negative
B two solutions and only one of them is negative
C four solutions two of which are negative
D no solution
Correct Answer
Option A
Solution

e4x+8e3x+13e2x−8ex+1=0e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0 Let ex=t\mathrm{e}^{\mathrm{x}}=\mathrm{t} Now, t4+8t3+13t2−8t+1=0\mathrm{t}^{4}+8 \mathrm{t}^{3}+13 \mathrm{t}^{2}-8 \mathrm{t}+1=0 Dividing equation by t2\mathrm{t}^{2}

t2+8t+13−8t+1t2=0t2+1t2+8(t−1t)+13=0(t−1t)2+2+8(t−1t)+13=0\begin{aligned} & t^{2}+8 t+13-\frac{8}{t}+\frac{1}{t^{2}}=0 \\\\ & t^{2}+\frac{1}{t^{2}}+8\left(t-\frac{1}{t}\right)+13=0 \\\\ & \left(t-\frac{1}{t}\right)^{2}+2+8\left(t-\frac{1}{t}\right)+13=0 \end{aligned}

Let t−1t=z\mathrm{t}-\dfrac{1}{\mathrm{t}}=\mathrm{z}

z2+8z+15=0z^{2}+8 z+15=0
(z+3)(z+5)=0\begin{aligned} & (z+3)(z+5)=0 \end{aligned}
z=−3 or z=−5z=-3 \text{ or } z=-5

So, t−1t=−3\mathrm{t}-\dfrac{1}{\mathrm{t}}=-3 or t−1t=−5\mathrm{t}-\dfrac{1}{\mathrm{t}}=-5 t2+3t−1=0t^{2}+3 t-1=0 or t2+5t−1=0t^{2}+5 t-1=0 t=−3±132\mathrm{t}=\dfrac{-3 \pm \sqrt{13}}{2} or t=−5±292\mathrm{t}=\dfrac{-5 \pm \sqrt{29}}{2} as t=ext=e^{x} so tt must be positive,

t=13−32 or 29−52t=\frac{\sqrt{13}-3}{2} \text{ or } \frac{\sqrt{29}-5}{2}

So, x=ln⁡(13−32)x=\ln \left(\dfrac{\sqrt{13}-3}{2}\right) or x=ln⁡(29−52)x=\ln \left(\dfrac{\sqrt{29}-5}{2}\right) Hence two solutions and both are negative.

Q70
Let S={x:x∈R and (3+2)x2−4+(3−2)x2−4=10}S = \left\{ {x:x \in \mathbb{R}\,\mathrm{and}\,{{(\sqrt 3 + \sqrt 2 )}^{{x^2} - 4}} + {{(\sqrt 3 - \sqrt 2 )}^{{x^2} - 4}} = 10} \right\}. Then n(S)n(S) is equal to
A 6
B 4
C 0
D 2
Correct Answer
Option B
Solution

Let (3+2)x2−4=t(\sqrt{3}+\sqrt{2})^{x^{2}-4}=t

t+1t=10⇒t2−10t+1=0⇒t=10±100−42=5±26\begin{aligned} & t+\frac{1}{t}=10 \\\\ \Rightarrow & t^{2}-10 t+1=0 \\\\ \Rightarrow & t=\frac{10 \pm \sqrt{100-4}}{2}=5 \pm 2 \sqrt{6} \end{aligned}

Case-I

t=5+26=(3+2)2⇒(3+2)x2−4=(3+2)2⇒x2−4=2⇒x2=6⇒x=±6\begin{aligned} & t=5+2 \sqrt{6} = (\sqrt{3}+\sqrt{2})^{2} \\\\ \Rightarrow & (\sqrt{3}+\sqrt{2})^{x^{2}-4}=(\sqrt{3}+\sqrt{2})^{2} \\\\ \Rightarrow & x^{2}-4=2 \Rightarrow x^{2}=6 \Rightarrow x=\pm \sqrt{6} \end{aligned}

Case-II : t=5−26t=5-2 \sqrt{6} = (3−2)2(\sqrt{3}-\sqrt{2})^{2} (3+2)x2−4=(3−2)2(\sqrt{3}+\sqrt{2})^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2} ⇒((3−2)−1)x2−4=(3−2)2\Rightarrow\left((\sqrt{3}-\sqrt{2})^{-1}\right)^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2} ⇒4−x2=2\Rightarrow 4-x^{2}=2 ⇒x2=2\Rightarrow x^{2}=2 ⇒x=±2\Rightarrow x=\pm \sqrt{2}

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