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Quadratic Equation and Inequalities

JEE Mathematics · 144 questions · Page 1 of 15 · Click an option or "Show Solution" to reveal answer

Q1
The number of real roots of the equation x∣x−2∣+3∣x−3∣+1=0x |x - 2| + 3|x - 3| + 1 = 0 is :
A 4
B 3
C 2
D 1
Correct Answer
Option D
Solution
\begin{aligned} & \text{ (I) } x

\begin{aligned} & \text { (II) } 2 \leq x

x=−1+322,−1−322\mathrm{x}=\frac{-1+\sqrt{32}}{2}, \frac{-1-\sqrt{32}}{2}

××\quad \quad \times \qquad \qquad \times 1 real roots

Q2
The value of 4+15+14+15+14+......∞4 + {1 \over {5 + {1 \over {4 + {1 \over {5 + {1 \over {4 + ......\infty }}}}}}}} is :
A 2 + 2530{2 \over 5}\sqrt {30}
B 2 + 4530{4 \over {\sqrt 5 }}\sqrt {30}
C 5 + 2530{2 \over 5}\sqrt {30}
D 4 + 4530{4 \over {\sqrt 5 }}\sqrt {30}
Correct Answer
Option A
Solution
y=4+15+1yy = 4 + {1 \over {5 + {1 \over y}}}
⇒y=4+y5y+1\Rightarrow y = 4 + {y \over {5y + 1}}
⇒5y2−20y−4=0\Rightarrow 5{y^2} - 20y - 4 = 0
⇒y=20±400+8010\Rightarrow y = {{20 \pm \sqrt {400 + 80} } \over {10}}
⇒y=20±43010,y>0\Rightarrow y = {{20 \pm 4\sqrt {30} } \over {10}},y > 0

∴\therefore

y=10+2305y = {{10 + 2\sqrt {30} } \over 5}
Q3
The sum of the roots of the equation x+1−2log⁡2(3+2x)+2log⁡4(10−2−x)=0x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0, is :
A log2 14
B log2 11
C log2 12
D log2 13
Correct Answer
Option B
Solution
x+1−2log⁡2(3+2x)+2log⁡4(10−2−x)=0x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0
log⁡2(2x+1)−log⁡2(3+2x)2+log⁡2(10−2−x)=0{\log _2}({2^{x + 1}}) - {\log _2}{(3 + {2^x})^2} + {\log _2}(10 - {2^{ - x}}) = 0
log2(2x+1.(10−2−x)(3+2x)2)=0lo{g_2}\left( {{{{2^{x + 1}}.(10 - {2^{ - x}})} \over {{{(3 + {2^x})}^2}}}} \right) = 0
2(10.2−x−1)(3+2x)2=1{{2({{10.2}^{ - x}} - 1)} \over {{{(3 + {2^x})}^2}}} = 1
⇒20.2x−2=9+22x+6.2x\Rightarrow {20.2^x} - 2 = 9 + {2^{2x}} + {6.2^x}

∴\therefore

(2x)2−14(2x)+11=0{({2^x})^2} - 14({2^x}) + 11 = 0

Roots are 2x1 & 2x2 ∴\therefore 2x1 . 2x2 = 11 x1 + x2 = log2(11)

Q4
The number of integral values of m for which the quadratic expression, (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), x ∈ \in R, is always positive, is :
A 7
B 8
C 3
D 6
Correct Answer
Option A
Solution

Expression is always positive it 2m + 1 > 0 ⇒\Rightarrow m > −-

12{1 \over 2}

& D < 0 ⇒\Rightarrow m2 −- 6m −- 3 < 0 3 −-

12\sqrt {12}

< m < 3 +

12\sqrt {12}

. . . . (iii) ∴\therefore Common interval is 3 −-

12\sqrt {12}

< m < 3 +

12\sqrt {12}

∴\therefore Intgral value of m {0, 1, 2, 3, 4, 5, 6}

Q5
Product of real roots of equation t2x2+∣x∣+9=0{t^2}{x^2} + \left| x \right| + 9 = 0
A is always positive
B is always negative
C does not exist
D none of these
Correct Answer
Option A
Solution

Product of real roots

=9t2>0,∀ t ∈R= {9 \over {{t^2}}} > 0,\forall \,t\, \in R

∴\therefore Product of real roots is always positive.

Q6
The value of aa for which the sum of the squares of the roots of the equation x2−(a−2)x−a−1=0{x^2} - \left( {a - 2} \right)x - a - 1 = 0 assume the least value is :
A 11
B 00
C 33
D 22
Correct Answer
Option A
Solution
x2−(a−2)x−a−1=0{x^2} - \left( {a - 2} \right)x - a - 1 = 0
⇒α+β=a−2;  αβ=−(a+1)\Rightarrow \alpha + \beta = a - 2;\,\,\alpha \beta = - \left( {a + 1} \right)
α2+β2=(α+β)2−2αβ{\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta
=a2−2a+6=(a−1)2+5= {a^2} - 2a + 6 = {\left( {a - 1} \right)^2} + 5

For min. value of

α2+β2{\alpha ^2} + {\beta ^2}

where α\alpha is an integer

⇒  a=1.\Rightarrow \,\,a = 1.
Q7
Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation
A x2−18x−16=0{x^2} - 18x - 16 = 0
B x2−18x+16=0{x^2} - 18x + 16 = 0
C x2+18x−16=0{x^2} + 18x - 16 = 0
D x2+18x+16=0{x^2} + 18x + 16 = 0
Correct Answer
Option B
Solution

Let two numbers be a and b then

a+b2=9{{a + b} \over 2} = 9

and

ab=4\sqrt {ab} = 4

∴\therefore Equation with roots

aa

and

bb

is

x2−(a+b)x+ab=0{x^2} - \left( {a + b} \right)x + ab = 0
⇒x2−18x+16=0\Rightarrow {x^2} - 18x + 16 = 0
Q8
If the roots of the equation x2−bx+c=0{x^2} - bx + c = 0 be two consecutive integers, then b2−4c{b^2} - 4c equals
A −2-2
B 33
C 22
D 11
Correct Answer
Option D
Solution

Let n and (n + 1) be the roots of x2 −- bx + c = 0.

Then, n + (n + 1) = b and n(n + 1) = c ∴\therefore b2 −- 4c = (2n + 1)2 −- 4n(n + 1) = 4n2 + 4n + 1 −- 4n2 −- 4n = 1

Q9
If (1−p)\left( {1 - p} \right) is a root of quadratic equation x2+px+(1−p)=0{x^2} + px + \left( {1 - p} \right) = 0 then its root are
A −1,2 - 1,2
B −1,1 - 1,1
C 0,−1 0,-1
D 0,10,1
Correct Answer
Option C
Solution

Let the second root be

α.\alpha .

Then

α+(1−p)=−p⇒α=−1\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1

Also

α.(1−p)=1−p\alpha .\left( {1 - p} \right) = 1 - p
⇒(α−1)(1−p)=0\Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0
⇒p=1\Rightarrow p = 1

[as

α=−1\alpha = - 1

] ∴\therefore Roots are

α=−1\alpha = - 1

and

p−1=0p-1=0
Q10
If the roots of the equation x2−bx+c=0{x^2} - bx + c = 0 be two consecutive integers, then b2−4c{b^2} - 4c equals
A −2-2
B 33
C 22
D 11
Correct Answer
Option D
Solution

Let

α,  α+1  \alpha ,\,\,\alpha + 1\,\,

be roots Then

α+α+1=b=\alpha + \alpha + 1 = b =

sum of - roots

α(α+1)=c\alpha \left( {\alpha + 1} \right) = c

== product of roots ∴\therefore

b2−4c{b^2} - 4c
=(2α+1)2−4α(α+1)= {\left( {2\alpha + 1} \right)^2} - 4\alpha \left( {\alpha + 1} \right)
=1.= 1.
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