Practice Start Test

Quadratic Equation and Inequalities

JEE Mathematics · 144 questions · Page 2 of 15 · Click an option or "Show Solution" to reveal answer

Q11
If pp and qq are the roots of the equation x2+px+q=0,{x^2} + px + q = 0, then
A p=1,  q=−2p = 1,\,\,q = - 2
B p=0,  q=1p = 0,\,\,q = 1
C p=−2,  q=0p = - 2,\,\,q = 0
D p=−2,  q=1p = - 2,\,\,q = 1
Correct Answer
Option A
Solution
p+q=−pp + q = - p

and

pq=q⇒q(p−1)=0pq = q \Rightarrow q\left( {p - 1} \right) = 0
⇒q=0\Rightarrow q = 0

or

p=1.p=1.

If

q=0,q = 0,

then

p=0.p=0.

i.e.

p=qp=q

∴\therefore

p=1p=1

and

q=−2.q=-2.
Q12
The value of 'aa' for which one root of the quadratic equation (a2−5a+3)x2+(3a−1)x+2=0\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0$ is twice as large as the other is
A −13 - {1 \over 3}
B 23 {2 \over 3}
C −23 - {2 \over 3}
D 13 {1 \over 3}
Correct Answer
Option B
Solution

Let the roots of given equation be α\alpha and

22

α\alpha then

α+2α=3α=1−3aa2−5a+3\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}

and

α.2α=2α2=2a2−5a+3\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}
⇒α=1−3a3(a2−5a+3)\Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}

∴\therefore

2[19(1−3a)2(a2−5a+3)2]2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]
=2a2−5a+3= {2 \over {{a^2} - 5a + 3}}
(1−3a)2(a2−5a+3)=9{{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9

or

9a2−6a+19{a^2} - 6a + 1
=9a2−45a+27= 9{a^2} - 45a + 27

or

39a=2639a = 26

or

a=23a = {2 \over 3}
Q13
If one root of the equation x2+px+12=0{x^2} + px + 12 = 0 is 4, while the equation x2+px+q=0{x^2} + px + q = 0 has equal roots, then the value of ′q′'q' is
A 4
B 12
C 3
D 494{{49} \over 4}
Correct Answer
Option D
Solution
44

is a root of

x2+px+12=0{x^2} + px + 12 = 0
⇒16+4p+12=0\Rightarrow 16 + 4p + 12 = 0
⇒p=−7\Rightarrow p = - 7

Now, the equation

x2+px+q=0{x^2} + px + q = 0

has equal roots. ∴\therefore

p2−4q=0{p^2} - 4q = 0
⇒q=p24=494\Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}
Q14
The number of real solutions of the equation x2−3∣x∣+2=0{x^2} - 3\left| x \right| + 2 = 0 is
A 3
B 2
C 4
D 1
Correct Answer
Option C
Solution
x2−3∣x∣+2=0{x^2} - 3\left| x \right| + 2 = 0
⇒∣x∣2−3∣x∣+2=0\Rightarrow {\left| x \right|^2} - 3\left| x \right| + 2 = 0
(∣x∣−2)(∣x∣−1)=0\left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 1} \right) = 0
∣x∣=1,2\left| x \right| = 1,2

or

x=±1,±2x = \pm 1, \pm 2

∴\therefore No. of solution

=4=4
Q15
If both the roots of the quadratic equation x2−2kx+k2+k−5=0{x^2} - 2kx + {k^2} + k - 5 = 0 are less than 5, then kk lies in the interval
A (5,6]\left( {5,6} \right]
B (6, ∞)\left( {6,\,\infty } \right)
C (−∞, 4)\left( { - \infty ,\,4} \right)
D [4, 5]\left[ {4,\,5} \right]
Correct Answer
Option C
Solution

both roots are less than

5,5,

then

(i)(i)

Discriminant

≥0\ge 0
(ii)                p(5)>0\left( {ii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p\left( 5 \right) > 0
(iii)               Sum  of  roots2<5\left( {iii} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2} < 5

Hence

(i)      4k2−4(k2+k−5)≥0\left( i \right)\,\,\,\,\,\,4{k^2} - 4\left( {{k^2} + k - 5} \right) \ge 0
4k2−4k2−4k+20≥04{k^2} - 4{k^2} - 4k + 20 \ge 0
4k≤20⇒k≤54k \le 20 \Rightarrow k \le 5
(ii)     f(5)>0;25−10k+k2+k−5>0\left( {ii} \right)\,\,\,\,\,f\left( 5 \right) > 0;25 - 10k + {k^2} + k - 5 > 0

or

k2−9k+20>0{k^2} - 9k + 20 > 0

or

k(k−4)−5(k−4)>0k\left( {k - 4} \right) - 5\left( {k - 4} \right) > 0

or

(k−5)(k−4)>0\left( {k - 5} \right)\left( {k - 4} \right) > 0
⇒k∈(−∞,4)∪(−∞,5)\Rightarrow k \in \left( { - \infty ,4} \right) \cup \left( { - \infty ,5} \right)
(iii)      Sum  of  roots2\left( {iii} \right)\,\,\,\,\,\,{{Sum\,\,of\,\,roots} \over 2}
=−b2a=2k2<5= - {b \over {2a}} = {{2k} \over 2} < 5

The intersection of

(i)(i)

,

(ii)(ii)

&

(iii)(iii)

gives

k∈(−∞,4).k \in \left( { - \infty ,4} \right).
Q16
If the roots of the quadratic equation x2+px+q=0{x^2} + px + q = 0 are tan⁡30∘\tan {30^ \circ } and tan⁡15∘\tan {15^ \circ }, respectively, then the value of 2+q−p2 + q - p is
A 2
B 3
C 0
D 1
Correct Answer
Option B
Solution
x2+px+q=0{x^2} + px + q = 0

Sum of roots

=tan⁡30∘+tan⁡15∘=−p= \tan {30^ \circ } + \tan {15^ \circ } = - p

Products of roots

=tan⁡30∘.tan⁡15∘=q= \tan {30^ \circ }.\tan {15^ \circ } = q
tan⁡45∘=tan⁡30∘+tan⁡15∘1−tan⁡30∘.tan⁡15∘\tan {45^ \circ } = {{\tan {{30}^ \circ } + \tan {{15}^ \circ }} \over {1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}}
                    =−p1−q=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{ - p} \over {1 - q}} = 1
⇒−p=1−q⇒q−p=1\Rightarrow - p = 1 - q \Rightarrow q - p = 1

∴\therefore

2+q−p=32 + q - p = 3
Q17
If the difference between the roots of the equation x2+ax+1=0{x^2} + ax + 1 = 0 is less than 5,\sqrt 5 , then the set of possible values of aa is
A (3,∞)\left( {3,\infty } \right)
B (−∞,−3)\left( { - \infty , - 3} \right)
C (−3,3)\left( { - 3,3} \right)
D (−3,∞)\left( { - 3,\infty } \right)
Correct Answer
Option C
Solution

Let α\alpha and β\beta are roots of the equation

x2+ax+1=0{x^2} + ax + 1 = 0

So,

α+β=−a\alpha + \beta = - a

and

αβ=1\alpha \beta = 1

given

∣α−β∣<5\left| {\alpha - \beta } \right| < \sqrt 5
⇒(α−β)2−4αβ<5\Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5

(as

(α−β)2=(α+β)2−4αβ{\left( {\alpha - \beta } \right)^2} = {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta

)

⇒a2−4<5\Rightarrow \sqrt {{a^2} - 4} < \sqrt 5
⇒a2−4<5\Rightarrow {a^2} - 4 < 5
⇒a2−9<0⇒a2<9\Rightarrow {a^2} - 9 < 0 \Rightarrow {a^2} < 9
⇒−3<a<3\Rightarrow - 3 < a < 3
⇒a∈(−3,3)\Rightarrow a \in \left( { - 3,3} \right)
Q18
If xx is real, the maximum value of 3x2+9x+173x2+9x+7{{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}} is
A 14{1 \over 4}
B 4141
C 11
D 177{17 \over 7}
Correct Answer
Option B
Solution
y=3x2+9x+173x2+9x+7y = {{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}
3x2(y−1)+9x(y−1)+7y−17=03{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0
D≥0D \ge 0

as

xx

is real

81(y−1)2−4×3(y−1)(7y−17)≥081{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \ge 0
⇒(y−1)(y−41)≤0\Rightarrow \left( {y - 1} \right)\left( {y - 41} \right) \le 0
⇒1≤y≤41\Rightarrow 1 \le y \le 41

∴\therefore Max value of

yy

is

4141
Q19
In a triangle PQR,    ∠R=π2.  If  tan⁡ (P2)PQR,\;\;\angle R = {\pi \over 2}.\,\,If\,\,\tan \,\left( {{P \over 2}} \right) and tan⁡(Q2) \tan \left( {{Q \over 2}} \right) are the roots of ax2+bx+c=0,  a≠0a{x^2} + bx + c = 0,\,\,a \ne 0 then
A a=b+ca = b + c
B c=a+bc = a + b
C b=cb = c
D b=a+cb = a + c
Correct Answer
Option B
Solution

∠\angleR = 90o ∴\therefore ∠\angleP + ∠\angleQ = 90o ⇒\Rightarrow

P2+Q2=902=45{P \over 2} + {Q \over 2} = {{90} \over 2} = 45

o

tan⁡(P2),tan⁡(Q2)\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)

are the roots of

ax2+bx+c=0a{x^2} + bx + c = 0

∴\therefore

tan⁡(P2)+tan⁡(Q2)=−ba,  \tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) = - {b \over a},\,\,

and

tan⁡(P2).tan⁡(Q2)=ca\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}
tan⁡(P2)+tan⁡(Q2)1−tan⁡(P2)tan⁡(Q2){{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}
=tan⁡(P2+Q2)= \tan \left( {{P \over 2} + {Q \over 2}} \right)

= tan 45o = 1

⇒−ba1−ca=1\Rightarrow {{ - {b \over a}} \over {1 - {c \over a}}} = 1
⇒−ba=aa−ca\Rightarrow - {b \over a} = {a \over a} - {c \over a}
⇒−b=a−c\Rightarrow - b = a - c

or

c=a+b.c = a + b.
Q20
If the roots of the equation bx2+cx+a=0b{x^2} + cx + a = 0 imaginary, then for all real values of xx, the expression 3b2x2+6bcx+2c23{b^2}{x^2} + 6bcx + 2{c^2} is :
A less than 4ab4ab
B greater than −4ab-4ab
C less than −4ab-4ab
D greater than 4ab4ab
Correct Answer
Option B
Solution

Given that roots of the equation

bx2+cx+a=0b{x^2} + cx + a = 0

are imaginary ∴\therefore

c2−4ab<0               ...(i){c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Let

y=3b2x2+6bc x+2c2y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}
⇒3b2x2+6bc x+2c2−y=0\Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0

As

xx

is real,

D≥0D \ge 0
⇒36b2c2−12b2(2c2−y)≥0\Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0
⇒12b2(3c2−2c2+y)≥0\Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0
⇒c2+y≥0\Rightarrow {c^2} + y \ge 0
⇒y≥−c2\Rightarrow y \ge - {c^2}

But from eqn.

(i),(i),
c2<4ab{c^2} < 4ab

or

−c2>−4ab- {c^2} > - 4ab

∴\therefore we get

y≥−c2>−4aby \ge - {c^2} > - 4ab
y>−4aby > - 4ab
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