# Free Objective Test 03 Practice Test - 11th and 12th

If every pair of the equations x2+px+qr=0 , x2+qx+rp=0 , x2+rx+pq=0 have a common root, then the sum of three common roots is

A.

(p+q+r)2

B.

p+q+r2

C.

-(p+q+r)

D.

-p+q+r

#### SOLUTION

Solution : A

Let the roots be α, β; β, γ and γ, α respectively.

∴ α + β = -p, β + γ = -q, γ + α = -r

Adding all, we get ∑α = (p+q+r)2 etc.

If x2hx21=0 , x23hx+35=0 (h>0) has a common root, then the value of h is equal to

A.

1

B.

2

C.

3

D.

4

#### SOLUTION

Solution : D

Subtracting, we get 2hx = 56 or hx = 28

Putting in any, x2=49

[28h]2 = 72 ⇒  h = 4(h > 0)

If  the two equations x2cx+d=0 and x2ax+b=0 have one common root and the second has equal roots, then 2(b+d)=)

A.

0

B.

a+c

C.

ac

D.

-ac

#### SOLUTION

Solution : C

Let roots of x2cx+d=0 be α, β then roots of x2ax+b=0 be α, α

∴ α + β = c, αβ = d, α + α = α, α2 = b

Hence 2(b+d)=2(α2+αβ)=2α(α+β)=ac

If the ratio of the roots of the equation ax2+bx+c=0 be p:q, then

A.

pqb2+(p+q)2ac=0

B.

pqb2(p+q)2ac=0

C.

pqa2(p+q)2bc=0

D.

pqb2(pq)2ac=0

#### SOLUTION

Solution : B

Let pα,qα be the roots of the given equation ax2+bx+c=0

Then pα + qα = ba and pα.qα=ca

From first relation, α = ba(p+q)

Substituting this value of α in second relation, we get

b2a2(p+q)2×pq=ca ⇒ b2pqac(p+q)2=0

Note : Students should remember this question as a fact.

If α and β are the roots of the equation x2a(x+1)b=0, then (α+1)(β+1)=

A.

b

B.

-b

C.

1-b

D.

b-1

#### SOLUTION

Solution : C

Given equation x2a(x+1)b=0

x2axab ⇒ α + β = α, αβ = -(a + b)

Now (α+1)(β+1)=αβ+α+β+1=(a+b)+a+1=1b

If α and β are the roots of the equation 2x23x+4=0, then the equation whose roots are α2 and β2 is

A.

4x2+7x+16=0

B.

4x2+7x+6=0

C.

4x2+7x+1=0

D.

4x27x+16=0

#### SOLUTION

Solution : A

α + β = 32 and αβ = 2

α2+β2=(α+β)22αβ=944=74

hence required equation x2(α2+β2)x+α2β2=0

x2+74x+4=0 ⇒ 4x2+7x+16=0

If α and β are the roots of the equation x2+6x+λ=0 and 3α+2β=20, then λ =

A.

-8

B.

-16

C.

16

D.

8

#### SOLUTION

Solution : B

α + β = -6    .....(i)

αβ = λ       ......(ii)

and given 3α + 2β = -20    ......(iii)

Solving (i) and (iii), we get β = 2, α = -8

Substituting these values in (ii), we get λ = -16

If the sum of the roots of the equation λx2+2x+3λ=0  be equal to their product, then λ =

A.

4

B.

-4

C.

6

D.

23

#### SOLUTION

Solution : D

Under condition 2λ=3  ⇒ λ=23

If 2+i3 is a root of the equation x2+px+q=0, where p and q are real, then (p, q) =

A.

(-4,7)

B.

(4,7)

C.

(4,7)

D.

(-4,-7)

#### SOLUTION

Solution : A

Since 2+i3 is a root, therefore 2i3 will be other root. Now sum of the roots = 4 = -p and product of roots = 7 = q. Hence (p, q) = (-4, 7).

If α and β be the roots of the equation 2x2+2(a+b)x+a2+b2=0, then the equation whose roots are (α+β)2 and (αβ)2) is

A.

x22abx(a2b2)2=0

B.

x24abx(a2b2)2=0

C.

x24abx+(a2b2)2=0

D.

None of these

#### SOLUTION

Solution : B

Sum of roots α + β = -(a + b) and αβ = a2+b22

⇒ (α+β)2=(a+b)2 and (αβ)2=α2+β22αβ

=2ab(a2+b2)=(ab)2

Now the required equation whose roots are

(α+β)2 and (αβ)2

x2{(α+β)2+(αβ)2}x+(α+β)2(αβ)2=0

⇒ x2{(a+b)2+(ab)2}x+(a+b)2(ab)2=0

⇒ x24abx(a2b2)2=0

If a root of the equation ax2+bx+c=0 be reciprocal of the equation then ax2+bx+c=0, then

A.

(ccaa)2=(bacb)(abbc).

B.

(bbaa)2=(cabc)(abbc).

C.

(ccaa)2=(bacb)(abbc).

D.

(cc+aa)2=(bacb)(abbc).

#### SOLUTION

Solution : A

Let α be a root of first equation, then 1α be a root of second equation.

therefore aα2+bα+c=0 and  a1α2+b1α+c=0 or cα2+bα+d=0

Hence α2babc=αccaa=1abbc

(ccaa)2=(bacb)(abbc).

If α,βare the roots of the equation ax2+bx+c=0 then the equation whose roots are α+1β and β+1α, is

A.

acx2+(a+c)bx+(a+c)2=0

B.

abx2+(a+c)bx+(a+c)2=0

C.

acx2+(a+b)cx+(a+c)2=0

D.

None of these

#### SOLUTION

Solution : A

Here α+β= baand αβ=ca

If roots are α+1β,β+1α, then sum of roots are

=(α+1β)+(β+1alpha)=(α+β)+α+βαβ=bac(a+c)

and product =(α+1β)(β+1α)

=αβ+1+1+1αβ=2+ca+ac

=2ac+c2+a2ac=(a+c)2ac

Hence required equation is given by

x2+bac(a+c)x+(a+c)2ac=0

acx2+(a+c)bx+(a+c)2=0

Trick : Let a = 1, b = -3, c = 2,then α = 1,β = 2

α+1β=32 and β+1α=3

required equation must be

(x3)(2x3)=0 i.e. 2x29x+9=0

Here (a) gives this equation on putting

a = 1, b = -3, c = 2

If the roots of the equation ax2+bx+c=0 beα and β,, then the roots of the equation cx2+bx+a=0 are

A.

α,β

B.

α,1β

C.

1α, 1β

D.

None of these

#### SOLUTION

Solution : C

α,β are roots of ax2+bx+c=0

α+β= ba and αβ=ca

Let the roots of cx2+bx+a=0 be α',β', then

α ' + β '=bc and α β=ac

but α+βαβ=baca=bc1α+1β=α+β

Hence α=1α and β=1β

If α and β are the roots of the equation 4x2+3x+7=0, then 1α+1β=

A.

37

B.

37

C.

35

D.

35

#### SOLUTION

Solution : A

Given equation 4x2+3x+7=0,

α+β=34 and αβ=74

Now 1α+1β=α+βαβ=3474=34×47=37

If one root of 5x2+13x+k=0 is reciprocal of the other, then k=

A.

0

B.

5

C.

16

D.

6

#### SOLUTION

Solution : B

Let first root =α and second root =1α

Product of roots =1
k5=1k=5