# Free Objective Test 01 Practice Test - 11th and 12th

If one root of the quadratic equation ax2 + bx + c = 0  is equal to the nth power of the other root, then the value of (acn)1n+1 + (anc)1n+1

A.

b

B.

-b

C.

b1n+1

D.

-b1n+1

#### SOLUTION

Solution : B

Let α, αn be two roots,

Then α+αn = -ba, α αnca

Eliminating α, we get (ca)1n+1 + (ca)nn+1 = -ba

a.a1n+1c1n+1a.ann+1cnn+1 = -b

or (anc)1n+1 + (acn)1n+1 = -b

If one root of the equation ax2 + bx + c = 0 be n times the other root, then

A.

na2 = bc(n+1)2

B.

nb2 = ac(n+1)2

C.

nc2 = ab(n+1)2

D.

nb3 = ac(n+1)2

#### SOLUTION

Solution : B

Let the roots be α and n α

Sum of roots, α + nα = -ba ⇒ α = -ba(n+1)    ......(i)

and product, α.n.α = ca ⇒ α2 = cna     ......(ii)

From (i) and (ii), we get

[ -ba(n+1) ]2 = cna ⇒ b2a2(n+1)2 = cna

nb2 = ac(n+1)2.

Note : Students should remember this question as a fact.

The coefficient of x in the equation x2 + px + q = 0 was taken as 17 in place of 13, its roots were found to be -2 and -15, the roots of the original equation are

A.

3, 10

B.

- 3, - 10

C.

- 5, - 18

D.

18,5

#### SOLUTION

Solution : B

Let the equation (in written form) be x2 + 17x + q = 0.
Roots are -2, -15.
So q = 30,
And correct equation is x2 + 13x + 30 = 0.
Hence roots are -3, -10.

The values of a for which( a21)x2 + 2(a-1)x + 2 is positive for any x are

A.

a ≥ 1

B.

a ≤ 1

C.

a > -3

D.

a < -3 or a > 1

#### SOLUTION

Solution : D

We know that the expression ax2 + bx + c > 0 for all x, if a > 0 and b2 < 4ac

(a21)x2 + 2(a-1)x + 2 is positive for all x, if a21 > 0 and 4(a1)28(a21) < 0

a21 > 0 and -4(a - 1)(a + 3) < 0

a21 > 0 and (a - 1)(a + 3) > 0

a2 > 1 and a < -3 or a > 1

⇒ a < -3 or a > 1

If the roots of the equation qx2 + px + q = 0 where p, q are real, are complex, then the roots of the equation x2 - 4qx + p2 = 0 are

A.

Real and unequal

B.

Real and equal

C.

Imaginary

D.

nothing can be said in particular

#### SOLUTION

Solution : A

The given equations are

qx2 + px + q = 0    ........(i)

and x2 - 4qx + p2 = 0    ........(ii)

Roots of (i) are complex, therefore p24q2 < 0

Now discriminant of (ii) is

16q24p2=4(p24q2) > 0

Hence, roots are real and unequal.

If x is real and k=x2x+1x2+x+1, then

A.

13k3

B.

k ≥ 5

C.

k ≤ 0

D.

k 4

#### SOLUTION

Solution : A

From k=x2x+1x2+x+1

We have x2(k1)+x(k+1)+k1=0

As given, x is real  ⇒ (k+1)24(k1)20

3k210k+30

Which is possible only when the value of k lies between the roots of the equation 3k210k+3=0

That is, when 13k3 {Since roots are 13 and 3}

If ax2+bx+c=0, then x =

A.

b±b24ac2a

B.

b±b2ac2a

C.

2cb±b24ac

D.

None of these

#### SOLUTION

Solution : D

Try making perfect square the given equation by adding and subtracting  b24a2
on solving it we'll find
x=b±b24ac2a

If the sum of two of the roots of x3+px2+qx+r=0 is zero, then pq =

A.

- r

B.

r

C.

2 r

D.

- 2 r

#### SOLUTION

Solution : B

Given that, α + β = 0

α + β + γ = -p ⇒ γ = -p

Substituting γ = -p in the given equation

p3+p3pq+r=0 ⇒ pq =r

If α, β, γ are the roots of the equation x3+4x+1=0,then (α+β)1+(β+γ)1+(γ+α)1=

A.

2

B.

3

C.

4

D.

5

#### SOLUTION

Solution : C

If α,β, γ are the roots of the equation

α+β+γ=0,αβ+βγ+γα=4,αβγ=1

therefore(α+β)1+(β+γ)1+(γ+α)1
=1γ+1α+1β
=(αβ+βγ+γααβγ)
=(41)=4

If α, β and γ are the roots of x3+8=0, then the equation whose roots are α2,β2 and γ2 is

A.

x38=0

B.

x316=0

C.

x3+64=0

D.

x364=0

#### SOLUTION

Solution : D

Let y=x2. then x=y

x3+8=0y32+8=0

y3=64 ⇒ y364=0

Thus the equation having roots α2,β2 and γ2 is x364=0

If two roots of the equation x33x+2=0 are same, then the roots will be

A.

2, 2, 3

B.

1, 1, -2

C.

- 2, 3, 3

D.

-2, -2, 1

#### SOLUTION

Solution : B

Given equation is x33x+2=0

x2(x1)+x(x1)2(x1)=0

(x1)(x2+x2)=0 ⇒ (x1)(x1)(x+2)=0

Hence roots are 1, 1, -2

One root of the following given equation 2x514x4+31x364x2+19x+130=0 is

A.

1

B.

3

C.

5

D.

7

#### SOLUTION

Solution : C

We can solve this problem using options given -
On putting x = 5, we get

2(5)514(5)4+31(5)364(5)2+19(5)+130=0

Hence x = 5 satisfies the given equation.

Thus 5 is a root of the equation.
Other options will not satisfy the given equation so the answer is x = 5.

If the roots of the equation 8x314x2+7x1=0 are in G.P., then the roots are

A.

1, 1214

B.

2, 4, 8

C.

3, 6, 12

D.

1,2,-4

#### SOLUTION

Solution : A

Let the roots be αβ,α,αβ,β0
Then the product of roots is α3=18=18
⇒ α=12 and hence β = 12
(Since, sum of roots =αβ+α+αβ=14/8),
so roots are 1,1214

Alternate method: By inspecting through options, we get the numbers 1,1214 satisfying the given equation.

If x2+2ax+103a>0 for all x ∈ R, then

A.

-5 < a < 2

B.

a < -5

C.

a > 5

D.

2 < a < 5

#### SOLUTION

Solution : A

According to given condition,

4a24(103a)<0  ⇒ a2+3a10<0

(a+5)(a2)<05<a<2.

The set of all real numbers x for which x2|x+2|+x>0, is

A.

(-∞, -2)∪ (2, ∞)

B.

(-∞, -2)∪(2, ∞)

C.

(-∞, -1)∪(1, ∞)

D.

(2, ∞)

#### SOLUTION

Solution : B

Case 1: When x + 2 ≥ 0 i.e/ x ≥ -2,

Then given inequality becomes

x2(x+2)+x>0  ⇒ x22>0  ⇒ |x|>2

⇒ therefore, in this case the part of the solution set is

[-2,-2) ∪ (2, ∞).

Case 2: When x + 2 < 0 i.e. x < -2,

Then given inequality becomes x2+(x+2)+x>0

x2+2x+2>0 ⇒ (x+1)2+1>0, which is true for all real x

Hence, the part of the solution set in this case is (-∞, -2). Combining the two cases, the solution set is (-∞, -2)∪[-2, -2)∪(2,)=(,2)∪(2, ∞)