Free Objective Test 01 Practice Test - 11th and 12th
Question 1
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
b
-b
b1n+1
-b1n+1
SOLUTION
Solution : B
Let α, αn be two roots,
Then α+αn = -ba, α αn = ca
Eliminating α, we get (ca)1n+1 + (ca)nn+1 = -ba
⇒ a.a−1n+1. c1n+1 + a.a−nn+1. cnn+1 = -b
or (anc)1n+1 + (acn)1n+1 = -b
Question 2
If one root of the equation ax2 + bx + c = 0 be n times the other root, then
na2 = bc(n+1)2
nb2 = ac(n+1)2
nc2 = ab(n+1)2
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.
Question 3
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
3, 10
- 3, - 10
- 5, - 18
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.
Question 4
The values of a for which( a2−1)x2 + 2(a-1)x + 2 is positive for any x are
a ≥ 1
a ≤ 1
a > -3
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
∴ (a2−1)x2 + 2(a-1)x + 2 is positive for all x, if a2−1 > 0 and 4(a−1)2−8(a2−1) < 0
⇒ a2−1 > 0 and -4(a - 1)(a + 3) < 0
⇒ a2−1 > 0 and (a - 1)(a + 3) > 0
⇒ a2 > 1 and a < -3 or a > 1
⇒ a < -3 or a > 1
Question 5
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
Real and unequal
Real and equal
Imaginary
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 p2−4q2 < 0
Now discriminant of (ii) is
16q2−4p2=−4(p2−4q2) > 0
Hence, roots are real and unequal.
Question 6
If x is real and k=x2−x+1x2+x+1, then
13≤k≤3
k ≥ 5
k ≤ 0
k ≤ 4
SOLUTION
Solution : A
From k=x2−x+1x2+x+1
We have x2(k−1)+x(k+1)+k−1=0
As given, x is real ⇒ (k+1)2−4(k−1)2≥0
⇒ 3k2−10k+3≤0
Which is possible only when the value of k lies between the roots of the equation 3k2−10k+3=0
That is, when 13≤k≤3 {Since roots are 13 and 3}
Question 7
If ax2+bx+c=0, then x =
b±√b2−4ac2a
−b±√b2−ac2a
2c−b±√b2−4ac
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±√b2−4ac2a
Question 8
If the sum of two of the roots of x3+px2+qx+r=0 is zero, then pq =
- r
r
2 r
- 2 r
SOLUTION
Solution : B
Given that, α + β = 0
α + β + γ = -p ⇒ γ = -p
Substituting γ = -p in the given equation
⇒ −p3+p3−pq+r=0 ⇒ pq =r
Question 9
If α, β, γ are the roots of the equation x3+4x+1=0,then (α+β)−1+(β+γ)−1+(γ+α)−1=
2
3
4
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
=1−γ+1−α+1−β
=−(αβ+βγ+γααβγ)
=−(4−1)=4
Question 10
If α, β and γ are the roots of x3+8=0, then the equation whose roots are α2,β2 and γ2 is
x3−8=0
x3−16=0
x3+64=0
x3−64=0
SOLUTION
Solution : D
Let y=x2. then x=√y
∴ x3+8=0 ⇒ y32+8=0
⇒ y3=64 ⇒ y3−64=0
Thus the equation having roots α2,β2 and γ2 is x3−64=0
Question 11
If two roots of the equation x3−3x+2=0 are same, then the roots will be
2, 2, 3
1, 1, -2
- 2, 3, 3
-2, -2, 1
SOLUTION
Solution : B
Given equation is x3−3x+2=0
⇒ x2(x−1)+x(x−1)−2(x−1)=0
⇒ (x−1)(x2+x−2)=0 ⇒ (x−1)(x−1)(x+2)=0
Hence roots are 1, 1, -2
Question 12
One root of the following given equation 2x5−14x4+31x3−64x2+19x+130=0 is
1
3
5
7
SOLUTION
Solution : C
We can solve this problem using options given -
On putting x = 5, we get2(5)5−14(5)4+31(5)3−64(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.
Question 13
If the roots of the equation 8x3−14x2+7x−1=0 are in G.P., then the roots are
1, 12, 14
2, 4, 8
3, 6, 12
1,2,-4
SOLUTION
Solution : A
Let the roots be αβ,α,αβ,β≠0
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 1,12, 14Alternate method: By inspecting through options, we get the numbers 1,12, 14 satisfying the given equation.
Question 14
If x2+2ax+10−3a>0 for all x ∈ R, then
-5 < a < 2
a < -5
a > 5
2 < a < 5
SOLUTION
Solution : A
According to given condition,
4a2−4(10−3a)<0 ⇒ a2+3a−10<0
⇒ (a+5)(a−2)<0⇒−5<a<2.
Question 15
The set of all real numbers x for which x2−|x+2|+x>0, is
(-∞, -2)∪ (2, ∞)
(-∞, -√2)∪(√2, ∞)
(-∞, -1)∪(1, ∞)
(√2, ∞)
SOLUTION
Solution : B
Case 1: When x + 2 ≥ 0 i.e/ x ≥ -2,
Then given inequality becomes
x2−(x+2)+x>0 ⇒ x2−2>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, ∞)