Free Objective Test 02 Practice Test - 11th and 12th
Question 1
If x is real and satisfies x + 2 > √x+4, then
x < -2
x >0
-3 < x < 0
-3 < x < 4
SOLUTION
Solution : B
Given, x + 2 > √x+4 ⇒ (x+2)2>(x+4) but inequality sign will be remained same if both sides are positive or negative.
R.H.S is positive and so L.H.S will also be positive.
Thus x>-2⇒x2+4x+4>x+4 ⇒x2+3x>0
⇒ x(x+3)>0⇒ x<−3 or x>0 ⇒ x>0
Question 2
The number of real values of x for which the equality |3x2+12x+6|=5x+16 holds good is
4
3
2
1
SOLUTION
Solution : C
Equation |3x2+12x+6|=5x+16 ......(i)
when 3x2+12x+6≥0 ⇔ x2+4x≥−2
⇔ |x+2|2≥4−2⇔ |x+2|≥(√2)2
⇔ x + 2 ≤ - √2 or x + 2 ≥ √2.......(ii)
Then (i) becomes 3x2+12x+6=5x+16
⇔ 3x2+7x−10=0 ⇒ x = 1, -103
But x = -103 does not satisfy (ii)
When 3x2+12x+6<0 ⇒ x2+4x>−2
⇒|x+2|≤√2⇒−√2−2≤x≤−2+√2....(iii)
⇒ 3x2+17x+22=0 ⇒ x=−2,−113
But x = - 113 does not satisfy (iii). So, 1 and -2 are the only solutions.
Question 3
If x, y, z are real and distinct, then x2+4y2+9z2−6yz−3zx−2xy =
Non-negative
Non-positive
Zero
positive only
SOLUTION
Solution : A
x, y, z, ∈ R and distinct.
Now, u = x2+4y2+9z2−6yz−3zx−2xy
=12(2x2+8y2+18z2−12yz−6zx−4xy)
=12{(x2−4xy+4y2)+(x2−6zx+9z2)+(4y2−12yz+9z2)}
=12{(x−2y)2+(x−3z)2+(2y−3z)2}
Since it is sum of squares. So U is always non-negative.
Question 4
If x is real, the expression x+22x2+3x+6 takes all value in the interval
(113,13)
(−113,13)
(−13,113)
[−13,113]
SOLUTION
Solution : B
If the given expression be y, then
y=2x2y+(3y−1)x+(6y−2)=0
If y ≠ 0 then Δ ≥ 0 for real x i.e. B2−4AC≥0
or −39y2+10y+1≥0 or (13y+1)(3y−1)≤0
⇒ −113≤y≤13
if y = 0 then x = -2 which is real and this value of y is included in the above range.
Question 5
If x is real, then the maximum and minimum values of expression x2+14x+9x2+2x+3will be
4, - 5
5, - 4
- 4, 5
- 4, - 5
SOLUTION
Solution : A
Let y=x2+14x+9x2+2x+3
⇒ y(x2+2x+3)−x2−14x−9=0
⇒ (y−1)x2+(2y−14)x+3y−9=0
For real x, its discriminant ≥ 0
i.e. 4(y−7)2−4(y−1)3(y−3)≥0
⇒ y2+y−20≤0 or (y−4)(y+5)≤0
Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:
Case 1: y−4≥0 or y≥4 y+5≤0 or y≤−5
This is not possible.
Case 2: y−4≤0 or y≤4 y+5≥0 or y≥−5 Both of these are satisfied if -5≤y≤4
hence maximum value of y is 4 and minimum value is -5
Question 6
If x2−3x+2 be a factor of x4−px2+q, then (p, q) =
(3, 4)
(4, 5)
(4, 3)
(5, 4)
SOLUTION
Solution : D
x2−3x+2 be factor of x4−px2+q=0
Hence (x2−3x+2)=0 ⇒ (x-2)(x-1) = 0
⇒ x = 2, 1, putting these values in given equation
so, 4p - q -16 = 0 ......(i)
and, p - q - 1 = 0 .......(ii)
Solving (i) and (ii), we get (p, q) = (5, 4)
Question 7
x2−11x+a and x2−14x+2a will have a common factor, if a =
24
0,24
3,24
0,3
SOLUTION
Solution : B
Exprssions are x2−11x+a and x2−14x+2a will have a common factor, then
⇒ x2−22a+14a=xa−2a=1−14+11
⇒ x2−8a=x−a=1−3 ⇒ x2=8a3 and x=a3
⇒ (a3)2=8a3 ⇒ a29=8a3 ⇒ a = 0, 24.
Alternate Method : We can check by putting the values of a from the options.
Question 8
If x2+ax+10=0 and x2+bx−10=0 have a common root, then a2−b2 is equal to
10
20
30
40
SOLUTION
Solution : D
Let α be a common root, then
α2+αα+10=0
and α2+bα−10=0
from (i) - (ii),
(a−b)α+20=0 ⇒ α = −20a−b
Substituting the value of α in (i), we get
(−20a−b)2+a(−20a−b)+10=0
⇒400 - 20 a(a - b) + 10(a−b)2 = 0
⇒40 -2a2 + 2ab + a2 + b2 -2ab = 0
⇒a2 - b2 = 40
Question 9
If the equation x2+px+q=0 and x2+qx+p=0, have a common root, then p + q + 1 =
0
1
2
-1
SOLUTION
Solution : A
Let α is the common root, so α2+pα+q=0
and α2+qα+p=0
⇒ (p - q) α + (q - p) = 0 ⇒ α=1
Put the value of α , p + q + 1 = 0
Question 10
If ax2+bx+c=0 and bx2+cx+a=0 have a common root
a ≠ 0, then a3+b3+c3abc=
1
2
3
4
SOLUTION
Solution : C
The condition for common roots gives (bc−a2)2=(ca−b2)(ab−c2) On simplification gives a3+b3+c3=3abc
Question 11
The number of real roots of the equation esin x−e−sin x−4=0 are
1
2
Infinite
0
SOLUTION
Solution : D
Given equation esin x−e−sin x−4=0
Let esin x=y, then given equation can be written as
y2−4y−1=0 ⇒ y=2±√5
But the value of y=esin x is always positive, so
y=2+√5 (∵ 2 < √5
⇒logey=loge(2+√5) ⇒ sin x = loge(2+√5)>1
no solution
Question 12
The number of real solutions of the equation |x|2−3|x|+2=0 are
1
2
3
4
SOLUTION
Solution : D
Given |x|2−3|x|+2=0
Here we consider two cases viz.x < 0 and x > 0
Case 1: x < 0 This gives x2+3x+2=0
⇒ (x+2)(x+1)=0 ⇒ x=−2,−1
Also x=−1,−2 satisfy x<0, so x=−1,−2 is solution in this case
Case 2: x > 0. This gives x2−3x+2=0
⇒ (x-2)(x-1) = 0 ⇒ x =2,1. so, x=2,1 is solution in this case. Hence the number5 of solutions are four i.e. x = -1, -2, 1, 2
Aliter : |x|2−3|x|+2=0
⇒ (|x|-1)(|x|-2)=0
⇒ |x| = 1 and |x| = 2 ⇒ x = ±1, x = ±2.
Question 13
The solution set of the equation xlogx(1−x)2=9 is
{- 2, 4}
{4}
{0, - 2, 4}
None of these
SOLUTION
Solution : B
xlogx(1−x)2=9
⇒ logx(9)=logx(1−x)2(;∵ ax=N ⇒ logaN=x)
⇒ 9=(1−x2) ⇒ 1+x2−2x−9=0
⇒ x2−2x−8=0 ⇒ (x + 2)(x - 4) = 0 ⇒ x = -2, 4
But x>0; ∴x=4
Question 14
If x2+y2=25,xy=12,then complete set of x =
{3, 4}
{3, -3}
{3, 4, -3, -4}
{-3, -3}
SOLUTION
Solution : C
x2+y2=25 and xy=12
⇒ x2+(12x)2=25 ⇒ x4+144−25x2=0
⇒ (x2−16)(x2−9)=0 ⇒ x2=16 and x2=9
⇒ x= ±4 and x = ±3
Question 15
The roots of the equation √3x+1+1 = √x are
0
1
0,1
roots doesnot exist
SOLUTION
Solution : D
given equation is √3x+1+1 = √x
⇒ √3x+1 = √x - 1
Squaring both the sides, we get 3x + 1 = x + 1 -2√x
⇒ 2√x + 2x = 0 (irrational function)
Thus x ≠ 0 and x ≠ 1, Since equation is non-quadratic equation.