# Free Objective Test 02 Practice Test - 11th and 12th

If x is real and satisfies x + 2 > x+4, then

A.

x < -2

B.

x >0

C.

-3 < x < 0

D.

-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+4x2+3x>0

x(x+3)>0x<3 or x>0  ⇒ x>0

The number of real values of x for which the equality |3x2+12x+6|=5x+16 holds good is

A.

4

B.

3

C.

2

D.

1

#### SOLUTION

Solution : C

Equation |3x2+12x+6|=5x+16     ......(i)

when 3x2+12x+60 ⇔ x2+4x2

|x+2|242|x+2|(2)2

⇔ x + 2 ≤ - 2 or x + 2 ≥ 2.......(ii)

Then (i) becomes 3x2+12x+6=5x+16

3x2+7x10=0 ⇒ x = 1, -103

But x = -103 does not satisfy (ii)

When 3x2+12x+6<0 ⇒ x2+4x>2

|x+2|222x2+2....(iii)

3x2+17x+22=0 ⇒ x=2,113

But x = - 113 does not satisfy (iii). So, 1 and -2 are the only solutions.

If x, y, z are real and distinct, then x2+4y2+9z26yz3zx2xy

A.

Non-negative

B.

Non-positive

C.

Zero

D.

positive only

#### SOLUTION

Solution : A

x, y, z, ∈ R and distinct.

Now, u = x2+4y2+9z26yz3zx2xy

=12(2x2+8y2+18z212yz6zx4xy)

=12{(x24xy+4y2)+(x26zx+9z2)+(4y212yz+9z2)}

=12{(x2y)2+(x3z)2+(2y3z)2}

Since it is sum of squares. So U is always non-negative.

If x is real, the expression x+22x2+3x+6 takes all value in the interval

A.

(113,13)

B.

(113,13)

C.

(13,113)

D.

[13,113]

#### SOLUTION

Solution : B

If the given expression be y, then

y=2x2y+(3y1)x+(6y2)=0

If y ≠ 0 then  Δ ≥ 0 for real x i.e. B24AC0

or 39y2+10y+10 or (13y+1)(3y1)0

⇒  113y13

if y = 0 then x = -2 which is real and this value of y is included in the above range.

If x is real, then the maximum and minimum values of expression x2+14x+9x2+2x+3will be

A.

4, - 5

B.

5,  - 4

C.

- 4, 5

D.

- 4, - 5

#### SOLUTION

Solution : A

Let y=x2+14x+9x2+2x+3

y(x2+2x+3)x214x9=0

(y1)x2+(2y14)x+3y9=0

For real x, its discriminant ≥ 0

i.e. 4(y7)24(y1)3(y3)0

y2+y200 or (y4)(y+5)0

Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:

Case 1: y40 or y4 y+50 or y5

This is not possible.

Case 2: y40 or y4 y+50 or y5 Both of these are satisfied if -5≤y≤4

hence maximum value of y is 4 and minimum value is -5

If x23x+2 be a factor of x4px2+q, then (p, q) =

A.

(3, 4)

B.

(4, 5)

C.

(4, 3)

D.

(5, 4)

#### SOLUTION

Solution : D

x23x+2 be factor of x4px2+q=0

Hence (x23x+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)

x211x+a and x214x+2a will have a common factor, if a =

A.

24

B.

0,24

C.

3,24

D.

0,3

#### SOLUTION

Solution : B

Exprssions are x211x+a and x214x+2a will have a common factor, then

x222a+14a=xa2a=114+11

x28a=xa=13 ⇒ 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.

If x2+ax+10=0 and x2+bx10=0 have a common root, then a2b2 is equal to

A.

10

B.

20

C.

30

D.

40

#### SOLUTION

Solution : D

Let α be a common root, then

α2+αα+10=0

and α2+bα10=0

from (i) - (ii),

(ab)α+20=0 ⇒ α = 20ab

Substituting the value of α in (i), we get

(20ab)2+a(20ab)+10=0

⇒400 - 20 a(a - b) + 10(ab)2 = 0

⇒40 -2a2 + 2ab + a2b2 -2ab = 0

a2b2 = 40

If the equation x2+px+q=0 and x2+qx+p=0, have a common root, then p + q + 1 =

A.

0

B.

1

C.

2

D.

-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

If ax2+bx+c=0 and bx2+cx+a=0 have a common root

a ≠ 0, then a3+b3+c3abc=

A.

1

B.

2

C.

3

D.

#### SOLUTION

Solution : C

The condition for common roots gives (bca2)2=(cab2)(abc2) On simplification gives a3+b3+c3=3abc

The number of real roots of the equation esin xesin x4=0 are

A.

1

B.

2

C.

Infinite

D.

0

#### SOLUTION

Solution : D

Given equation esin xesin x4=0

Let  esin x=y, then given equation can be written as

y24y1=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

The number of real solutions of the equation |x|23|x|+2=0 are

A.

1

B.

2

C.

3

D.

4

#### SOLUTION

Solution : D

Given |x|23|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 x23x+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|23|x|+2=0

⇒ (|x|-1)(|x|-2)=0

⇒ |x| = 1 and |x| = 2 ⇒ x = ±1, x = ±2.

The solution set of the equation xlogx(1x)2=9 is

A.

{- 2, 4}

B.

{4}

C.

{0, - 2, 4}

D.

None of these

#### SOLUTION

Solution : B

xlogx(1x)2=9

⇒ logx(9)=logx(1x)2(;∵ ax=N ⇒ logaN=x)

⇒ 9=(1x2) ⇒ 1+x22x9=0

⇒ x22x8=0 ⇒ (x + 2)(x - 4) = 0 ⇒ x = -2, 4
But x>0; x=4

If x2+y2=25,xy=12,then complete set of  x =

A.

{3, 4}

B.

{3, -3}

C.

{3, 4, -3, -4}

D.

{-3, -3}

#### SOLUTION

Solution : C

x2+y2=25 and xy=12

⇒ x2+(12x)2=25 ⇒ x4+14425x2=0

⇒ (x216)(x29)=0 ⇒ x2=16 and x2=9

⇒ x= ±4 and x = ±3

The roots of the equation 3x+1+1 = x are

A.

0

B.

1

C.

0,1

D.

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 -2x

⇒ 2x + 2x = 0            (irrational function)

Thus x ≠ 0 and x ≠ 1, Since equation is non-quadratic equation.