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

If x +  1x = 2cosθ, then x31x3

A.

cos3θ

B.

2cos3θ

C.

12cos3θ

D.

13cos3θ

#### SOLUTION

Solution : B

We have x +  1x = 2cosθ,

Now x31x3 = (x+1x)3 - 3x1x(x+1x)

= (2cosθ)33(2cosθ)=8cos3θ6cosθ

=2(4cos3θ3cosθ)=2cos3θ.

Trick: Put x = 1 θ=0.

Then x31x3 = 2 = 2cos3θ.

If cos(θα) = a, sin(θβ) = b,

then cos2(αβ) + 2ab sin(αβ) is equal to

A.

4a2b2

B.

a2b2

C.

a2+b2

D.

-a2b2

#### SOLUTION

Solution : C

We have sin(αβ) = sin(θβ¯¯¯¯¯¯¯¯¯¯¯¯¯θα)

= sin(θβ)cos(θα)cos(θβ)sin(θα)

= ba - 1b21a2

And cos(αβ)=cos(θβ¯¯¯¯¯¯¯¯¯¯¯¯¯θα)

= cos(θβ)cos(θα)+sin(θβ)sin(θα)

= a1b2+b1a2

∴ Given expression is cos2(αβ)+2absin(αβ)

= (a1b2+b1a2)2  + 2ab{ab1a21b2}

= a2+b2.

Trick:  Put α=30, β=60 and θ=90,

then a =  12, b =  12

cos2(αβ)+2absin(αβ)3412 × (- 12) =  12

which is given by option (c).

If acos3α+3acosαsin2α=m and

asin3α+3acos2αsinα=n, Then (m+n)23+(mn)23

is equal to

A.

2a2

B.

2a13

C.

2a23

D.

2a3

#### SOLUTION

Solution : C

Adding and subtracting the given relation,

we get (m+n)=acos3α+3acosαsin2α

+ 3acos2α.sinα+asin3α

= a(cosα+sinα)3

and similarly (mn)=a(cosαsinα)3

Thus, (m+n)23+(mn)23
=a23 {(cosα+sinα)2+(cosαsinα)2}
=a23 [2(cos2α+sin2α)] = 2a23.

The minimum value of 3sinθ+4cosθ is

A.

5

B.

1

C.

3

D.

-5

#### SOLUTION

Solution : D

Minimum value of (3sinθ+4cosθ) is -32+42 i.e., -5.

The minimum value of 9tan2θ+4cot2θ is

A.

13

B.

9

C.

6

D.

12

#### SOLUTION

Solution : D

A.M.  G.M.

9tan2θ+4cot2θ2 4cot2θ.9tan2θ

9tan2θ+4cot2θ  12

Therefore, the minimum value is 12.

If secθ = 54, then tan θ2

A.

13

B.

34

C.

14

D.

19

#### SOLUTION

Solution : A

Given that secθ54

secθ1+tan2(θ2)1tan2(θ2)   541+tan2(θ2)1tan2(θ2)

55tan2(θ2)=4+4tan2(θ2)

9tan2(θ2) = 1 tan(θ2)13.

cot2151cot215+1

A.

12

B.

32

C.

334

D.

3

#### SOLUTION

Solution : B

cot2151cot215+1cos215sin2151cos215sin215+1

= cos215sin215cos215+sin215 = cos(30)32

If sin A =  45 and cos B = - 1213, where A and B lie in first

and third quadrant respectively, then cos(A + B) =

A.

5665

B.

- 5665

C.

1665

D.

- 1665

#### SOLUTION

Solution : D

We have sin A =  45 and cos B = - 1213

Now, cos(A + B) = cos A cos B - sin A sin B

= 11625(1213)45(1144169)

= - 35 × 121345(513) = - 1665

(Since A lies in first quadrant and B lies in third quadrant).

If A + B = 225, then  cotA1+cotA. cotB1+cotB =

A.

1

B.

-1

C.

0

D.

12

#### SOLUTION

Solution : D

cotA1+cotA. cotB1+cotB1(1+tanA)(1+tanB)

1tanA+tanB+1+tanAtanB

[ ∵ tan(A + B) = tan225]

tanA + tan B = 1 - tan A tan B

11tanAtanB+1+tanAtanB12.

If tan A = - 12 and tan B = -  13, then A + B =

[IIT 1967; MNR 1987; MP PET 1989]

A.

π4

B.

3π4

C.

5π4

D.

None of these

#### SOLUTION

Solution : B

We have tan A = - 12 and tan B = - 13

Now, tan(A + B) =  tanA+tanB1tanAtanB1213112.13 = -1

tan(A + B) = tan 3π4. Hence, A + B =  3π4.

If tanθsinαcosαsinα+cosα, then sinα+cosα and

sinαcosα must be equal to

A.

2cosθ,2sinθ

B.

2sinθ,2cosθ

C.

2sinθ,2sinθ

D.

2cosθ,2cosθ

#### SOLUTION

Solution : A

We have tanθsinαcosαsinα+cosα

tanθsin(απ4)cos(απ4) tanθ = tan(απ4)

θ = απ4  α = θπ4

Hence, sinα+cosα = sin(θ+π4)+cos(θ+π4)

= 2cosθ

And sinαcosα = sin(θ+π4) - cos(θ+π4)

=  12 sinθ12 cosθ12 cosθ12 sinθ

22 sinθ = 2sinθ=2sinθ.

If cos6α+sin6α+K sin22α=1, then K=

A.

43

B.

34

C.

12

D.

2

#### SOLUTION

Solution : B

Since cos6α+sin6α+K sin22α=1 using formula a3+b3=(a+b)33ab(a+b) and on solving, we get the required result i.e., K=34

tan α+2tan 2α+4tan 4α+8cot 8α=

[IIT 1988; MP PET 1991]

A.

tanα

B.

tan2α

C.

cotα

D.

cot2α

#### SOLUTION

Solution : C

=tan α+2tan 2α+4tan 4α+8cot 8α
tan α+2tan 2α+4[sin 4αcos 4α+2cos 8αsin 8α]

=tan α+2tan 2α+4[cos 4α cos 8α+sin 4α sin 8α+cos 4α cos 8αsin 8α cos 4α]

=tan α+2tan 2α+4[cos 4α+cos 4α cos 8αsin 8α cos 4α]

=tan α+2tan 2α+4[cos 4α(1+cos 8α)cos 4α sin 8α]

=tan α+2tan 2α+4[2cos2 4α2sin 4α cos 4α]

=tan α+2tan 2α+4cot 4α
=tan α+2(tan 2α+2cot 4α)
=tan α+2[sin 2αcos 2α+2cos 4αsin 4α]
=tan α+2[cos 2α(1+cos 4α)sin 4α cos 2α]
=tan α+2cot 2α=sin αcos α+2cos 2αsin 2α
=cos α+cos α cos 2αsin 2α cos α
=1+cos 2αsin 2α=2cos2α2sin α cos α=cot α

cos248sin212=

A.

514

B.

5+18

C.

314

D.

3+122

#### SOLUTION

Solution : B

cos2Asin2B = cos(A + B). Cos(A - B)

cos248sin212=cos60.cos 36

=12(5+14)=5+18

If 3π4<α<π, then cosec2α+2cotα is equal to

A.

1+cotα

B.

1cotα

C.

1cotα

D.

1+cotα

#### SOLUTION

Solution : C

cosec2α+2cotα=1+cot2α+2cotα=|1+cotα|
But 3π4<α<πcotα<11+cotα<0
Hence, |1+cotα|=(1+cotα)