Free Objective Test 01 Practice Test - 11th and 12th
Question 1
If x=sin(2tan−12),y=sin(12tan−143) then
A.
x = 1 – y
B.
x2=1–y
C.
x2=1+y
D.
y2=1−x
SOLUTION
Solution : D
Let tan−12=α⇒x=sin2α=45y=sin(β2)=√1−cosβ2=√1−352=1√5
Question 2
sin−1|cosx|−cos−1|sinx|=a has at least one solution if a ∈
A.
0
B.
[0,π2]
C.
[π2,3π2]
D.
(0,π)
SOLUTION
Solution : A
sin−1|cosx|−cos−1|sinx|=π2−cos−1|cosx|−π2+sin−1|sinx|=a⇒sin−1|sinx|−cos−1|cosx|=a⇒a=0∀x
Question 3
sin−1(sin10)=
A.
10
B.
3π−10
C.
π+6
D.
2π−6
SOLUTION
Solution : B
3π<10<3π+π2∴10 rad ∈Q3∴−π2<3π−10<0∴3π−10∈Q4Also sin(3π−10)=sin10Hence sin−1(sin10)=sin−1sin(3π−10)=3π−10
Question 4
cos−1x=tan−1√1−x2x, then:
A.
X∈R
B.
−≤x≤1,x≠0
C.
0<x≤1
D.
None of these
SOLUTION
Solution : C
Putting θ=cos−1x in R.H.S., we have
R.H.S =tan−1√1−x2x=tan−1√1−cos2θcosθ(0≤θ≤π)=tan−1sinθcosθ=tan−1tanθ=θ=cos−1x when −π2<θ<π2
i.e., when 0≤θ<π2
i.e., when 0<cosθ≤1 i.e., 0<x≤1.
Question 5
The value of x for which sin(cot−1(1+x))=cos(tan−1x) is:
A.
12
B.
1
C.
0
D.
−12
SOLUTION
Solution : D
sin(cot−1(1+x))=cos(tan−1x)⇒cot−1(1+x)=[π2±tan−1x]⇒π2−tan−1(1+x)=π2±tan−1x⇒1+x=±x
x=−12 Which, on verification, satisfies the equation.
Question 6
3cos−1x−πx−π2=0 has :
A.
One solution
B.
Infinite solutions
C.
No solution
D.
None of these
SOLUTION
Solution : A
3cos−1x−πx+π2
Clearly graphs of y=3cos−1x and y=πx+π2 in the domain of cos−1x i.e., in [-1, 1] intersect only once, therefore there is only one solution of the given equation.
Question 7
If cos−1x+cos−1y+cos−1z=3π, then xy+yz+zx=
A.
0
B.
1
C.
3
D.
-3
SOLUTION
Solution : C
Given cos−1x+cos−1y+cos−1z=3π
∵0≤cos−1x≤π
∴0≤cos−1y≤π and 0≤cos−1z≤π
Here cos−1x=cos−1y=cos−1z=π
⇒x=y=z=cosπ=−1
∴xy+yz+zx=(−1)(−1)+(−1)(−1)+(−1)(−1)
=1+1+1=3.
Question 8
The value of tan(tan−112−tan−113)=
A.
56
B.
76
C.
16
D.
17
SOLUTION
Solution : D
tan[tan−112−tan−113]=tan[tan−112−131+16]
=tan tan−1(16×67)=17
Question 9
tan[2tan−1(15)−π4]=
A.
177
B.
−177
C.
717
D.
−717
SOLUTION
Solution : D
tan[2tan−1(15)−π4]=tan[tan−1251−125−tan−1(1)]
=tan[tan−1512−tan−1(1)]=tan tan−1(512−11+512)=−717
Question 10
12cos−1(1−x1+x)=
A.
cot−1√x
B.
tan−1√x
C.
tan−1x
D.
cot−1x
SOLUTION
Solution : B
Let x=tan2θ⇒θ=tan−1√x
Now, 12cos−1(1−x1+x)
=12cos−1(1−tan2θ1+tan2θ)
=12cos−1cos2θ=2θ2=θ=tan−1√x.
Question 11
4tan−115−tan−11239 is equal to
A.
π
B.
π2
C.
π3
D.
π4
SOLUTION
Solution : D
Since 2tan−1x=tan−12x1−x2
∴4tan−115=2[2tan−115]=2tan−1251−125
=2tan−11024=tan−120241−100576=tan−1120119
So, 4tan−115−tan−11239=tan−1120119−tan−11239
=tan−1120119−12391+120119.1239=tan−1(120×239)−119(119×239)+120
⇒tan−11=π4.
Question 12
If 3sin−12x1+x2−4cos−11−x21+x2+2tan−12x1+x2=π3 then x=
A.
√3
B.
1√3
C.
1
D.
None of these
SOLUTION
Solution : B
3sin−12x1+x2−4cos−11−x21+x2+2tan−12x1+x2=π3
Putting x=tanθ
3sin−1(2tanθ1+tan2θ)−4cos−1(1−tan2θ1+tan2θ)
+2tan−1(2tanθ1−tan2θ)=π3
⇒3sin−1(sin2θ)−4cos−1(cos2θ)
+2tan−1(tan2θ)=π3
⇒3(2θ)−4(2θ)+2(2θ)=π3⇒6θ−8θ+4θ=π3
⇒θ=π6⇒tan−1x=π6⇒x=tanπ6=1√3
Question 13
The value of sin[2tan−1(13)]+cos[tan−1(2√2)]=
A.
1615
B.
1415
C.
1215
D.
1115
SOLUTION
Solution : B
sin[2tan−1(13)]+cos[tan−1(2√2)]
=sin[tan−1231−19]+cos[tan−1(2√2)]
=sin[tan−134]+cos[tan−12√2]
=tan−12√2=cos−113
Also tan−134=sin−135
=35+13=1415.
Question 14
sin(12cos−145)=
A.
1√10
B.
−1√10
C.
110
D.
−110
SOLUTION
Solution : A
Let 0≤x≤π
So, cos−145=x⇒cos x=45 ....(i)
Now sin(12cos−145)=sin(x2) ....(ii)
From (i), cos x=45⇒1−2sin2x2=45
⇒2sin2x2=1−45=15⇒sinx2=√110.
Question 15
The value of cos−1(cos12)−sin−1(sin14) is
A.
−2
B.
8π−26
C.
4π+2
D.
None of these
SOLUTION
Solution : D
cos−1(cos12)−sin−1(sin14)⇒4π−12+5π−14=9π−26