12th Grade > Mathematics
INVERSE TRIGONOMETRIC FUNCTIONS MCQs
Total Questions : 60
| Page 5 of 6 pages
Answer: Option D. -> None of these
:
D
cos−1(cos12)−sin−1(sin14)⇒4π−12+5π−14=9π−26
:
D
cos−1(cos12)−sin−1(sin14)⇒4π−12+5π−14=9π−26
Answer: Option D. -> π4
:
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.
:
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.
Answer: Option C. -> 3
:
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.
:
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.
Answer: Option A. -> 1√10
:
A
Let 0≤x≤π
So,cos−145=x⇒cosx=45 ....(i)
Now sin(12cos−145)=sin(x2) ....(ii)
From (i), cosx=45⇒1−2sin2x2=45
⇒2sin2x2=1−45=15⇒sinx2=√110.
:
A
Let 0≤x≤π
So,cos−145=x⇒cosx=45 ....(i)
Now sin(12cos−145)=sin(x2) ....(ii)
From (i), cosx=45⇒1−2sin2x2=45
⇒2sin2x2=1−45=15⇒sinx2=√110.
Answer: Option C. -> 0
:
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.
:
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.
Answer: Option A. -> 1
:
A
(sin−1y)2+cos−1x=nπ24
(sin−1y)2−cos−1x=π216
⇒(sin−1y)2=(4n+1)π232,cos−1x=π2(4n−1)32
⇒0≤(4n+1)32π2≤π24,0≤(4n−1)32π2≤π
⇒−14≤n≤74,14,≤n≤8π+14
:
A
(sin−1y)2+cos−1x=nπ24
(sin−1y)2−cos−1x=π216
⇒(sin−1y)2=(4n+1)π232,cos−1x=π2(4n−1)32
⇒0≤(4n+1)32π2≤π24,0≤(4n−1)32π2≤π
⇒−14≤n≤74,14,≤n≤8π+14
Answer: Option D. -> 12
:
D
Let α=cos−1√p:β=cos−1√1−p
and γ=cos−1√1−q or cosα=√p:cosβ=√1−p
and cosγ=√1−q.
Therefore sinα=√1−p,sinβ=√p and sinγ=√q.
The given equation may be written as α+β+γ=3π4orα+β=3π4−γ or cos(α+β)=cos(3π4−γ)
⇒cosαcosβ−sinαsinβ = cos{π−(π4+γ)}=−cos(π4+γ)
⇒√p√1−p−√1−p√p=−(1√2√1−q−1√2.√q)
⇒0=√1−q−√q⇒1−q=q⇒q=12.
:
D
Let α=cos−1√p:β=cos−1√1−p
and γ=cos−1√1−q or cosα=√p:cosβ=√1−p
and cosγ=√1−q.
Therefore sinα=√1−p,sinβ=√p and sinγ=√q.
The given equation may be written as α+β+γ=3π4orα+β=3π4−γ or cos(α+β)=cos(3π4−γ)
⇒cosαcosβ−sinαsinβ = cos{π−(π4+γ)}=−cos(π4+γ)
⇒√p√1−p−√1−p√p=−(1√2√1−q−1√2.√q)
⇒0=√1−q−√q⇒1−q=q⇒q=12.
Answer: Option A. -> 0
:
A
cos−1√124=cos−1√32=π6,sec−1√2=π4
sin−1√2−√34=sin−1√4−2√38=sin−1√3−12√2=π12
∴ sin−1√2−√34+cos−1√124+sec−1√2=π2
and sin−1(cotπ2)=sin−10=0
:
A
cos−1√124=cos−1√32=π6,sec−1√2=π4
sin−1√2−√34=sin−1√4−2√38=sin−1√3−12√2=π12
∴ sin−1√2−√34+cos−1√124+sec−1√2=π2
and sin−1(cotπ2)=sin−10=0
Answer: Option C. -> ϕ
:
C
x2−px+sin−1(sin4)>0 for all real x.
⇒x2−px+sin−1(sin4)>0
⇒x2−px+(π−4)>0∀xϵR
⇒D=p2−4(π−4)<0 ⇒p2+16−4π<0
Since 16−4π>0,p2+16−4π cannot be negative for any value of pϵR.
∴ Set of values of p=ϕ
:
C
x2−px+sin−1(sin4)>0 for all real x.
⇒x2−px+sin−1(sin4)>0
⇒x2−px+(π−4)>0∀xϵR
⇒D=p2−4(π−4)<0 ⇒p2+16−4π<0
Since 16−4π>0,p2+16−4π cannot be negative for any value of pϵR.
∴ Set of values of p=ϕ
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