Lakshya Education MCQs

Question: 3cos1xπxπ2=0 has :

 
Options:
A.One solution
B.Infinite solutions
C.No solution
D.None of these
Answer: Option A
: A



3cos1xπx+π2
Clearly graphs of y=3cos1x and y=πx+π2 in the domain of cos1x i.e., in [-1, 1] intersect only once, therefore there is only one solution of the given equation.

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More Questions on This Topic :

Question 1. 12cos1(1x1+x)=
  1.    cot−1√x
  2.    tan−1√x
  3.    tan−1x
  4.    cot−1x
Answer: Option B
: B

Let x=tan2θθ=tan1x
Now, 12cos1(1x1+x)
=12cos1(1tan2θ1+tan2θ)
=12cos1cos2θ=2θ2=θ=tan1x.
Question 2. 4tan115tan11239 is equal to
  1.    π
  2.    π2
  3.    π3
  4.    π4
Answer: Option D
: D

Since 2tan1x=tan12x1x2
4tan115=2[2tan115]=2tan1251125
=2tan11024=tan120241100576=tan1120119
So, 4tan115tan11239=tan1120119tan11239
=tan112011912391+120119.1239=tan1(120×239)119(119×239)+120
tan11=π4.
Question 3. If cos1x+cos1y+cos1z=3π, then xy+yz+zx=

 
Answer: Option D
: C

Given cos1x+cos1y+cos1z=3π
0cos1xπ
0cos1yπ and 0cos1zπ
Here cos1x=cos1y=cos1z=π
x=y=z=cosπ=1
xy+yz+zx=(1)(1)+(1)(1)+(1)(1)
=1+1+1=3.
Question 4. The value of cos1(cos12)sin1(sin14) is
  1.    −2
  2.    8π−26
  3.    4π+2
  4.    None of these
Answer: Option D
: D

cos1(cos12)sin1(sin14)4π12+5π14=9π26
Question 5. The value of sin[2tan1(13)]+cos[tan1(22)]=
  1.    1615
  2.    1415
  3.    1215
  4.    1115
Answer: Option B
: B

sin[2tan1(13)]+cos[tan1(22)]
=sin[tan123119]+cos[tan1(22)]
=sin[tan134]+cos[tan122]
=tan122=cos113
Alsotan134=sin135
=35+13=1415.
Question 6. If 3sin12x1+x24cos11x21+x2+2tan12x1+x2=π3 then x=
  1.    √3
  2.    1√3
  3.    1
  4.    None of these
Answer: Option B
: B

3sin12x1+x24cos11x21+x2+2tan12x1+x2=π3
Putting x=tanθ
3sin1(2tanθ1+tan2θ)4cos1(1tan2θ1+tan2θ)
+2tan1(2tanθ1tan2θ)=π3
3sin1(sin2θ)4cos1(cos2θ)
+2tan1(tan2θ)=π3
3(2θ)4(2θ)+2(2θ)=π36θ8θ+4θ=π3
θ=π6tan1x=π6x=tanπ6=13

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