Relations and Functions - Online Test

Answer : Option D
Explaination / Solution:

domain of the function fg is equals to {x ∈R : g (x) ≠ 0 } = R – { x ∈R : g (x) = 0 }.

Answer : Option D
Explaination / Solution:

This function exists only if cosx−1≥0 ⇒cosx≥1 ⇒cosx>1 OR cosx=1 since maximum value of cosine function is 1 so, cos x >1 is not possible

Answer : Option A
Explaination / Solution:

f(x)=|x|−x−−−−−−√

this function exists only if |x|−x≥0

⇒|x|≥x

∀ x∈R  it is true 

so domain of f(x) is R

Answer : Option D
Explaination / Solution:

this function exists only if 1−cosx≥0 ⇒cos⁡x≤1 it is possible ∀x∈R

Answer : Option B
Explaination / Solution:

this exists only if

sinx−1≥0sinx≥1sinx>1,notpossible∴sinx=1

⇒sinx=sin(2nπ+π2)⇒x=2nπ+π2(n∈I)

Answer : Option D
Explaination / Solution:

Answer : Option A
Explaination / Solution:

f(x)=x;g(x)=cosxDf=RDg=RDf+g=Df∩Dg$\begin{array}{l}f(x)=x;g(x)=\cos x\\ D_f=R\\ D_g=R\\ D_{f+g}=D_f\cap D_g\end{array}$

so,

Df+g=R

SO given function is defined for all reals

Answer : Option B
Explaination / Solution:

this function is defined only if 

x−|x|>0⇒x>|x|

which is not possible because

|x|≥x

so,  x∈ϕ

Answer : Option C
Explaination / Solution:

gof(x)=g(f(x))⇒gof(x)=g(tanx)=3−tan2x−−−−−−−−√

which is defined only if

3−tan2x≥0⇒3≥tan2x⇒−3–√≤tanx≤3–√⇒−π3≤x≤π3

so domain of gof is [−π3,π3]

Answer : Option B
Explaination / Solution:

fog(x)=f(g(x))=f(1−x2−−−−−√)⇒fog(x)=tan1−x2−−−−−√

and it is defined if

1−x2≥0⇒x2≤1⇒−1≤x≤1