CBSE 11TH MATHEMATICS - Online Test

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

this function is defined only if 

x−|x|>0⇒x>|x|

which is not possible because

|x|≥x

so,  x∈ϕ

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]

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

f(x)−−−−√  is defined only if

f(x)will be non negative.

Hence ∴−f(x)−−−−−√is defined only for 0.