InterviewSolution
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InterviewSolution
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Let `f(x)=x^2a n dg(x)=sinxfora l lx in Rdot`Then the set of all `x`satisfying `(fogogof)(x)=(gogof)(x),w h e r e(fog)(x)=f(g(x)),`is`+-sqrt(npi),n in {0,1,2, dot}``+-sqrt(npi),n in {1,2, dot}``pi/2+2npi,n in { ,-2,-1,0,1,2}``2npi,n in { ,-2,-1,0,1,2, }`A. `pm sqrt(n pi), n in {0,1,2, …..}`B. `pm sqrt(n pi), n in {1,2,…}`C. `pi//2+2n pi, n in {…, -2,-1,0,1,2, …}`D. `2n pi, n in {…, -2,-1,0,1,2, … }` |
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Answer» Correct Answer - B `f(x)=x^(2),g(x)=sinx` `(gof)(x)=sin x^(2)` `go(gof)(x)=sin(sinx^(2))` `(fogogof)(x)=(sin(sin x^(2)))^(2) " …(i)" ` Again, `(gof)(x)=sin x^(2)` `(gogof)(x)=sin(sin x^(2)) " …(ii)" ` Given, `(fogogof) (x)=(gogof)(x)` `rArr (sin(sin^(2)))^(2)=sin(sin x^(2))` `rArr sin(sin x^(2)) {sin (sin x^(2))-1}=0` `implies sin(sin x^(2) )=0 or sin(sin x^(2))=1` `rArr sin x^(2)=0 or sin x^(2)=(pi)/(2)` ` therefore x^(2) = n pi ` `[ sin x^(2)=(pi)/(2) " is not possible as " -1 le sin theta le 1]` ` x =pm sqrt(n pi)` |
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