If f(x)=x+sinx;g(x)=e−x;u=√c+1−√c;v=√c−√c−1;(c>1), then
A
f∘g(u)<f∘g(v)
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B
g∘f(u)<g∘f(v)
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C
g∘f(u)>g∘f(v)
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D
f∘g(u)>f∘g(v)
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Solution
The correct option is Cg∘f(u)>g∘f(v) u=√c+1−√c u=1√c+1+√c and v=1√c−1+√c Clearly, u<v and they lie in (0,1) f′(x)=1+cosx For x∈(0,1) f′(x)>0, so f(x) is increasing . Therefore, f(u)<f(v) We know that g(x) is decreasing function, so g∘f(u)>g∘f(v)