If $f(x)=x+\sin x, g(x)=e^{-x}, u=\sqrt{c+1}-\sqrt{c}$, $v=\sqrt{c}-\sqrt{c-1},(c > 1)$, then

JEE MATHEMATICS MCQ: INCREASING & DECREASING

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f(x)=x+\sin x, g(x)=e^{-x}, u=\sqrt{c+1}-\sqrt{c}

v=\sqrt{c}-\sqrt{c-1},(c > 1)

, then

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Correct option: C

g o f(u) > g \circ f(v)

Option C (g o f(u) > g o f(v)) is correct because f is increasing, g is decreasing, and u < v, implying f(u) < f(v) and thus g(f(u)) > g(f(v)).

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f\left( x \right) = \frac{{K\sin x + 2\cos x}}{{\sin x + \cos x}}

x

f\left( x \right) = {x^2} + ax + 1

a

f(x)=

\sin x-\cos x-a x+b

f(x)=x e^{x(1-x)}, x \in R

f

g

fog

f\left( x \right) = {x^3} + b{x^2} + cx + d

0 < {b^2} < c

\left( { - \infty ,\;\infty } \right)