Question

Let $f$ and $g$ be increasing and decreasing functions respectively from $(0,\infty )$ to $(0,\infty )$ and let $h\left(x\right)=f\left[g\left(x\right)\right]$. If $h\left(0\right)=0$ then $h\left(x\right)-h\left(1\right)$ is

• A. always zero
• B. always negative
• C. always positive
• D. strictly increasing
• E. None of these

Answer 1

The correct option is E None of these

Let $F\left(x\right)=h\left(x\right)-h\left(1\right)=f\left(g\left(x\right)\right)-h\left(1\right)$ $F^{\prime }\left(x\right)=f^{\prime }\left(g\left(x\right)\right).g^{\prime }\left(x\right)=(+)(-)=-ive.$ (As f is increasing function $f^{\prime }\left(g\left(x\right)\right)$ is +ive and as g is decreasing function $g^{\prime }\left(x\right)$ is-ive.)
Since F'(x) is-ive therefore F(x)i.e.h(x)-h(1) is decreasing function.
Now split the interval $I=[0,\infty ]$ into two intervals $I_1,0\le x<1and{I}_2,1\le x<\infty ,$
Apply the definition of decreasing function on $h\left(x\right)-h\left(1\right):$ on $I_1,0\le x<1,\underset{\left(Big\right)}{h\left(x\right)}-\underset{\left(Less\right)}{h\left(1\right)}=+ive$ On $I_2,1\le x<\infty ,\underset{\left(Less\right)}{h\left(x\right)}-\underset{\left(Big\right)}{h\left(1\right)}=-ive$ Hence for $I,h\left(x\right)-h\left(1\right)$ is neither always zero nor always +ive nor always-ive,nor strictly increasing throughout.
Hence (v) is the correct answer.