Question:
Let \(f\) and \(g\) be two differentiable functions on \(R\) such that \(f'(x)>0\) and \(g'(x)<0\) for all \(x\in R\). Then for all \(x\):
Let \(f\) and \(g\) be two differentiable functions on \(R\) such that \(f'(x)>0\) and \(g'(x)<0\) for all \(x\in R\). Then for all \(x\):
Show Hint
\(f\) is strictly increasing and \(g\) is strictly decreasing. Compare the inner arguments first, then apply the monotonicity of the outer function.
Updated On: Nov 11, 2026
- \(f(g(x))>f(g(x-1))\)
- \(f(g(x))>f(g(x+1))\)
- \(g(f(x))>g(f(x-1))\)
- \(g(f(x))<g(f(x+1))\)
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The Correct Option is B
Solution and Explanation
\(f'>0\Rightarrow f\) strictly increasing; \(g'<0\Rightarrow g\) strictly decreasing.
(1) \(g(x)<g(x-1)\) so \(f(g(x))<f(g(x-1))\). Option (1) is false.
(2) \(g(x)>g(x+1)\) so \(f(g(x))>f(g(x+1))\) (f increasing). Option (2) is true.
(3) \(f(x)>f(x-1)\) so \(g(f(x))<g(f(x-1))\) (g decreasing). Option (3) is false.
(4) \(f(x+1)>f(x)\) so \(g(f(x+1))<g(f(x))\), i.e. \(g(f(x))>g(f(x+1))\). Option (4) is false.
Answer: (2) \(f(g(x))>f(g(x+1))\)
(1) \(g(x)<g(x-1)\) so \(f(g(x))<f(g(x-1))\). Option (1) is false.
(2) \(g(x)>g(x+1)\) so \(f(g(x))>f(g(x+1))\) (f increasing). Option (2) is true.
(3) \(f(x)>f(x-1)\) so \(g(f(x))<g(f(x-1))\) (g decreasing). Option (3) is false.
(4) \(f(x+1)>f(x)\) so \(g(f(x+1))<g(f(x))\), i.e. \(g(f(x))>g(f(x+1))\). Option (4) is false.
Answer: (2) \(f(g(x))>f(g(x+1))\)
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