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Multiplication of Integers

Given that ...

Question

Given that $f^{'} (x) > g^{'} (x)$ for all real $x,$ and $f (0) = g (0),$ then $f (x) < g (x)$ for all $n$ belonging to

(0, \infty)

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(- \infty, 0)

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(- \infty, \infty)

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None of these

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Solution

The correct option is D $(- \infty, 0)$
Let $h (x) = f (x) - g (x)$
$\Rightarrow h^{'} (x) = f^{'} (x) - g^{'} (x) > 0 \forall x \in R$
$h (x)$ is an increasing function and $h (0) = f (0) - g (0) = 0.$
Therefore, $h (x) > 0 \forall x \in (0, \infty)$ and $h (x) < 0 \forall x \in (- \infty, 0)$

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Similar questions

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Given that f′(x)>g′(x) for all real x, and f(0)=g(0), then f(x)<g(x) for all n belonging to

The correct option is D (−∞,0)Let h(x)=f(x)−g(x)⇒h′(x)=f′(x)−g′(x)>0∀x∈Rh(x) is an increasing function and h(0)=f(0)−g(0)=0.Therefore, h(x)>0∀x∈(0,∞) and h(x)<0∀x∈(−∞,0)

Given that $f^{'} (x) > g^{'} (x)$ for all real $x,$ and $f (0) = g (0),$ then $f (x) < g (x)$ for all $n$ belonging to

The correct option is D $(- \infty, 0)$
Let $h (x) = f (x) - g (x)$
$\Rightarrow h^{'} (x) = f^{'} (x) - g^{'} (x) > 0 \forall x \in R$
$h (x)$ is an increasing function and $h (0) = f (0) - g (0) = 0.$
Therefore, $h (x) > 0 \forall x \in (0, \infty)$ and $h (x) < 0 \forall x \in (- \infty, 0)$