Given hx-fx-gx- If h-x - 0- fx- gx-

If h’(x) ≥ 0, then we can say f’(x) g’(x) ≥ 0 This would mean f’(x) ≥ g’(x) We can’t say f(x) ≥ g(x) from this. For example, if f(x) = 1 and g(x) = 2, we ge ...

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Byju's Answer

Standard XII

Mathematics

Use of Monotonicity for Proving Inequalities

Given hx=fx-g...

Question

Given $h (x) = f (x) - g (x) . I f h^{'} (x) \geq 0, f (x) \geq g (x)$ .

True

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False

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Solution

The correct option is B
False

If h’(x) $\geq$ 0, then we can say f’(x) - g’(x) $\geq$ 0

This would mean f’(x) $\geq$ g’(x)

We can’t say f(x) $\geq$ g(x) from this. For example, if f(x) = 1 and g(x) = 2, we get h(x) = 1-2 = -1

Now, h’(x) = 0. So the given conditions are satisfied.

But we can’t say f(x) $\geq$ g(x) from this.

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Similar questions

Given $h (x) = f (x) - g (x) . I f h^{'} (x) \geq 0, f (x) \geq g (x)$ .

Q. Let

f (x) = x | x |, g (x) = sin (x)

and

h (x) = (g \circ f) (x)

. Then

lim x \to a^{-} f (x) = l

lim x \to a^{-} g (x) = 0

lim x \to a^{-} h (x) = m

,then

lim x \to a^{-} (f (x) - g (x) + h (x)) =

$\frac{(f + g) x}{(f + h) x} = \frac{f (x) + g (x)}{f (x) + h (x)}$

Q. Assertion :Let

f (x) = tan x

and

g (x) = x^{2}

then

f (x) + g (x)

is neither even nor odd function. Reason: If

h (x) = f (x) + g (x)

, then

h (x)

does not satisfy the condition

h (- x) = h (x)

and

h (- x) = - h (x)

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Given h(x)=f(x)−g(x). If h′(x)≥0,f(x)≥g(x).

Given $h (x) = f (x) - g (x) . I f h^{'} (x) \geq 0, f (x) \geq g (x)$ .