Suppose f is differentiable on R and a-f-x- b where x- R where a- b - 0- If f 0 -0- then

The correct option is D ax≤ f(x)≤ bx For x gt; 0 . Applying Lagrange's theorem on. (0,x] we have c∈ ( 0,x ) such that f(x)/x= f(x ...

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Standard XII

Mathematics

Conditional Probability

Suppose f i...

Question

Suppose $f$ is differentiable on $R$ and $a \leq f^{'} (x) \leq b$ where $x \in R$ where $a, b > 0$ . If $f (0) = 0$ , then

f (x) \leq m i n (a x, b x)

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f (x) \geq m i n (a x, b x)

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a \leq f (x) \leq b

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a x \leq f (x) \leq b x

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Solution

The correct option is D $a x \leq f (x) \leq b x$
For $x > 0$ . Applying Lagrange's theorem on. $(0, x]$ we have $c \in (0, x)$ such that
$\frac{f (x)}{x} = \frac{f (x) - f (0)}{x - 0} = f^{'} (c)$
But,
$a \leq f^{'} (c) \leq b$ so $a \leq \frac{f (x)}{x} \leq b \Rightarrow a x \leq f (x) \leq b x$ , $x > 0$ .
Similarly for $x < 0$ applying Lagrange's theorem for $(x, 0]$ , we have $a x \leq f (x) \leq b x$ .

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Suppose f is differentiable on R and a≤f′(x)≤b where x∈R where a,b>0. If f(0)=0, then

The correct option is D ax≤f(x)≤bxFor x>0. Applying Lagrange's theorem on. (0,x] we have c∈(0,x) such thatf(x)x=f(x)−f(0)x−0=f′(c) But,a≤f′(c)≤b so a≤f(x)x≤b⇒ax≤f(x)≤bx, x>0. Similarly for x<0 applying Lagrange's theorem for (x,0], we have ax≤f(x)≤bx.

Suppose $f$ is differentiable on $R$ and $a \leq f^{'} (x) \leq b$ where $x \in R$ where $a, b > 0$ . If $f (0) = 0$ , then