SATHEE: Application of Derivatives 2 Question 16

Application of Derivatives 2 Question 16

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

$(1987, 2 M)$

(a) always negative

(b) always positive

(d) None of these

Objective Questions II

(One or more than one correct option)

Show Answer

Answer:

Correct Answer: 16. (c)

Solution:

Let $F (x) = h (x) - h (1) = f g (x) - f g (1)$

$\begin{aligned} ∴ F^{'} (x) & = f^{'} g (x) \cdot g^{'} (x) \\ = (+) (\to) = - ve \end{aligned}$

[since, $f (x)$ is an increasing function $f^{'} (g (x))$ is + ve and $g (x)$ is decreasing function $g^{'} (f (x))$ is $\to$ ve ]

Since, $f^{'} (x)$ is -ve.

$∴ f (x)$ is decreasing function.

When

$0 \leq x < 1$

$\Rightarrow$

$h (x) - h (1) = + ve$

When

$x \geq 1$ ,

$\Rightarrow$

$h (x) - h (1) = - ve$

Hence, for $x > 0$ ,

$h (x) - h (1)$ is neither always positive nor always negative, so it is not strictly increasing throughout.

Therefore, option (d) is the answer.

Application of Derivatives 2 Question 16

Objective Questions II

Answer:

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Application of Derivatives 2 Question 16

Objective Questions II

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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