JEE Advanced 2000

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Consider the following statments in $S$ and $R$ \\ $S$ : Both $sin x$ and $cos x$ are decreasing functions in the interval $(\frac{π}{2}, π)$ \\ $R$ : If a differentiable function decreases in an interval $(a, b)$ , then its derivative also decreases in $(a, b)$ . \\ Which of the following is true?

Q: Why does $f'(x) < 0$ not automatically mean $f''(x) < 0$?

That is a fantastic question! A function $f(x)$ is decreasing if its slope $f'(x)$ is negative. However, the *rate* at which that slope changes is determined by the second derivative $f''(x)$. If $f''(x) > 0$, the slope is becoming *less* negative (or more positive), which means the slope is actually *increasing*, even though the function itself is still decreasing. It is a subtle but crucial distinction!

Q: Is it possible for a function to be decreasing but have an increasing derivative?

Absolutely! Consider the function $f(x) = \frac{1}{x}$ for $x > 0$. Its derivative is $f'(x) = -\frac{1}{x^2}$. Since $x^2$ is always positive, $f'(x)$ is always negative, so the function is decreasing. But if you look at the second derivative, $f''(x) = \frac{2}{x^3}$, which is positive for $x > 0$. This means the derivative $f'(x)$ is increasing. The slope is getting flatter as $x$ increases.

Select Answer:

Visualized Solution

Defining the Interval

Consider the interval $x \in (\frac{π}{2}, π)$ .
This corresponds to the Second Quadrant in trigonometry.
We need to evaluate the monotonicity of $sin x$ and $cos x$ here.

Behavior of $sin x$

At $x = \frac{π}{2}$ , $sin x = 1$ .
At $x = π$ , $sin x = 0$ .
The curve continuously drops from $1$ to $0$ .

Monotonicity of $sin x$

Since the value decreases as $x$ increases, $sin x$ is a decreasing function in $(\frac{π}{2}, π)$ .

Behavior of $cos x$

At $x = \frac{π}{2}$ , $cos x = 0$ .
At $x = π$ , $cos x = - 1$ .
The curve continuously drops from $0$ to $- 1$ .

Monotonicity of $cos x$

Since the value decreases as $x$ increases, $cos x$ is also a decreasing function in $(\frac{π}{2}, π)$ .

Evaluating Statement $S$

Statement $S$ claims both $sin x$ and $cos x$ are decreasing in $(\frac{π}{2}, π)$ .
Based on our analysis, Statement $S$ is Correct.

Decoding Statement $R$

Statement $R$ : If $f (x)$ decreases, then $f^{'} (x)$ also decreases.
Mathematically, $f (x)$ decreasing implies $f^{'} (x) < 0$ .
$f^{'} (x)$ decreasing implies $f^{''} (x) < 0$ .

The Counter-Example Setup

Let's test Statement $R$ with a standard function: $f (x) = \frac{1}{x}$ for $x > 0$ .
We will check if its derivative decreases as the function decreases.

First Derivative of $f (x)$

$f (x) = \frac{1}{x} ⟹ f^{'} (x) = - \frac{1}{x ^{2}}$
Since $x^{2} > 0$ , $f^{'} (x)$ is always negative.
Therefore, $f (x)$ is indeed a decreasing function.

Visualizing the Derivative

Let's draw tangents at $x = 1$ and $x = 2$ .
At $x = 1$ , slope $m = - 1$ .
At $x = 2$ , slope $m = - 0.25$ .

Second Derivative of $f (x)$

Let's find $f^{''} (x)$ to check if $f^{'} (x)$ is decreasing.
$f^{'} (x) = - x^{- 2} ⟹ f^{''} (x) = 2 x^{- 3} = \frac{2}{x ^{3}}$
For $x > 0$ , $f^{''} (x) > 0$ .

Conclusion for Statement $R$

Since $f^{''} (x) > 0$ , the derivative $f^{'} (x)$ is increasing, not decreasing.
This directly contradicts Statement $R$ .
Therefore, Statement $R$ is Wrong.

Final Answer

Statement $S$ is Correct.
Statement $R$ is Wrong.
The correct option is: $S$ is correct and $R$ is wrong.

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The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Setup

Imagine you are standing on the unit circle at the angle

x = \frac{π}{2}

, looking towards

x = π

. This region corresponds to the second quadrant.

As you move along the circle, your height, defined by

f (x) = sin x

, descends from

1

0

. Simultaneously, your horizontal position, defined by

g (x) = cos x

, descends from

0

- 1

Since both functions are strictly moving downward in this interval, Statement $S$ is correct.

The Trap of Statement $R$

Statement

R

suggests that if a function is decreasing, its rate of change (the derivative) must also be decreasing. This is a common misconception that requires rigorous testing.

Consider the function

f (x) = \frac{1}{x}

. Its derivative is given by:

f^{'} (x) = - \frac{1}{x ^{2}}

Since

f^{'} (x) < 0

for all

x > 0

, the function

f (x)

is strictly decreasing.

Testing the Derivative's Trend

Let us examine the behavior of the slope

f^{'} (x)

at different points:

x = 1

, the slope is

f^{'} (1) = - 1

x = 2

, the slope is

f^{'} (2) = - 0.25

Because

- 0.25 > - 1

, the slope is becoming less negative. In mathematical terms, the derivative is increasing, which implies

f^{''} (x) > 0

Final Conclusion

The logic of Statement

R

fails because a function can be decreasing (

f^{'} (x) < 0

) while its derivative is increasing (

f^{''} (x) > 0

). This occurs whenever the function is concave up.

The derivative

f^{'} (x)

being negative only confirms that the function is decreasing; it provides no information regarding the trend of the derivative itself.

By using the counter-example

f (x) = \frac{1}{x}

, we have proven that Statement $S$ is correct and Statement $R$ is false.

Common Traps & FAQs

Why does

f^{'} (x) < 0

not automatically mean

f^{''} (x) < 0

That is a fantastic question! A function

f (x)

is decreasing if its slope

f^{'} (x)

is negative. However, the *rate* at which that slope changes is determined by the second derivative

f^{''} (x)

. If

f^{''} (x) > 0

, the slope is becoming *less* negative (or more positive), which means the slope is actually *increasing*, even though the function itself is still decreasing. It is a subtle but crucial distinction!

Is it possible for a function to be decreasing but have an increasing derivative?

Absolutely! Consider the function

f (x) = \frac{1}{x}

for

x > 0

. Its derivative is

f^{'} (x) = - \frac{1}{x ^{2}}

. Since

x^{2}

is always positive,

f^{'} (x)

is always negative, so the function is decreasing. But if you look at the second derivative,

f^{''} (x) = \frac{2}{x ^{3}}

, which is positive for

x > 0

. This means the derivative

f^{'} (x)

is increasing. The slope is getting flatter as

x

increases.

How do I visualize the slope of a curve getting 'less negative'?

Imagine a steep downhill slope. As you move forward, the slope becomes less steep—it is still downhill, but you are not dropping as fast as before. That is a 'less negative' slope. Mathematically, this corresponds to a positive second derivative, which indicates that the curve is concave up.

Select Answer:

Visualized Solution

Defining the Interval

Behavior of $sin x$

Monotonicity of $sin x$

Behavior of $cos x$

Monotonicity of $cos x$

Evaluating Statement $S$

Decoding Statement $R$

The Counter-Example Setup

First Derivative of $f (x)$

Visualizing the Derivative

Second Derivative of $f (x)$

Conclusion for Statement $R$

Final Answer

The Sigma Insight: Monotonicity

Analyzing the Setup

The Trap of Statement $R$

Testing the Derivative's Trend

Final Conclusion

Similar Questions

For the function $f (x) = sin x + 3 x - \frac{2}{π} (x^{2} + x)$ , where $x \in [0, \frac{π}{2}]$ , consider the following two statements : (I) $f$ is increasing in $(0, \frac{π}{2})$ . (II) $f^{'}$ is decreasing in $(0, \frac{π}{2})$ . Between the above two statements,

For the function $f (x) = (cos x) - x + 1, x \in R$ , between the following two statements (S1) $f (x) = 0$ for only one value of $x$ in $[0, π]$ . (S2) $f (x)$ is decreasing in $[0, \frac{π}{2}]$ and increasing in $[\frac{π}{2}, π]$ .

If $f (x) = \frac{x}{s i n x}$ and $g (x) = \frac{x}{t a n x}$ , where $0 < x \leq 1$ , then in this interval

Let $f (x) = x cos^{- 1} (- sin ∣ x ∣)$ , $x \in [- \frac{π}{2}, \frac{π}{2}]$ , then which of the following is true?

Let $g : (0, \infty) \to R$ be a differentiable function such that $\int (\frac{x ( c o s x - s i n x )}{e ^{x} + 1} + \frac{g ( x ) ( e ^{x} + 1 - x e ^{x} )}{( e ^{x} + 1 ) ^{2}}) d x = \frac{xg ( x )}{e ^{x} + 1} + c$ , for all $x > 0$ , where $c$ is an arbitrary constant. Then

In this questions there are entries in columns I and II. Each entry in column I is related to exactly one entry in column II. Write the correct letter from column II against the entry number in column I in your answer book. Let the functions defined in column I have domain $(- π /2, π /2)$

List-I

List-II

Let $f (x) = x cos^{- 1} (sin (- ∣ x ∣)), x \in (- \frac{π}{2}, \frac{π}{2})$ , then which of the following is true ?

Consider the following three statements for the function $f : (0, \infty) \to R$ defined by $f (x) = ∣ lo g_{e} x ∣ - ∣ x - 1∣$ : (I) $f$ is differentiable at all $x > 0$ . (II) $f$ is increasing in $(0, 1)$ . (III) $f$ is decreasing in $(1, \infty)$ . Then.

Select Answer:

Visualized Solution

Defining the Interval

Behavior of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinx

Monotonicity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinx

Behavior of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosx

Monotonicity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosx

Evaluating Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Decoding Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

The Counter-Example Setup

First Derivative of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Visualizing the Derivative

Second Derivative of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Conclusion for Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Final Answer

The Sigma Insight: Monotonicity

Analyzing the Setup

The Trap of Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Testing the Derivative's Trend

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=sinxx​ and g(x)=tanxx​, where 0<x≤1, then in this interval

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=xcos−1(−sin∣x∣), x∈[−2π​,2π​], then which of the following is true?

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=xcos−1(sin(−∣x∣)),x∈(−2π​,2π​), then which of the following is true ?

Behavior of $sin x$

Monotonicity of $sin x$

Behavior of $cos x$

Monotonicity of $cos x$

Evaluating Statement $S$

Decoding Statement $R$

First Derivative of $f (x)$

Second Derivative of $f (x)$

Conclusion for Statement $R$

The Trap of Statement $R$

If $f (x) = \frac{x}{s i n x}$ and $g (x) = \frac{x}{t a n x}$ , where $0 < x \leq 1$ , then in this interval

Let $f (x) = x cos^{- 1} (- sin ∣ x ∣)$ , $x \in [- \frac{π}{2}, \frac{π}{2}]$ , then which of the following is true?

Let $f (x) = x cos^{- 1} (sin (- ∣ x ∣)), x \in (- \frac{π}{2}, \frac{π}{2})$ , then which of the following is true ?