JEE Advanced 2014

LEVELJEE Advanced

Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $f : [0, 4 π] \to [0, π]$ be defined by $f (x) = cos^{- 1} (cos x)$ . The number of points $x \in [0, 4 π]$ satisfying the equation $f (x) = \frac{10 - x}{10}$ is

Q: Wait, why isn't $\cos^{-1}(\cos x)$ just $x$ everywhere? If I just set $x = 1 - \frac{x}{10}$, I get $x = \frac{10}{11}$. Why is that wrong?

That is a classic trap! The identity $\cos^{-1}(\cos x) = x$ is only valid when $x$ is within the principal value branch, which is $[0, \pi]$. Outside this interval, the function $\cos^{-1}(\cos x)$ behaves like a periodic wave. You must define the function piecewise for each interval of length $\pi$ to account for these reflections.

Q: How do I determine the equations for the sawtooth wave without memorizing them?

You don't need to memorize them! Just look at the graph. In $[0, \pi]$, it's a line through $(0,0)$ and $(\pi, \pi)$, so $f(x) = x$. In $[\pi, 2\pi]$, it's a line through $(\pi, \pi)$ and $(2\pi, 0)$, which has a slope of $-1$ and passes through $(2\pi, 0)$, giving $f(x) = 2\pi - x$. You can derive each segment using the point-slope form of a line.

Q: Why do we stop checking after $x=10$?

Great observation! The function $f(x) = \cos^{-1}(\cos x)$ has a range of $[0, \pi]$, meaning it is always non-negative. The line $g(x) = 1 - \frac{x}{10}$ becomes negative for any $x > 10$. Since a non-negative function cannot equal a negative value, there can be no intersection points for $x > 10$.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Do not be fooled by the identity

cos^{- 1} (cos x) = x

. This is only true for

x \in [0, π]

. Outside this interval, the function reflects to maintain its range

[0, π]

. Visualize the graph of

f (x) = cos^{- 1} (cos x)

as a periodic sawtooth wave oscillating between

0

and

π

Hint 2

Hint 2 (The Arsenal): To solve this, partition the domain

[0, 4 π]

into four segments:

[0, π], [π, 2 π], [2 π, 3 π],

and

[3 π, 4 π]

. In each segment, the function

f (x)

is a linear expression with a slope of either

1

- 1

Hint 3

Hint 3 (The Execution): For each segment, set the linear expression equal to

1 - \frac{x}{10}

. Solve for

x

and verify if the solution lies within the current segment's boundaries. Remember, the line

g (x) = 1 - \frac{x}{10}

crosses the x-axis at

x = 10

, and for

x > 10

g (x) < 0

, which is impossible for the range of

cos^{- 1}

Visualized Solution

Understanding $f (x) = cos^{- 1} (cos x)$

Given function: $f (x) = cos^{- 1} (cos x)$
Domain: $x \in [0, 4 π]$
Range: $y \in [0, π]$

Graphing the Sawtooth Wave

The function $f (x)$ is periodic with a period of $2 π$ .
It forms a continuous "sawtooth" wave.

Piecewise Definition of $f (x)$

For $x \in [0, π]$ , $f (x) = x$
For $x \in [π, 2 π]$ , $f (x) = 2 π - x$
For $x \in [2 π, 3 π]$ , $f (x) = x - 2 π$
For $x \in [3 π, 4 π]$ , $f (x) = 4 π - x$

The Linear Equation $g (x)$

Equation to solve: $f (x) = \frac{10 - x}{10}$
Let the right hand side be $g (x) = 1 - \frac{x}{10}$

Graphing the Line $g (x)$

At $x = 0$ , $g (0) = 1$
At $x = 10$ , $g (10) = 0$
Note: $3 π \approx 9.42$ , so $x = 10$ is slightly after $3 π$ .

Intersection in $[0, π]$

In the first branch: $f (x) = x$
Set $x = 1 - \frac{x}{10}$
$⟹ x + \frac{x}{10} = 1 ⟹ \frac{11 x}{10} = 1$
$⟹ x = \frac{10}{11} \approx 0.91$

Intersection in $[π, 2 π]$

In the second branch: $f (x) = 2 π - x$
Set $2 π - x = 1 - \frac{x}{10}$
$⟹ x - \frac{x}{10} = 2 π - 1$
$⟹ \frac{9 x}{10} = 2 π - 1 ⟹ x \approx 5.87$

Intersection in $[2 π, 3 π]$

In the third branch: $f (x) = x - 2 π$
Set $x - 2 π = 1 - \frac{x}{10}$
$⟹ x + \frac{x}{10} = 2 π + 1$
$⟹ \frac{11 x}{10} = 2 π + 1 ⟹ x \approx 6.62$

Checking the Interval $[3 π, 4 π]$

For $x > 10$ , the line $g (x) = 1 - \frac{x}{10} < 0$
The function $f (x) \geq 0$ for all $x$ .
Therefore, no intersection is possible in $[3 π, 4 π]$ .

Final Count of Solutions

The graphs intersect at exactly 3 points.
The solutions are $x \approx 0.91, 5.87, 6.62$ .
Final Answer: 3

00:00 / 00:00

The Sigma Insight: Properties of Inverse Trigonometric Functions

Solution Diagram

Analyzing the Setup

The problem asks us to find the number of points

x \in [0, 4 π]

where the function

f (x) = cos^{- 1} (cos x)

intersects the line

g (x) = \frac{10 - x}{10}

Many students mistakenly assume

cos^{- 1} (cos x) = x

for all

x

. However, the range of

cos^{- 1}

is strictly

[0, π]

, which forces the function to behave as a periodic sawtooth wave.

Visualizing the Sawtooth

Because the function reflects to remain within the range

[0, π]

, we define

f (x)

across the domain

[0, 4 π]

as follows:

f (x) = ⎩ ⎨ ⎧ x 2 π - x x - 2 π 4 π - x x \in [0, π] x \in [π, 2 π] x \in [2 π, 3 π] x \in [3 π, 4 π]

The Linear Intruder

The line

g (x) = 1 - \frac{x}{10}

starts at

g (0) = 1

and terminates at the

x

-axis when

g (10) = 0

Since

3 π \approx 9.42

and

4 π \approx 12.56

, the point

x = 10

falls within the final interval

[3 π, 4 π]

. We must determine how many times this line intersects the sawtooth branches.

The Systematic Hunt

We evaluate the intersections branch by branch:

1. In $[0, π]$ : Set

x = 1 - \frac{x}{10} \Rightarrow \frac{11 x}{10} = 1 \Rightarrow x = \frac{10}{11} \approx 0.91

. Since

0.91 \in [0, π]

, this is a valid intersection.

2. In $[π, 2 π]$ : Set

2 π - x = 1 - \frac{x}{10} \Rightarrow \frac{9 x}{10} = 2 π - 1 \Rightarrow x = \frac{10 ( 2 π - 1 )}{9} \approx 5.87

. Since

5.87 \in [π, 2 π]

, this is a valid intersection.

3. In $[2 π, 3 π]$ : Set

x - 2 π = 1 - \frac{x}{10} \Rightarrow \frac{11 x}{10} = 2 π + 1 \Rightarrow x = \frac{10 ( 2 π + 1 )}{11} \approx 6.62

. Since

6.28 < 6.62 < 9.42

, this is a valid intersection.

4. In $[3 π, 4 π]$ : The line

g (x)

crosses the

x

-axis at

x = 10

. For any

x > 10

g (x) < 0

, while

f (x) \geq 0

for all

x

. Therefore, no further intersections exist in this interval.

Final Conclusion

By systematically analyzing each branch of the sawtooth wave, we have identified exactly 3 points of intersection.

Visualizing the function allows us to bypass complex algebraic traps and clearly observe the geometric reality of the problem.

Common Traps & FAQs

Wait, why isn't

cos^{- 1} (cos x)

just

x

everywhere? If I just set

x = 1 - \frac{x}{10}

, I get

x = \frac{10}{11}

. Why is that wrong?

That is a classic trap! The identity

cos^{- 1} (cos x) = x

is only valid when

x

is within the principal value branch, which is

[0, π]

. Outside this interval, the function

cos^{- 1} (cos x)

behaves like a periodic wave. You must define the function piecewise for each interval of length

π

to account for these reflections.

How do I determine the equations for the sawtooth wave without memorizing them?

You don't need to memorize them! Just look at the graph. In

[0, π]

, it's a line through

(0, 0)

and

(π, π)

, so

f (x) = x

. In

[π, 2 π]

, it's a line through

(π, π)

and

(2 π, 0)

, which has a slope of

- 1

and passes through

(2 π, 0)

, giving

f (x) = 2 π - x

. You can derive each segment using the point-slope form of a line.

Why do we stop checking after

x = 10

Great observation! The function

f (x) = cos^{- 1} (cos x)

has a range of

[0, π]

, meaning it is always non-negative. The line

g (x) = 1 - \frac{x}{10}

becomes negative for any

x > 10

. Since a non-negative function cannot equal a negative value, there can be no intersection points for

x > 10

Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $f : [0, 4 π] \to [0, π]$ be defined by $f (x) = cos^{- 1} (cos x)$ . The number of points $x \in [0, 4 π]$ satisfying the equation $f (x) = \frac{10 - x}{10}$ is

Enter Numerical Value:

Visualized Solution

Understanding $f (x) = cos^{- 1} (cos x)$

Graphing the Sawtooth Wave

Piecewise Definition of $f (x)$

The Linear Equation $g (x)$

Graphing the Line $g (x)$

Intersection in $[0, π]$

Intersection in $[π, 2 π]$

Intersection in $[2 π, 3 π]$

Checking the Interval $[3 π, 4 π]$

Final Count of Solutions

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

Visualizing the Sawtooth

The Linear Intruder

The Systematic Hunt

Final Conclusion

Similar Questions

For $x \in (- 1, 1]$ , the number of solutions of the equation $sin^{- 1} x = 2 tan^{- 1} x$ is equal to _________.

Let $tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2})$ , for $x \in R$ . Then the number of real solutions of the equation $1 + cos (2 x) = 2 tan^{- 1} (tan x)$ in the set $(- \frac{3 π}{2}, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{2}) \cup (\frac{π}{2}, \frac{3 π}{2})$ is equal to

Let $f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x})), 0 < x < 1$ . Then :

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

Let $S = {x \in R : 0 < x < 1 and 2 tan^{- 1} (\frac{1 - x}{1 + x}) = cos^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})}$ . If $n (S)$ denotes the number of elements in $S$ then :

For any $y \in R$ , let $cot^{- 1} (y) \in (0, π)$ and $tan^{- 1} (y) \in (- \frac{π}{2}, \frac{π}{2})$ . Then the sum of all the solutions of the equation $tan^{- 1} (\frac{6 y}{9 - y ^{2}}) + cot^{- 1} (\frac{9 - y ^{2}}{6 y}) = \frac{2 π}{3}$ for $0 < ∣ y ∣ < 3$ , is equal to

If the sum of all the solutions of $tan^{- 1} (\frac{2 x}{1 - x ^{2}}) + cot^{- 1} (\frac{1 - x ^{2}}{2 x}) = \frac{π}{3}$ , $- 1 < x < 1, x \neq = 0$ , is $α - \frac{4}{3}$ , then $α$ is equal to

The sum of the absolute maximum and absolute minimum values of the function $f (x) = tan^{- 1} (sin x - cos x)$ in the interval $[0, π]$ is

If $cos^{- 1} (\frac{2}{3 x}) + cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2} (x > \frac{3}{4})$ , then x is equal to :

Enter Numerical Value:

Visualized Solution

Understanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)=cos−1(cosx)

Graphing the Sawtooth Wave

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

The Linear Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Graphing the Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Intersection in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,π]

Intersection in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [π,2π]

Intersection in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [2π,3π]

Checking the Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [3π,4π]

Final Count of Solutions

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

Visualizing the Sawtooth

The Linear Intruder

The Systematic Hunt

Final Conclusion

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=cos(2tan−1sin(cot−1x1−x​​)),0<x<1. Then :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of all the solutions of tan−1(1−x22x​)+cot−1(2x1−x2​)=3π​, −1<x<1,x=0, is α−3​4​, then α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the absolute maximum and absolute minimum values of the function f(x)=tan−1(sinx−cosx) in the interval [0,π] is

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Understanding $f (x) = cos^{- 1} (cos x)$

Piecewise Definition of $f (x)$

The Linear Equation $g (x)$

Graphing the Line $g (x)$

Intersection in $[0, π]$

Intersection in $[π, 2 π]$

Intersection in $[2 π, 3 π]$

Checking the Interval $[3 π, 4 π]$

For $x \in (- 1, 1]$ , the number of solutions of the equation $sin^{- 1} x = 2 tan^{- 1} x$ is equal to _________.

Let $f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x})), 0 < x < 1$ . Then :

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

Let $S = {x \in R : 0 < x < 1 and 2 tan^{- 1} (\frac{1 - x}{1 + x}) = cos^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})}$ . If $n (S)$ denotes the number of elements in $S$ then :

If the sum of all the solutions of $tan^{- 1} (\frac{2 x}{1 - x ^{2}}) + cot^{- 1} (\frac{1 - x ^{2}}{2 x}) = \frac{π}{3}$ , $- 1 < x < 1, x \neq = 0$ , is $α - \frac{4}{3}$ , then $α$ is equal to

The sum of the absolute maximum and absolute minimum values of the function $f (x) = tan^{- 1} (sin x - cos x)$ in the interval $[0, π]$ is

If $cos^{- 1} (\frac{2}{3 x}) + cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2} (x > \frac{3}{4})$ , then x is equal to :