JEE Main 2021 (26 Aug Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x})), 0 < x < 1$ . Then :

Q: Why can't we just differentiate the whole expression directly using the chain rule? It seems like a lot of work to simplify it first.

That is a classic trap! While theoretically possible, differentiating a nested function like $\cos(2 \tan^{-1} \sin(\dots))$ directly would lead to a monstrous algebraic expression that is almost impossible to simplify into the given options. In JEE Advanced, the 'simplification first' strategy is almost always the intended path. Always look for a way to reduce the function to its simplest form before applying calculus.

Q: How do we know that $\cos(\cos^{-1}(y)) = y$? What about the domain?

Excellent observation! You are right to be cautious. The identity $\cos(\cos^{-1}(y)) = y$ holds true only when $y$ is in the domain $[-1, 1]$. Since we are given $0 < x < 1$, the expression $\frac{1-x}{1+x}$ will always lie between $0$ and $1$, which is well within the valid domain. Therefore, the simplification is mathematically rigorous.

Q: I got stuck at $2 \tan^{-1} \sqrt{x}$. How did you know to convert it to $\cos^{-1}$ instead of $\sin^{-1}$ or $\tan^{-1}$?

Look at the outer function! We had $f(x) = \cos(2 \tan^{-1} \sqrt{x})$. If we convert the inner part to a $\cos^{-1}$ form, the $\cos$ and $\cos^{-1}$ will cancel each other out perfectly. Always choose the inverse identity that matches the outer trigonometric function to create this 'cancellation' effect.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Do not be intimidated by the 'Russian Doll' structure of the function. In problems involving nested inverse trigonometric functions, the key is to start from the innermost term and work your way outwards. Treat the innermost expression as a single variable, say

θ

Hint 2

Hint 2: Once you have defined

θ = cot^{- 1} \frac{1 - x}{x}

, visualize it geometrically. Construct a right-angled triangle where the ratio of the base to the perpendicular is defined by your expression. This will allow you to convert the inverse trigonometric function into a simple algebraic ratio.

Hint 3

Hint 3: After simplifying the inner part to

sin θ = x

, you will be left with

f (x) = cos (2 tan^{- 1} x)

. Recall the standard identity

2 tan^{- 1} A = cos^{- 1} (\frac{1 - A ^{2}}{1 + A ^{2}})

. This is the bridge that will collapse the entire expression into a simple rational function.

Visualized Solution

$A na l y z in g t h e I nn er m os tT er m$

Given: $f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x}))$
Let's isolate the innermost expression: $θ = cot^{- 1} \frac{1 - x}{x}$
This implies $cot θ = \frac{1 - x}{x}$

$G eo m e t r i c R e p r ese n t a t i o n$

We know $cot θ = \frac{Base}{Perpendicular}$
Let Base $= 1 - x$
Let Perpendicular $= x$

$F in d in g t h eH y p o t e n u se$

Using Pythagoras theorem: $Hypotenuse^{2} = Base^{2} + Perpendicular^{2}$
$Hypotenuse^{2} = (1 - x)^{2} + (x)^{2}$
$Hypotenuse^{2} = 1 - x + x = 1$
Therefore, $Hypotenuse = 1$

$E v a l u a t in g t h e S in e T er m$

From the triangle, $sin θ = \frac{Perpendicular}{Hypotenuse}$
$sin θ = \frac{x}{1} = x$
Thus, $sin (cot^{- 1} \frac{1 - x}{x}) = x$

$U p d a t in g t h e M ain F u n c t i o n$

Substitute the simplified term back into the original function.
$f (x) = cos (2 tan^{- 1} x)$
Now we need to simplify $2 tan^{- 1} x$ .

$A ppl y in g I n v er se T r i g I d e n t i t y$

Recall the identity: $2 tan^{- 1} A = cos^{- 1} (\frac{1 - A ^{2}}{1 + A ^{2}})$
Here, $A = x$
$2 tan^{- 1} x = cos^{- 1} (\frac{1 - ( x ) ^{2}}{1 + ( x ) ^{2}})$

$F ina l F or m o f f (x)$

$2 tan^{- 1} x = cos^{- 1} (\frac{1 - x}{1 + x})$
Substitute this back: $f (x) = cos (cos^{- 1} (\frac{1 - x}{1 + x}))$
Since $0 < x < 1$ , the domain is valid, so $f (x) = \frac{1 - x}{1 + x}$

$A ppl y in g t h e Q u o t i e n tR u l e$

We need to find $f^{'} (x)$ for the differential equation.
$f (x) = \frac{1 - x}{1 + x}$
Using Quotient Rule: $(\frac{u}{v})^{'} = \frac{v u ^{'} - u v ^{'}}{v ^{2}}$
$f^{'} (x) = \frac{( 1 + x ) \frac{d}{d x} ( 1 - x ) - ( 1 - x ) \frac{d}{d x} ( 1 + x )}{( 1 + x ) ^{2}}$

$E v a l u a t in g f^{'} (x)$

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

$M a t c hin g w i t h t h e O pt i o n s$

We have $f^{'} (x) = \frac{- 2}{( 1 + x ) ^{2}}$ and $f (x) = \frac{1 - x}{1 + x}$
Let's look at the options. They involve $(1 - x)^{2} f^{'} (x)$ or $(1 + x)^{2} f^{'} (x)$ .
Let's multiply $f^{'} (x)$ by $(1 - x)^{2}$ :
$(1 - x)^{2} f^{'} (x) = (1 - x)^{2} (\frac{- 2}{( 1 + x ) ^{2}})$

$T h e F ina l D i f f er e n t ia l E q u a t i o n$

Notice that $(\frac{1 - x}{1 + x})^{2} = (f (x))^{2}$
So, $(1 - x)^{2} f^{'} (x) = - 2 (\frac{1 - x}{1 + x})^{2} = - 2 (f (x))^{2}$
Rearranging gives: $(1 - x)^{2} f^{'} (x) + 2 (f (x))^{2} = 0$
This perfectly matches Option 3.

00:00 / 00:00

The Sigma Insight: Properties of Inverse Trigonometric Functions

Solution Diagram

Analyzing the Setup

The given function is:

f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x}))

At first glance, this appears to be a nightmare of nested functions. However, in the context of JEE Advanced, such complexity is often a mask for simplicity. We treat this as a "Russian Doll" problem, peeling it layer by layer starting from the innermost term.

The Geometric Core

Let us define the innermost term as

θ = cot^{- 1} \frac{1 - x}{x}

. This implies:

cot θ = \frac{1 - x}{x}

Now, visualize a right-angled triangle where the base is

1 - x

and the perpendicular is

x

. By the Pythagorean theorem, the hypotenuse is:

h = (1 - x)^{2} + (x)^{2} = 1 - x + x = 1

Since the hypotenuse is

1

, we find that

sin θ = \frac{Perpendicular}{Hypotenuse} = x

. The inner expression has now simplified significantly, reducing our function to:

f (x) = cos (2 tan^{- 1} x)

The Transformation

We now address the term

2 tan^{- 1} x

. We recall the standard trigonometric identity:

2 tan^{- 1} A = cos^{- 1} (\frac{1 - A ^{2}}{1 + A ^{2}})

Substituting

A = x

into this identity, we obtain:

2 tan^{- 1} x = cos^{- 1} (\frac{1 - x}{1 + x})

Consequently, our function becomes:

f (x) = cos (cos^{- 1} (\frac{1 - x}{1 + x}))

Given the domain

0 < x < 1

, the inverse trigonometric functions cancel out perfectly. This leaves us with the simplified algebraic form:

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

The Calculus Finale

To find the derivative

f^{'} (x)

, we apply the quotient rule:

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

To relate this to the required differential equation, we observe the structure of the result. By manipulating the expression, we arrive at the final relationship:

(1 - x)^{2} f^{'} (x) + 2 (f (x))^{2} = 0

The beast is conquered. The function simplifies to a rational expression, and its derivative follows directly from standard calculus rules.

Common Traps & FAQs

Why can't we just differentiate the whole expression directly using the chain rule? It seems like a lot of work to simplify it first.

That is a classic trap! While theoretically possible, differentiating a nested function like

cos (2 tan^{- 1} sin (\dots))

directly would lead to a monstrous algebraic expression that is almost impossible to simplify into the given options. In JEE Advanced, the 'simplification first' strategy is almost always the intended path. Always look for a way to reduce the function to its simplest form before applying calculus.

How do we know that

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

? What about the domain?

Excellent observation! You are right to be cautious. The identity

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

holds true only when

y

is in the domain

[- 1, 1]

. Since we are given

0 < x < 1

, the expression

\frac{1 - x}{1 + x}

will always lie between

0

and

1

, which is well within the valid domain. Therefore, the simplification is mathematically rigorous.

I got stuck at

2 tan^{- 1} x

. How did you know to convert it to

cos^{- 1}

instead of

sin^{- 1}

tan^{- 1}

Look at the outer function! We had

f (x) = cos (2 tan^{- 1} x)

. If we convert the inner part to a

cos^{- 1}

form, the

cos

and

cos^{- 1}

will cancel each other out perfectly. Always choose the inverse identity that matches the outer trigonometric function to create this 'cancellation' effect.

Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x})), 0 < x < 1$ . Then :

Select Answer:

Visualized Solution

$A na l y z in g t h e I nn er m os tT er m$

$G eo m e t r i c R e p r ese n t a t i o n$

$F in d in g t h eH y p o t e n u se$

$E v a l u a t in g t h e S in e T er m$

$U p d a t in g t h e M ain F u n c t i o n$

$A ppl y in g I n v er se T r i g I d e n t i t y$

$F ina l F or m o f f (x)$

$A ppl y in g t h e Q u o t i e n tR u l e$

$E v a l u a t in g f^{'} (x)$

$M a t c hin g w i t h t h e O pt i o n s$

$T h e F ina l D i f f er e n t ia l E q u a t i o n$

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Geometric Core

The Transformation

The Calculus Finale

Similar Questions

$1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

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

Prove that $cos tan^{- 1} sin cot^{- 1} x = \frac{x ^{2} + 1}{x ^{2} + 2}$ .

If $y (x) = cot^{- 1} (\frac{1 + s i n x + 1 - s i n x}{1 + s i n x - 1 - s i n x})$ , $x \in (\frac{π}{2}, π)$ , then $\frac{d y}{d x}$ at $x = \frac{5 π}{6}$ is:

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 $tan^{- 1} y = tan^{- 1} x + tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $∣ x ∣ < \frac{1}{3}$ . Then a value of $y$ is :

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Animated Solution for Mathematics - Inverse Trigonometric Functions: .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 :

Select Answer:

Visualized Solution

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

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

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

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

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

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

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

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

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

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

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

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Geometric Core

The Transformation

The Calculus Finale

Similar Questions

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Animated Solution for Mathematics - Inverse Trigonometric Functions: Let $f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x})), 0 < x < 1$ . Then :

$A na l y z in g t h e I nn er m os tT er m$

$G eo m e t r i c R e p r ese n t a t i o n$

$F in d in g t h eH y p o t e n u se$

$E v a l u a t in g t h e S in e T er m$

$U p d a t in g t h e M ain F u n c t i o n$

$A ppl y in g I n v er se T r i g I d e n t i t y$

$F ina l F or m o f f (x)$

$A ppl y in g t h e Q u o t i e n tR u l e$

$E v a l u a t in g f^{'} (x)$

$M a t c hin g w i t h t h e O pt i o n s$

$T h e F ina l D i f f er e n t ia l E q u a t i o n$

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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

Prove that $cos tan^{- 1} sin cot^{- 1} x = \frac{x ^{2} + 1}{x ^{2} + 2}$ .

If $y (x) = cot^{- 1} (\frac{1 + s i n x + 1 - s i n x}{1 + s i n x - 1 - s i n x})$ , $x \in (\frac{π}{2}, π)$ , then $\frac{d y}{d x}$ at $x = \frac{5 π}{6}$ is:

Let $tan^{- 1} y = tan^{- 1} x + tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $∣ x ∣ < \frac{1}{3}$ . Then a value of $y$ is :

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

If $(sin^{- 1} x)^{2} - (cos^{- 1} x)^{2} = a; 0 < x < 1, a \neq = 0$ , then the value of $2 x^{2} - 1$ is :

The value of $cot^{- 1} (\frac{1 + t a n ^{2} ( 2 ) - 1}{t a n ( 2 )}) - cot^{- 1} (\frac{1 + t a n ^{2} ( \frac{1}{2} ) + 1}{t a n ( \frac{1}{2} )})$ 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