JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be a twice differentiable function such that $(sin x cos y) (f (2 x + 2 y) - f (2 x - 2 y)) = (cos x sin y) (f (2 x + 2 y) + f (2 x - 2 y))$ for all $x, y \in R$ . If $f^{'} (0) = 1/2$ , then the value of $24 f^{''} (5 π /3)$ is:

Q: Why did the negative sign disappear when we used the sine identity?

Great catch! Remember that $\sin(\theta)$ is an odd function, meaning $\sin(-\theta) = -\sin(\theta)$. When we had $\sin(y-x)$ in the denominator, we rewrote it as $-\sin(x-y)$. This negative sign in the denominator perfectly cancelled out the negative sign we obtained from the Componendo and Dividendo step on the left side. It is all about keeping track of those signs!

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Observe the structure of the functional equation. It involves

f (2 x + 2 y)

and

f (2 x - 2 y)

on both sides. This is a classic setup for a functional equation that can be reduced to a simpler form by separating the variables

x

and

y

. Focus on isolating the function terms on one side and the trigonometric terms on the other.

Hint 2

Hint 2 (The Arsenal): Recall the algebraic tool of Componendo and Dividendo: if

\frac{A - B}{A + B} = \frac{C}{D}

, then

\frac{A}{B} = \frac{C + D}{D - C}

. Also, keep your compound angle identities for sine and cosine ready—they are the key to simplifying the trigonometric expression on the right-hand side.

Hint 3

Hint 3 (The Execution): Start by rearranging the equation to

\frac{f ( 2 x + 2 y ) - f ( 2 x - 2 y )}{f ( 2 x + 2 y ) + f ( 2 x - 2 y )} = \frac{t a n y}{t a n x}

. Apply Componendo and Dividendo to both sides, and then use the identity

sin (A + B) = sin A cos B + cos A sin B

to simplify the resulting expression into a form that allows you to identify the function

f (x)

Visualized Solution

Analyzing the Functional Equation

Given: $(sin x cos y) (f (2 x + 2 y) - f (2 x - 2 y)) = (cos x sin y) (f (2 x + 2 y) + f (2 x - 2 y))$
Goal: Find $24 f^{''} (\frac{5 π}{3})$ given $f^{'} (0) = \frac{1}{2}$ .

Separating Variables

Group $f$ terms on one side and trigonometric terms on the other.
$\frac{f ( 2 x + 2 y ) - f ( 2 x - 2 y )}{f ( 2 x + 2 y ) + f ( 2 x - 2 y )} = \frac{c o s x s i n y}{s i n x c o s y}$

Simplifying the Trigonometric Ratio

Rewrite the right side using $tan$ :
$\frac{c o s x s i n y}{s i n x c o s y} = (\frac{s i n y}{c o s y}) \cdot (\frac{c o s x}{s i n x})$
$\frac{f ( 2 x + 2 y ) - f ( 2 x - 2 y )}{f ( 2 x + 2 y ) + f ( 2 x - 2 y )} = \frac{t a n y}{t a n x}$

Applying Componendo and Dividendo

Apply Componendo and Dividendo: $\frac{A - B}{A + B} = \frac{C}{D} ⟹ \frac{A}{B} = \frac{C + D}{C - D}$
Apply to: $\frac{f ( 2 x + 2 y ) - f ( 2 x - 2 y )}{f ( 2 x + 2 y ) + f ( 2 x - 2 y )} = \frac{t a n y}{t a n x}$
$\frac{( f ( 2 x + 2 y ) - f ( 2 x - 2 y )) + ( f ( 2 x + 2 y ) + f ( 2 x - 2 y ))}{( f ( 2 x + 2 y ) - f ( 2 x - 2 y )) - ( f ( 2 x + 2 y ) + f ( 2 x - 2 y ))} = \frac{t a n y + t a n x}{t a n y - t a n x}$

Simplifying the Left Side

Numerator: $2 f (2 x + 2 y)$
Denominator: $- 2 f (2 x - 2 y)$
Left side becomes: $- \frac{f ( 2 x + 2 y )}{f ( 2 x - 2 y )}$
Equation: $- \frac{f ( 2 x + 2 y )}{f ( 2 x - 2 y )} = \frac{t a n y + t a n x}{t a n y - t a n x}$

Simplifying the Right Side

Right side: $\frac{t a n y + t a n x}{t a n y - t a n x}$
Convert to sine and cosine: $\frac{\frac{s i n y}{c o s y} + \frac{s i n x}{c o s x}}{\frac{s i n y}{c o s y} - \frac{s i n x}{c o s x}}$
Multiply numerator and denominator by $cos x cos y$ :
$\frac{s i n y c o s x + c o s y s i n x}{s i n y c o s x - c o s y s i n x}$

Applying Compound Angle Formulas

Numerator: $sin y cos x + cos y sin x = sin (y + x)$
Denominator: $sin y cos x - cos y sin x = sin (y - x)$
Right side becomes: $\frac{s i n ( y + x )}{s i n ( y - x )}$
Equation: $- \frac{f ( 2 x + 2 y )}{f ( 2 x - 2 y )} = \frac{s i n ( x + y )}{s i n ( y - x )}$

Absorbing the Negative Sign

We have: $- \frac{f ( 2 x + 2 y )}{f ( 2 x - 2 y )} = \frac{s i n ( x + y )}{s i n ( y - x )}$
Use the property of sine: $sin (y - x) = - sin (x - y)$
Substitute this into the denominator: $\frac{s i n ( x + y )}{- s i n ( x - y )}$
The negative signs on both sides cancel out:
$\frac{f ( 2 x + 2 y )}{f ( 2 x - 2 y )} = \frac{s i n ( x + y )}{s i n ( x - y )}$

Substituting Variables

Let $u = 2 x + 2 y$ and $v = 2 x - 2 y$ .
Then $x + y = \frac{u}{2}$ and $x - y = \frac{v}{2}$ .
Substitute these into the equation:
$\frac{f ( u )}{f ( v )} = \frac{s i n ( \frac{u}{2} )}{s i n ( \frac{v}{2} )}$

Deducing the Function $f (x)$

Rearrange: $\frac{f ( u )}{s i n ( \frac{u}{2} )} = \frac{f ( v )}{s i n ( \frac{v}{2} )}$
Since $u$ and $v$ are independent, each side must equal a constant $k$ .
$\frac{f ( x )}{s i n ( \frac{x}{2} )} = k ⟹ f (x) = k sin (\frac{x}{2})$

Finding the Constant $k$

We are given $f^{'} (0) = \frac{1}{2}$ .
First, find the derivative $f^{'} (x)$ :
$f^{'} (x) = \frac{d}{d x} (k sin (\frac{x}{2})) = k \cdot \frac{1}{2} cos (\frac{x}{2})$
Substitute $x = 0$ : $f^{'} (0) = \frac{k}{2} cos (0) = \frac{k}{2}$

Solving for $k$

Equate to the given value: $\frac{k}{2} = \frac{1}{2}$
Solve for $k$ : $k = 1$
Therefore, the exact function is $f (x) = sin (\frac{x}{2})$

Calculating the Second Derivative

We need $f^{''} (x)$ .
We know $f^{'} (x) = \frac{1}{2} cos (\frac{x}{2})$ (since $k = 1$ ).
Differentiate again with respect to $x$ :
$f^{''} (x) = \frac{1}{2} \cdot (- sin (\frac{x}{2})) \cdot \frac{1}{2} = - \frac{1}{4} sin (\frac{x}{2})$

Evaluating at $x = \frac{5 π}{3}$

Substitute $x = \frac{5 π}{3}$ into $f^{''} (x)$ :
$f^{''} (\frac{5 π}{3}) = - \frac{1}{4} sin (\frac{1}{2} \cdot \frac{5 π}{3})$
$f^{''} (\frac{5 π}{3}) = - \frac{1}{4} sin (\frac{5 π}{6})$

Final Calculation

Calculate $sin (\frac{5 π}{6}) = sin (π - \frac{π}{6}) = sin (\frac{π}{6}) = \frac{1}{2}$
So, $f^{''} (\frac{5 π}{3}) = - \frac{1}{4} \cdot \frac{1}{2} = - \frac{1}{8}$
We need $24 f^{''} (\frac{5 π}{3}) = 24 \cdot (- \frac{1}{8}) = - 3$
Final Answer: $- 3$

00:00 / 00:00

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The given functional equation is:

(sin x cos y) (f (2 x + 2 y) - f (2 x - 2 y)) = (cos x sin y) (f (2 x + 2 y) + f (2 x - 2 y))

Our objective is to determine the value of

24 f^{''} (5 π /3)

given the initial condition

f^{'} (0) = 1/2

The Art of Separation

To simplify, we group the function terms on the left and the trigonometric terms on the right. By cross-multiplying, we obtain:

\frac{f ( 2 x + 2 y ) - f ( 2 x - 2 y )}{f ( 2 x + 2 y ) + f ( 2 x - 2 y )} = \frac{cos x sin y}{sin x cos y}

The right side simplifies to the product of two ratios:

(\frac{s i n y}{c o s y}) \cdot (\frac{c o s x}{s i n x})

, which is equivalent to

\frac{t a n y}{t a n x}

. Thus, the equation becomes:

\frac{f ( 2 x + 2 y ) - f ( 2 x - 2 y )}{f ( 2 x + 2 y ) + f ( 2 x - 2 y )} = \frac{tan y}{tan x}

The Algebraic Key

The left side of the equation follows the form

\frac{A - B}{A + B} = \frac{C}{D}

. Applying the property of Componendo and Dividendo, we use the identity

\frac{A}{B} = \frac{C + D}{D - C}

(or equivalently,

\frac{A}{B} = \frac{D + C}{D - C}

with sign adjustments).

Applying this to our expression, the twos cancel out, yielding:

- \frac{f ( 2 x + 2 y )}{f ( 2 x - 2 y )} = \frac{tan y + tan x}{tan y - tan x}

The Trigonometric Collapse

We expand the tangents on the right side into sines and cosines:

\frac{tan y + tan x}{tan y - tan x} = \frac{\frac{s i n y}{c o s y} + \frac{s i n x}{c o s x}}{\frac{s i n y}{c o s y} - \frac{s i n x}{c o s x}}

Multiplying the numerator and denominator by

cos x cos y

results in the compound angle expansion:

\frac{sin y cos x + cos y sin x}{sin y cos x - cos y sin x} = \frac{sin ( x + y )}{sin ( y - x )}

Since

sin (y - x) = - sin (x - y)

, the negative signs on both sides of the equation cancel out. We are left with:

\frac{f ( 2 x + 2 y )}{f ( 2 x - 2 y )} = \frac{sin ( x + y )}{sin ( x - y )}

The Reveal

Let us introduce the substitution

u = 2 x + 2 y

and

v = 2 x - 2 y

. Consequently,

x + y = u /2

and

x - y = v /2

Substituting these into our equation gives:

\frac{f ( u )}{f ( v )} = \frac{sin ( u /2 )}{sin ( v /2 )}

This implies that

\frac{f ( u )}{s i n ( u /2 )} = \frac{f ( v )}{s i n ( v /2 )}

. Since

u

and

v

are independent variables, both sides must equal a constant

k

. Therefore, the function takes the form:

f (x) = k sin (x /2)

Final Calculation

We are given

f^{'} (0) = 1/2

. Differentiating

f (x) = k sin (x /2)

yields

f^{'} (x) = \frac{k}{2} cos (x /2)

x = 0

f^{'} (0) = k /2

. Equating this to

1/2

, we find

k = 1

, so

f (x) = sin (x /2)

The second derivative is

f^{''} (x) = - \frac{1}{4} sin (x /2)

. Evaluating at

x = 5 π /3

f^{''} (5 π /3) = - \frac{1}{4} sin (5 π /6) = - \frac{1}{4} \cdot \frac{1}{2} = - \frac{1}{8}

Finally, we calculate the requested value:

24 f^{''} (5 π /3) = 24 \cdot (- \frac{1}{8}) = - 3

The final answer is -3.

Common Traps & FAQs

Wait, why can we just assume

f (x) = k sin (x /2)

? Is that the only solution?

That is a brilliant question! When we arrive at the equation

\frac{f ( u )}{s i n ( u /2 )} = \frac{f ( v )}{s i n ( v /2 )}

, we have a function of

u

equal to a function of

v

. Since

u

and

v

are independent variables, the only way this equality can hold for all

u

and

v

is if both sides are equal to the same constant,

k

. This is a standard technique in functional equations. While there could theoretically be other solutions, the problem specifies that

f

is a twice-differentiable function, which restricts the possibilities to this specific form.

Why did the negative sign disappear when we used the sine identity?

Great catch! Remember that

sin (θ)

is an odd function, meaning

sin (- θ) = - sin (θ)

. When we had

sin (y - x)

in the denominator, we rewrote it as

- sin (x - y)

. This negative sign in the denominator perfectly cancelled out the negative sign we obtained from the Componendo and Dividendo step on the left side. It is all about keeping track of those signs!

How do I know when to use Componendo and Dividendo in these types of problems?

Look for the structure! Whenever you see a fraction involving a sum and a difference of the same two terms—like

\frac{A - B}{A + B}

—Componendo and Dividendo is almost always the intended path. It is designed to 'clean up' that specific structure, often leading to a much simpler ratio that reveals the underlying function.

Select Answer:

Visualized Solution

Analyzing the Functional Equation

Separating Variables

Simplifying the Trigonometric Ratio

Applying Componendo and Dividendo

Simplifying the Left Side

Simplifying the Right Side

Applying Compound Angle Formulas

Absorbing the Negative Sign

Substituting Variables

Deducing the Function $f (x)$

Finding the Constant $k$

Solving for $k$

Calculating the Second Derivative

Evaluating at $x = \frac{5 π}{3}$

Final Calculation

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The Art of Separation

The Algebraic Key

The Trigonometric Collapse

The Reveal

Final Calculation

Similar Questions

If $f (x) = {x^{3} sin (\frac{1}{x}), 0, x \neq = 0 x = 0$ then

Let $f (x) = \frac{s i n x + c o s x - 2}{s i n x - c o s x}, x \in [0, π] - {\frac{π}{4}}$ , then $f (\frac{7 π}{12}) f^{''} (\frac{7 π}{12})$ is equal to

Let $f : S \to S$ where $S = (0, \infty)$ be a twice differentiable function such that $f (x + 1) = x f (x)$ . If $g : S \to R$ be defined as $g (x) = lo g_{e} f (x)$ , then the value of $∣ g^{''} (5) - g^{''} (1) ∣$ is equal to :

Let $f (x) = x^{3} + x^{2} f^{'} (1) + 2 x f^{''} (2) + f^{'''} (3), x \in R$ . Then the value of $f^{'} (5)$ is:

If $y = y (x)$ and it follows the relation $x cos y + y cos x = π$ then $y^{''} (0) =$

Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equal :

Let $f$ and $g$ be twice differentiable functions on $R$ such that $f^{''} (x) = g^{''} (x) + 6 x$ , $f^{'} (1) = 4 g^{'} (1) - 3 = 9$ , $f (2) = 3 g (2) = 12$ . Then which of the following is NOT true?

If $y (θ) = \frac{2 c o s θ + c o s 2 θ}{c o s 3 θ + 4 c o s 2 θ + 5 c o s θ + 2}$ , then at $θ = \frac{π}{2}$ , $y^{''} + y^{'} + y$ is equal to :

Let $g (x) = lo g (f (x))$ where $f (x)$ is twice differentiable positive function on $(0, \infty)$ such that $f (x + 1) = x f (x)$ . Then, for $N = 1, 2, 3, \dots\dots\dots$ $g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =$

Let $f$ be a twice differentiable function such that $f^{''} (x) = - f (x)$ , and $f^{'} (x) = g (x), h (x) = [f (x)]^{2} + [g (x)]^{2}$ . Find $h (10)$ if $h (5) = 11$ .

Select Answer:

Visualized Solution

Analyzing the Functional Equation

Separating Variables

Simplifying the Trigonometric Ratio

Applying Componendo and Dividendo

Simplifying the Left Side

Simplifying the Right Side

Applying Compound Angle Formulas

Absorbing the Negative Sign

Substituting Variables

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

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

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

Calculating the Second Derivative

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=35π​

Final Calculation

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The Art of Separation

The Algebraic Key

The Trigonometric Collapse

The Reveal

Final Calculation

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Deducing the Function $f (x)$

Finding the Constant $k$

Solving for $k$

Evaluating at $x = \frac{5 π}{3}$

If $f (x) = {x^{3} sin (\frac{1}{x}), 0, x \neq = 0 x = 0$ then

Let $f (x) = \frac{s i n x + c o s x - 2}{s i n x - c o s x}, x \in [0, π] - {\frac{π}{4}}$ , then $f (\frac{7 π}{12}) f^{''} (\frac{7 π}{12})$ is equal to

Let $f (x) = x^{3} + x^{2} f^{'} (1) + 2 x f^{''} (2) + f^{'''} (3), x \in R$ . Then the value of $f^{'} (5)$ is:

If $y = y (x)$ and it follows the relation $x cos y + y cos x = π$ then $y^{''} (0) =$

Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equal :

Let $f$ and $g$ be twice differentiable functions on $R$ such that $f^{''} (x) = g^{''} (x) + 6 x$ , $f^{'} (1) = 4 g^{'} (1) - 3 = 9$ , $f (2) = 3 g (2) = 12$ . Then which of the following is NOT true?

If $y (θ) = \frac{2 c o s θ + c o s 2 θ}{c o s 3 θ + 4 c o s 2 θ + 5 c o s θ + 2}$ , then at $θ = \frac{π}{2}$ , $y^{''} + y^{'} + y$ is equal to :

Let $g (x) = lo g (f (x))$ where $f (x)$ is twice differentiable positive function on $(0, \infty)$ such that $f (x + 1) = x f (x)$ . Then, for $N = 1, 2, 3, \dots\dots\dots$ $g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =$

Let $f$ be a twice differentiable function such that $f^{''} (x) = - f (x)$ , and $f^{'} (x) = g (x), h (x) = [f (x)]^{2} + [g (x)]^{2}$ . Find $h (10)$ if $h (5) = 11$ .