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JEE Mathematics 2025 Question with Solution

If the function f(x)=tan⁡(tan⁡x)−sin⁡(sin⁡x)tan⁡x−sin⁡xf(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} is continuous at x=0x = 0, then f(0)f(0) is equal to:

Answer

Correct answer:2

Step-by-step solution

Taylor Series Method

Given:

f(x)=tan⁡(tan⁡x)−sin⁡(sin⁡x)tan⁡x−sin⁡xf(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}

and the function is continuous at x=0x = 0.

Find: f(0)f(0)

Since the function is continuous at x=0x = 0,

f(0)=lim⁡x→0tan⁡(tan⁡x)−sin⁡(sin⁡x)tan⁡x−sin⁡xf(0) = \lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}

Now use Taylor expansions near x=0x = 0:

tan⁡x=x+x33+O(x5)\tan x = x + \frac{x^3}{3} + O(x^5) sin⁡x=x−x36+O(x5)\sin x = x - \frac{x^3}{6} + O(x^5)

Let

u=tan⁡x=x+x33+O(x5)u = \tan x = x + \frac{x^3}{3} + O(x^5)

Then

tan⁡(tan⁡x)=tan⁡(ν)=ν+ν33+O(ν5)\tan(\tan x) = \tan(\nu) = \nu + \frac{\nu^3}{3} + O(\nu^5)

So,

tan⁡(tan⁡x)=x+x33+x33+O(x5)=x+2x33+O(x5)\tan(\tan x) = x + \frac{x^3}{3} + \frac{x^3}{3} + O(x^5) = x + \frac{2x^3}{3} + O(x^5)

Similarly, let

v=sin⁡x=x−x36+O(x5)v = \sin x = x - \frac{x^3}{6} + O(x^5)

Then

sin⁡(sin⁡x)=sin⁡(v)=v−v36+O(v5)\sin(\sin x) = \sin(v) = v - \frac{v^3}{6} + O(v^5)

Hence,

sin⁡(sin⁡x)=x−x36−x36+O(x5)=x−x33+O(x5)\sin(\sin x) = x - \frac{x^3}{6} - \frac{x^3}{6} + O(x^5) = x - \frac{x^3}{3} + O(x^5)

Also,

tan⁡x−sin⁡x=(x+x33+O(x5))−(x−x36+O(x5))\tan x - \sin x = \left(x + \frac{x^3}{3} + O(x^5)\right) - \left(x - \frac{x^3}{6} + O(x^5)\right) =x32+O(x5)= \frac{x^3}{2} + O(x^5)

Now the numerator becomes

tan⁡(tan⁡x)−sin⁡(sin⁡x)=(x+2x33+O(x5))−(x−x33+O(x5))\tan(\tan x) - \sin(\sin x) = \left(x + \frac{2x^3}{3} + O(x^5)\right) - \left(x - \frac{x^3}{3} + O(x^5)\right) =x3+O(x5)= x^3 + O(x^5)

Therefore,

lim⁡x→0f(x)=lim⁡x→0x3+O(x5)x32+O(x5)=2\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x^3 + O(x^5)}{\frac{x^3}{2} + O(x^5)} = 2

So,

f(0)=2f(0) = 2

Therefore, the required value is 22.

Using L'Hôpital's Rule

Given:

f(x)=tan⁡(tan⁡x)−sin⁡(sin⁡x)tan⁡x−sin⁡xf(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}

Find: f(0)f(0)

Because the function is continuous at x=0x = 0,

f(0)=lim⁡x→0tan⁡(tan⁡x)−sin⁡(sin⁡x)tan⁡x−sin⁡xf(0) = \lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}

At x=0x = 0, both numerator and denominator are 00, so the form is 00\frac{0}{0}.

Differentiate once:

ddx[tan⁡(tan⁡x)−sin⁡(sin⁡x)]=sec⁡2(tan⁡x)sec⁡2x−cos⁡(sin⁡x)cos⁡x\frac{d}{dx}\left[\tan(\tan x) - \sin(\sin x)\right] = \sec^2(\tan x)\sec^2 x - \cos(\sin x)\cos x ddx[tan⁡x−sin⁡x]=sec⁡2x−cos⁡x\frac{d}{dx}\left[\tan x - \sin x\right] = \sec^2 x - \cos x

Substituting x=0x = 0 again gives 00\frac{0}{0}, so the indeterminate form persists.

The provided working states that after differentiating again and simplifying, the limit evaluates to 22. Therefore,

f(0)=2f(0) = 2

So the required value is 22.

Common mistakes

  • Using only the first-order approximations tan⁡x≈x\tan x \approx x and sin⁡x≈x\sin x \approx x makes both numerator and denominator zero, which gives no information. Expand up to the x3x^3 terms because the leading non-zero behavior appears there.

  • Expanding tan⁡(tan⁡x)\tan(\tan x) as if it were just tan⁡x\tan x is incorrect. This is a composite function, so first expand the inner function and then substitute into the series for the outer function.

  • While using L'Hôpital's Rule, stopping after one differentiation is wrong because the form is still 00\frac{0}{0} at x=0x = 0. Differentiate again or switch to Taylor series to resolve the indeterminate form completely.

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