JEE Main 2026 — Mathematics Calculus

Given the function:

$$f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t \, dt$$f(x)=∫0x​tan(t−x)dt−∫0x​f(t)tantdt

First, we simplify the first integral. Let $u = x - t$u=x−t, then $du = -dt$du=−dt. When $t = 0$t=0, $u = x$u=x; when $t = x$t=x, $u = 0$u=0.

$$\int_{0}^{x} \tan(t-x) dt = \int_{x}^{0} \tan(-u) (-du) = \int_{x}^{0} \tan u \, du = -\int_{0}^{x} \tan u \, du = -[\ln(\sec u)]_{0}^{x} = -\ln(\sec x) = \ln(\cos x)$$∫0x​tan(t−x)dt=∫x0​tan(−u)(−du)=∫x0​tanudu=−∫0x​tanudu=−[ln(secu)]0x​=−ln(secx)=ln(cosx)

So, the equation becomes:

$$f(x) = \ln(\cos x) - \int_{0}^{x} f(t) \tan t \, dt$$f(x)=ln(cosx)−∫0x​f(t)tantdt

Differentiating both sides with respect to $x$x using the Leibniz rule:

$$f'(x) = \dfrac{-\sin x}{\cos x} - f(x) \tan x$$f′(x)=cosx−sinx​−f(x)tanx

$$f'(x) + f(x) \tan x = -\tan x$$f′(x)+f(x)tanx=−tanx

This is a linear first-order differential equation. The integrating factor (IF) is:

$$\text{IF} = e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x$$IF=e∫tanxdx=eln(secx)=secx

Multiplying the differential equation by $\sec x$secx:

$$f'(x) \sec x + f(x) \sec x \tan x = -\sec x \tan x$$f′(x)secx+f(x)secxtanx=−secxtanx

$$\dfrac{d}{dx} (f(x) \sec x) = -\sec x \tan x$$dxd​(f(x)secx)=−secxtanx

Integrating both sides with respect to $x$x:

$$f(x) \sec x = -\sec x + C$$f(x)secx=−secx+C

To find $C$C, we use the initial condition. From $\displaystyle f(x) = \ln(\cos x) - \int_{0}^{x} f(t) \tan t \, dt$f(x)=ln(cosx)−∫0x​f(t)tantdt, substituting $x = 0$x=0 gives $f(0) = \ln(1) - 0 = 0$f(0)=ln(1)−0=0.

$$0 \cdot \sec 0 = -\sec 0 + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$$0⋅sec0=−sec0+C⇒0=−1+C⇒C=1

Thus, $f(x) \sec x = 1 - \sec x$f(x)secx=1−secx

$$f(x) = \cos x - 1$$f(x)=cosx−1

Now, we find the required derivatives:

$$f'(x) = -\sin x$$f′(x)=−sinx

$$f''(x) = -\cos x$$f′′(x)=−cosx

Evaluating these at the given points:

$$f\left(\dfrac{\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right) - 1 = \dfrac{\sqrt{3}}{2} - 1$$f(6π​)=cos(6π​)−1=23​​−1

$$f'\left(-\dfrac{\pi}{6}\right) = -\sin\left(-\dfrac{\pi}{6}\right) = \sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$$f′(−6π​)=−sin(−6π​)=sin(6π​)=21​

$$f''\left(\dfrac{\pi}{6}\right) = -\cos\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2}$$f′′(6π​)=−cos(6π​)=−23​​

Finally, substituting these values into the required expression:

$$f''\left(\dfrac{\pi}{6}\right) + 12f'\left(-\dfrac{\pi}{6}\right) + f\left(\dfrac{\pi}{6}\right) = -\dfrac{\sqrt{3}}{2} + 12\left(\dfrac{1}{2}\right) + \left(\dfrac{\sqrt{3}}{2} - 1\right)$$f′′(6π​)+12f′(−6π​)+f(6π​)=−23​​+12(21​)+(23​​−1)

$$= -\dfrac{\sqrt{3}}{2} + 6 + \dfrac{\sqrt{3}}{2} - 1 = 5$$=−23​​+6+23​​−1=5

Answer: $5$5