JEE Main 2026 — Mathematics Calculus

Given $\displaystyle f(x) = \int_1^x f(t)\,dt + (1-x)(\log_e x - 1) + e$f(x)=∫1x​f(t)dt+(1−x)(loge​x−1)+e

Substituting $x = 1$x=1 in the given equation:

$$f(1) = \int_1^1 f(t)\,dt + (1-1)(\log_e 1 - 1) + e$$f(1)=∫11​f(t)dt+(1−1)(loge​1−1)+e

$$f(1) = 0 + 0 + e = e$$f(1)=0+0+e=e

Differentiating the given equation with respect to $x$x using the Leibniz rule:

$$f'(x) = f(x) + \dfrac{d}{dx}[(1-x)(\log_e x - 1)]$$f′(x)=f(x)+dxd​[(1−x)(loge​x−1)]

$$f'(x) = f(x) + (-1)(\log_e x - 1) + (1-x)\left(\dfrac{1}{x}\right)$$f′(x)=f(x)+(−1)(loge​x−1)+(1−x)(x1​)

$$f'(x) = f(x) - \log_e x + 1 + \dfrac{1}{x} - 1$$f′(x)=f(x)−loge​x+1+x1​−1

$$f'(x) - f(x) = \dfrac{1}{x} - \log_e x$$f′(x)−f(x)=x1​−loge​x

This is a linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$dxdy​+Py=Q, where $P = -1$P=−1 and $Q = \dfrac{1}{x} - \log_e x$Q=x1​−loge​x.

Integrating Factor (IF) $\displaystyle = e^{\int -1 \,dx} = e^{-x}$=e∫−1dx=e−x

Multiplying the differential equation by the IF:

$$e^{-x} f'(x) - e^{-x} f(x) = e^{-x} \left(\dfrac{1}{x} - \log_e x\right)$$e−xf′(x)−e−xf(x)=e−x(x1​−loge​x)

$$\dfrac{d}{dx}[e^{-x} f(x)] = e^{-x} \dfrac{1}{x} - e^{-x} \log_e x$$dxd​[e−xf(x)]=e−xx1​−e−xloge​x

Notice that $\dfrac{d}{dx}[e^{-x} \log_e x] = -e^{-x} \log_e x + e^{-x} \dfrac{1}{x}$dxd​[e−xloge​x]=−e−xloge​x+e−xx1​.

Thus, $\dfrac{d}{dx}[e^{-x} f(x)] = \dfrac{d}{dx}[e^{-x} \log_e x]$dxd​[e−xf(x)]=dxd​[e−xloge​x]

Integrating both sides with respect to $x$x:

$$e^{-x} f(x) = e^{-x} \log_e x + C$$e−xf(x)=e−xloge​x+C

$$f(x) = \log_e x + C e^x$$f(x)=loge​x+Cex

Using the initial condition $f(1) = e$f(1)=e:

$$e = \log_e 1 + C e^1$$e=loge​1+Ce1

$$e = 0 + C e \Rightarrow C = 1$$e=0+Ce⇒C=1

Therefore, the function is $f(x) = \log_e x + e^x$f(x)=loge​x+ex.

We need to find $f(f(1))$f(f(1)):

Since $f(1) = e$f(1)=e, we evaluate $f(e)$f(e):

$$f(e) = \log_e e + e^e = 1 + e^e$$f(e)=loge​e+ee=1+ee

Answer: $(1 + e^e)$(1+ee)