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Maths Question 19 – JEE-MAIN 2026

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Let f:[1,∞)→R be a differentiable function defined as f(x)=∫1xf(t)dt+(1−x)(loge⁡x−1)+e. Then the value of f(f(1)) is:

Evaluate the given function f(x) at x=1 to find the value of f(1). Remember that ∫aag(t)dt=0 and loge⁡1=0.

Step 1: Evaluate f(1) and Differentiate f(x)✦ Active

First, substitute x=1 into the given definition of f(x) to find f(1):

f(1)=∫11f(t)dt+(1−1)(loge⁡1−1)+e

Since ∫11f(t)dt=0 and (1−1)=0 and loge⁡1=0, we get:

f(1)=0+0⋅(−1)+e⟹f(1)=e

Next, differentiate the given equation for f(x) with respect to x. Using the Leibniz integral rule ddx∫axg(t)dt=g(x) and the product rule for (1−x)(loge⁡x−1):

f′(x)=f(x)+ddx[(1−x)(loge⁡x−1)]+ddx(e)

Applying the product rule, ddx[(1−x)(loge⁡x−1)]=(−1)(loge⁡x−1)+(1−x)(1x)=−loge⁡x+1+1x−1=−loge⁡x+1x. Thus:

f′(x)=f(x)−loge⁡x+1x
💡 Teacher's Secret Hint

Remember to apply both the Leibniz rule for the integral and the product rule for the algebraic term carefully.

Step 2: Formulate and Solve the Differential Equation○ Expand

Rearrange the differentiated equation to form a first-order linear differential equation:

f′(x)−f(x)=1x−loge⁡x

This is of the form f′(x)+P(x)f(x)=Q(x), where P(x)=−1. The integrating factor (I.F.) is e∫P(x)dx:

I.F.=e∫−1dx=e−x

Multiply the differential equation by the integrating factor:

e−xf′(x)−e−xf(x)=e−x(1x−loge⁡x)

The left side is the derivative of a product, ddx(e−xf(x)). So, integrate both sides:

ddx(e−xf(x))=e−x(1x−loge⁡x)

To integrate the right side, consider ∫e−x(1x−loge⁡x)dx. This integral can be solved by recognizing the form ∫eax(f(x)+f′(x))dx or by integration by parts. Let's use integration by parts for ∫e−xloge⁡xdx. Let u=loge⁡x, dv=e−xdx. Then du=1xdx, v=−e−x:

∫e−xloge⁡xdx=−e−xloge⁡x−∫(−e−x)1xdx=−e−xloge⁡x+∫e−xxdx

Substitute this back into the integral for e−xf(x):

e−xf(x)=∫e−xxdx−(−loge⁡xe−x+∫e−xxdx)+C

The ∫e−xxdx terms cancel out:

e−xf(x)=e−xloge⁡x+C

Divide by e−x to find f(x):

f(x)=loge⁡x+Cex
💡 Teacher's Secret Hint

The integral ∫e−x(1x−loge⁡x)dx is a standard form. Recognize that ddx(loge⁡x)=1x. This hints at using integration by parts or a specific integral identity.

Step 3: Apply Initial Condition and Calculate f(f(1))○ Expand

Use the initial condition f(1)=e (from Step 1) to find the constant C:

f(1)=loge⁡1+Ce1

Since loge⁡1=0:

e=0+Ce⟹C=1

So, the explicit form of f(x) is:

f(x)=loge⁡x+ex

Finally, we need to find f(f(1)). We know f(1)=e, so we need to calculate f(e):

f(e)=loge⁡e+ee

Since loge⁡e=1:

f(e)=1+ee

Thus, f(f(1))=1+ee.

💡 Teacher's Secret Hint

Ensure you correctly substitute f(1) into f(x) to find f(f(1)). Don't confuse e with ee.

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