Verified Solution • 2025–26 Session moderate

QJEEAdv18Ma3039: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Ma3039 Solution: Let f : [0, ) R be a continuous function such that for all x…

Let f : [0, ∞\infty ) →\to R be a continuous function such that

f(x)=1−2x+∫0xex−tf(t)dtf(x) = 1 - 2x + \int_0^x {{e^{x - t}}f(t)dt} for all x ∈\in [0, ∞\infty ). Then, which of the following statement(s) is (are) TRUE?

Multiple Choice Options

  • A The curve y = f(x) passes through the point (1, 2)
  • B The curve y = f(x) passes through the point (2, −- 1)
  • C The area of the region {(x,y)∈[0,1]×R:f(x)≤y≤1−x2}\{ (x,y) \in [0,1] \times R:f(x) \le y \le \sqrt {1 - {x^2}} \} is π−24{{\pi - 2} \over 4}
  • D The area of the region {(x,y)∈[0,1]×R:f(x)≤y≤1−x2}\{ (x,y) \in [0,1] \times R:f(x) \le y \le \sqrt {1 - {x^2}} \} is π−14{{\pi - 1} \over 4}

Step-by-Step Verified Solution

We have, f(x)=1−2x+∫0xex−tf(t)dtf(x) = 1 - 2x + \int_0^x {{e^{x - t}}f(t)dt}

On multiplying e −- x both sides, we get

e−xf(x)=e−x−2xe−x+∫0xex−tf(t)dt{e^{ - x}}f(x) = {e^{ - x}} - 2x{e^{ - x}} + \int_0^x {{e^{x - t}}f(t)dt}

On differentiating both side w.r.t. x, we get

e−xf′(x)−e−xf(x)=−e−x−2e−x+2e−x+e−xf(x){e^{ - x}}f'(x) - {e^{ - x}}f(x) = - {e^{ - x}} - 2{e^{ - x}} + 2{e^{ - x}} + {e^{ - x}}f(x)

⇒\Rightarrow f'(x) −- 2f (x) = 2x −- 3

[dividing both sides by e −- x]

Let f(x) = y

⇒\Rightarrow f′(x)=dydxf'(x) = {{dy} \over {dx}}

∴\therefore dydx−2y=2x−3{{dy} \over {dx}} - 2y = 2x - 3

which is linear differential equation of the form dydx+Py=Q{{dy} \over {dx}} + Py = Q . Here, P = −- 2 and Q = 2x −- 3

Now, IF=e∫P dx=e∫−2 dx=e−2xIF = {e^{\int {P\,dx} }} = {e^{\int { - 2\,dx} }} = {e^{ - 2x}}

∴\therefore Solution of the given differential equation is

y.e−2x=∫(2x−3)e−2xdx+Cy.{e^{ - 2x}} = \int {(2x - 3){e^{ - 2x}}} dx + C

y.e−2x=−(2x−3).e−2x2+2∫e−2x2dx+Cy.{e^{ - 2x}} = {{ - (2x - 3).{e^{ - 2x}}} \over 2} + 2\int {{{{e^{ - 2x}}} \over 2}dx + C}

[by using integration by parts]

⇒\Rightarrow y.e−2x=−(2x−3) e−2x2−e−2x2+Cy.{e^{ - 2x}} = {{ - (2x - 3)\,{e^{ - 2x}}} \over 2} - {{{e^{ - 2x}}} \over 2} + C

⇒\Rightarrow y = (1 −- x) + Ce2x

On putting x = 0 and y = 1, we get

1 = 1 + C ⇒\Rightarrow C = 0

∴\therefore y = 1 −- x

y = 1 −- x passes through (2, −- 1)

Now, area of region bounded by curve y = 1−x2\sqrt {1 - {x^2}} and y = 1 −- x is shows as

JEE Advanced 2018 Paper 1 Offline Mathematics - Application of Integration Question 11 English Explanation

∴\therefore Area of shaded region

= Area of 1st quadrant of a circle −- Area of Δ\Delta OAB

= π4(1)2−12×1×1{\pi \over 4}{(1)^2} - {1 \over 2} \times 1 \times 1

=π4−12=π−24= {\pi \over 4} - {1 \over 2} = {{\pi - 2} \over 4}

Hence, options b and c are correct.

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