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Definite Integration question

2022 · 29 Jun · Shift 2 · Q27
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  5. /2022 · 29 Jun · Shift 2 · Q27

Definite Integration question

2022 · 29 Jun · Shift 2 · Q27

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f be a real valued continuous function on [0, 1] and f(x)=x+∫01(x−t)f(t)dtf(x) = x + \int\limits_0^1 {(x - t)f(t)dt}f(x)=x+0∫1​(x−t)f(t)dt. Then, which of the following points (x, y) lies on the curve y = f(x) ?
  1. A
    (2, 4)
  2. B
    (1, 2)
  3. C
    (4, 17)
  4. D
    (6, 8)
View written solutionFree

Correct answer: D

  1. Given integral equation

We have

f(x)=x+∫01(x−t)f(t) dt.f(x)=x+\int_0^1 (x-t)f(t)\,dt.f(x)=x+∫01​(x−t)f(t)dt.

We must find the curve y=f(x)y=f(x)y=f(x) and then check which given point lies on it.

  1. Simplify the integral

Since the integration variable is ttt, treat xxx as a constant inside the integral:

∫01(x−t)f(t) dt=x∫01f(t) dt−∫01tf(t) dt.\int_0^1 (x-t)f(t)\,dt = x\int_0^1 f(t)\,dt - \int_0^1 t f(t)\,dt.∫01​(x−t)f(t)dt=x∫01​f(t)dt−∫01​tf(t)dt.

Let

A=∫01f(t) dt,B=∫01tf(t) dt.A=\int_0^1 f(t)\,dt, \qquad B=\int_0^1 t f(t)\,dt.A=∫01​f(t)dt,B=∫01​tf(t)dt.

Then

f(x)=x+xA−B=(1+A)x−B.f(x)=x+xA-B=(1+A)x-B.f(x)=x+xA−B=(1+A)x−B.

So f(x)f(x)f(x) must be a linear function. Write

f(x)=px+q.f(x)=px+q.f(x)=px+q.

Comparing, we have

p=1+A,q=−B.p=1+A, \qquad q=-B.p=1+A,q=−B.
  1. Compute AAA and BBB in terms of p,qp,qp,q

If

f(t)=pt+q,f(t)=pt+q,f(t)=pt+q,

then

A=∫01(pt+q) dt=p2+q,A=\int_0^1 (pt+q)\,dt=\frac p2+q,A=∫01​(pt+q)dt=2p​+q,

and

B=∫01t(pt+q) dt=∫01(pt2+qt) dt=p3+q2.B=\int_0^1 t(pt+q)\,dt=\int_0^1 (pt^2+qt)\,dt=\frac p3+\frac q2.B=∫01​t(pt+q)dt=∫01​(pt2+qt)dt=3p​+2q​.

Using p=1+Ap=1+Ap=1+A and q=−Bq=-Bq=−B:

p=1+p2+q,p=1+\frac p2+q,p=1+2p​+q, q=−(p3+q2).q=-\left(\frac p3+\frac q2\right).q=−(3p​+2q​).
  1. Solve for ppp and qqq

From

p=1+p2+q,p=1+\frac p2+q,p=1+2p​+q,

we get

p2=1+q⇒p=2+2q.\frac p2=1+q \quad \Rightarrow \quad p=2+2q.2p​=1+q⇒p=2+2q.

From

q=−p3−q2,q=-\frac p3-\frac q2,q=−3p​−2q​,

we get

3q2=−p3⇒9q=−2p⇒p=−9q2.\frac{3q}{2}=-\frac p3 \quad \Rightarrow \quad 9q=-2p \quad \Rightarrow \quad p=-\frac{9q}{2}.23q​=−3p​⇒9q=−2p⇒p=−29q​.

Equating the two expressions for ppp:

2+2q=−9q2.2+2q=-\frac{9q}{2}.2+2q=−29q​.

Multiply by 222:

4+4q=−9q,4+4q=-9q,4+4q=−9q, 13q=−4,13q=-4,13q=−4, q=−413.q=-\frac{4}{13}.q=−134​.

Then

p=2+2q=2−813=1813.p=2+2q=2-\frac{8}{13}=\frac{18}{13}.p=2+2q=2−138​=1318​.

Hence

f(x)=1813x−413.f(x)=\frac{18}{13}x-\frac{4}{13}.f(x)=1318​x−134​.
  1. Check the options

We test whether each point (x,y)(x,y)(x,y) satisfies

y=1813x−413.y=\frac{18}{13}x-\frac{4}{13}.y=1318​x−134​.
  • A: (2,4)(2,4)(2,4)

f(2)=\frac{36}{13}-\frac{4}{13}=\frac{32}{13}\ne 4.

Not on the curve. - **B: $(1,2)$**

f(1)=\frac{18}{13}-\frac{4}{13}=\frac{14}{13}\ne 2.

Not on the curve. - **C: $(4,17)$**

f(4)=\frac{72}{13}-\frac{4}{13}=\frac{68}{13}\ne 17.

Not on the curve. - **D: $(6,8)$**

f(6)=\frac{108}{13}-\frac{4}{13}=\frac{104}{13}=8.

Thisliesonthecurve.−−−6.∗∗Conclusion∗∗ThecorrectoptionisThis lies on the curve. --- 6. **Conclusion** The correct option isThisliesonthecurve.−−−6.∗∗Conclusion∗∗Thecorrectoptionis

\boxed{\text{D: }(6,8)}.

Thismatchesthestoredcorrectanswer. This matches the stored correct answer.Thismatchesthestoredcorrectanswer.
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