JEE Advanced Problem 2

Calculus Level 5

Suppose f ( x ) f(x) and g ( x ) g(x) are 2 continuous functions defined for 0 x 1 0 \leq x \leq 1 .

f ( x ) = 0 1 e x + t . f ( t ) d t \displaystyle f(x) = \int_0^1 e^{x + t} . f(t) dt and g ( x ) = 0 1 e x + t . g ( t ) d t + x \displaystyle g(x) = \int_0^1 e^{x + t} . g(t) dt + x

A A . The value of f ( 1 ) f(1) is

1) 0 0

2) 1 1

3) e 1 e^{-1}

4) e e

B B . The value of g ( 0 ) g(0) - f ( 0 ) f(0) is

5) 2 3 e 2 \frac{2}{3 - e^{2}}

6) 2 e 2 2 \frac{2}{ e^{2} - 2}

7) 2 e 2 1 \frac{2}{ e^{2} - 1}

8) 0 0

C C . The value of g ( o ) g ( 2 ) \frac{g(o)}{g(2)} is

9) 0 0

10) 1 3 \frac{1}{3}

11) 1 e 2 \frac{1}{ e^{2}}

12) 2 e 2 \frac{2}{ e^{2}}

Details for entering the answer:

If your answer comes as option 1 for A, option 7 for B and option 12 for C, then write your answer as 1712.


The answer is 1510.

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1 solution

Prakhar Gupta
Mar 31, 2015

Let's observe f ( x ) f(x) . f ( x ) = 0 1 e x + t . f ( t ) d t f(x) = \int_{0}^{1}e^{x+t}.f(t)dt f ( x ) = e x 0 1 e t . f ( t ) d t f(x) =e^{x} \int_{0}^{1} e^{t}.f(t)dt Let's define a number α \alpha as α = 0 1 e t . f ( t ) d t \alpha = \int_{0}^{1}e^{t}.f(t)dt Hence we can write that:- f ( x ) = α e x f(x) = \alpha e^{x} Hence we can write that α = 0 1 e t . α e t d t \alpha = \int_{0}^{1} e^{t} . \alpha e^{t} dt We will be urged to cancel α \alpha and get a false expression.

It means only one thing that it will be wrong to cancel α \alpha because α = 0 \alpha = 0 .

Hence we get f ( x ) = 0 f(x) = 0 .

Now Let's observe g ( x ) g(x) . g ( x ) = e x 0 1 e t . g ( t ) d t + x g(x) = e^{x} \int_{0}^{1} e^{t}.g(t)dt + x Again let us define β \beta as follows:- β = 0 1 e t . g ( t ) d t \beta = \int_{0}^{1} e^{t}.g(t)dt Hence we can write :- g ( x ) = β e x + x g(x) = \beta e^{x}+x Again we can write that:- β = 0 1 e t ( β e t + t ) d t \beta = \int_{0}^{1} e^{t}(\beta e^{t}+t) dt β = β 0 1 e 2 t d t + 0 1 e t t d t \beta = \beta \int_{0}^{1}e^{2t}dt + \int_{0}^{1}e^{t}tdt β = β ( e 2 1 2 ) + 1 \beta = \beta \Big( \dfrac{e^{2}-1}{2} \Big) +1 β = 2 3 e 2 \beta = \dfrac{2}{3-e^{2}} Hence we get that:- g ( x ) = 2 e x 3 e 2 + x g(x) =\dfrac{2e^{x}}{3-e^{2}}+x Now we can get our desired answers.

Did hundred percent same way as prakhar.Nicely explained solution!

rahul saxena - 5 years, 7 months ago

nice solution , to a nice question +1 !!!

Rudraksh Sisodia - 4 years, 7 months ago

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