JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 6)
Let f : [0, $$\infty $$) $$ \to $$ R be a continuous function such that
$$f(x) = 1 - 2x + \int_0^x {{e^{x - t}}f(t)dt} $$ for all x $$ \in $$ [0, $$\infty $$). Then, which of the following statement(s) is (are) TRUE?
$$f(x) = 1 - 2x + \int_0^x {{e^{x - t}}f(t)dt} $$ for all x $$ \in $$ [0, $$\infty $$). Then, which of the following statement(s) is (are) TRUE?
The curve y = f(x) passes through the point (1, 2)
The curve y = f(x) passes through the point (2, $$-$$1)
The area of the region $$\{ (x,y) \in [0,1] \times R:f(x) \le y \le \sqrt {1 - {x^2}} \} $$ is $${{\pi - 2} \over 4}$$
The area of the region $$\{ (x,y) \in [0,1] \times R:f(x) \le y \le \sqrt {1 - {x^2}} \} $$ is $${{\pi - 1} \over 4}$$
Explanation
We have, $$f(x) = 1 - 2x + \int_0^x {{e^{x - t}}f(t)dt} $$
On multiplying e$$-$$x both sides, we get
$${e^{ - x}}f(x) = {e^{ - x}} - 2x{e^{ - x}} + \int_0^x {{e^{x - t}}f(t)dt} $$
On differentiating both side w.r.t. x, we get
$${e^{ - x}}f'(x) - {e^{ - x}}f(x) = - {e^{ - x}} - 2{e^{ - x}} + 2{e^{ - x}} + {e^{ - x}}f(x)$$
$$ \Rightarrow $$ f'(x) $$-$$ 2f (x) = 2x $$-$$ 3
[dividing both sides by e$$-$$x]
Let f(x) = y
$$ \Rightarrow $$ $$f'(x) = {{dy} \over {dx}}$$
$$ \therefore $$ $${{dy} \over {dx}} - 2y = 2x - 3$$
which is linear differential equation of the form $${{dy} \over {dx}} + Py = Q$$. Here, P = $$-$$2 and Q = 2x $$-$$ 3
Now, $$IF = {e^{\int {P\,dx} }} = {e^{\int { - 2\,dx} }} = {e^{ - 2x}}$$
$$ \therefore $$ Solution of the given differential equation is
$$y.{e^{ - 2x}} = \int {(2x - 3){e^{ - 2x}}} dx + C$$
$$y.{e^{ - 2x}} = {{ - (2x - 3).{e^{ - 2x}}} \over 2} + 2\int {{{{e^{ - 2x}}} \over 2}dx + C} $$
[by using integration by parts]
$$ \Rightarrow $$ $$y.{e^{ - 2x}} = {{ - (2x - 3)\,{e^{ - 2x}}} \over 2} - {{{e^{ - 2x}}} \over 2} + C$$
$$ \Rightarrow $$ 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 = $$\sqrt {1 - {x^2}} $$ and y = 1 $$-$$ x is shows as
$$ \therefore $$ Area of shaded region
= Area of 1st quadrant of a circle $$-$$ Area of $$\Delta $$OAB
= $${\pi \over 4}{(1)^2} - {1 \over 2} \times 1 \times 1$$
$$ = {\pi \over 4} - {1 \over 2} = {{\pi - 2} \over 4}$$
Hence, options b and c are correct.
On multiplying e$$-$$x both sides, we get
$${e^{ - x}}f(x) = {e^{ - x}} - 2x{e^{ - x}} + \int_0^x {{e^{x - t}}f(t)dt} $$
On differentiating both side w.r.t. x, we get
$${e^{ - x}}f'(x) - {e^{ - x}}f(x) = - {e^{ - x}} - 2{e^{ - x}} + 2{e^{ - x}} + {e^{ - x}}f(x)$$
$$ \Rightarrow $$ f'(x) $$-$$ 2f (x) = 2x $$-$$ 3
[dividing both sides by e$$-$$x]
Let f(x) = y
$$ \Rightarrow $$ $$f'(x) = {{dy} \over {dx}}$$
$$ \therefore $$ $${{dy} \over {dx}} - 2y = 2x - 3$$
which is linear differential equation of the form $${{dy} \over {dx}} + Py = Q$$. Here, P = $$-$$2 and Q = 2x $$-$$ 3
Now, $$IF = {e^{\int {P\,dx} }} = {e^{\int { - 2\,dx} }} = {e^{ - 2x}}$$
$$ \therefore $$ Solution of the given differential equation is
$$y.{e^{ - 2x}} = \int {(2x - 3){e^{ - 2x}}} dx + C$$
$$y.{e^{ - 2x}} = {{ - (2x - 3).{e^{ - 2x}}} \over 2} + 2\int {{{{e^{ - 2x}}} \over 2}dx + C} $$
[by using integration by parts]
$$ \Rightarrow $$ $$y.{e^{ - 2x}} = {{ - (2x - 3)\,{e^{ - 2x}}} \over 2} - {{{e^{ - 2x}}} \over 2} + C$$
$$ \Rightarrow $$ 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 = $$\sqrt {1 - {x^2}} $$ and y = 1 $$-$$ x is shows as
$$ \therefore $$ Area of shaded region
= Area of 1st quadrant of a circle $$-$$ Area of $$\Delta $$OAB
= $${\pi \over 4}{(1)^2} - {1 \over 2} \times 1 \times 1$$
$$ = {\pi \over 4} - {1 \over 2} = {{\pi - 2} \over 4}$$
Hence, options b and c are correct.
Comments (0)
