Let f:[0, infty) rightarrow R be a continuous | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = 1 - 2x + e^x \int_0^x e^{-t} f(t) dt$.Dividing by $e^x$,we get $e^{-x} f(x) = (1 - 2x)e^{-x} + \int_0^x e^{-t} f(t) dt$.Differentiat

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(C) Given $f(x) = 1 - 2x + e^x \int_0^x e^{-t} f(t) dt$.Dividing by $e^x$,we get $e^{-x} f(x) = (1 - 2x)e^{-x} + \int_0^x e^{-t} f(t) dt$.Differentiating with respect to $x$ using the Leibniz rule:$e^{-x} f'(x) - e^{-x} f(x) = -2e^{-x} - (1 - 2x)e^{-x} + e^{-x} f(x)$.Simplifying,$f'(x) - f(x) = -2 - 1 + 2x + f(x)$,so $f'(x) - 2f(x) = 2x - 3$.This is a linear differential equation with integrating factor $I.F. = e^{\int -2 dx} = e^{-2x}$.Multiplying by $I.F.$,$\frac{d}{dx} (f(x) e^{-2x}) = (2x - 3)e^{-2x}$.Integrating both sides: $f(x) e^{-2x} = \int (2x - 3)e^{-2x} dx = (2x - 3) \frac{e^{-2x}}{-2} - \int 2 \frac{e^{-2x}}{-2} dx = -x e^{-2x} + \frac{3}{2} e^{-2x} + \int e^{-2x} dx = -x e^{-2x} + \frac{3}{2} e^{-2x} - \frac{1}{2} e^{-2x} + C = -x e^{-2x} + e^{-2x} + C$.Thus,$f(x) = -x + 1 + C e^{2x}$.From the original equation,at $x=0$,$f(0) = 1 - 0 + 0 = 1$. Substituting into $f(x)$,$1 = 0 + 1 + C$,so $C = 0$.Therefore,$f(x) = 1 - x$.Check options:$(A)$ $f(1) = 1 - 1 = 0 
eq 2$. $(A)$ is false.$(B)$ $f(2) = 1 - 2 = -1$. $(B)$ is true.$(C)$ Area $= \int_0^1 (\sqrt{1-x^2} - (1-x)) dx = [\frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1} x - x + \frac{x^2}{2}]_0^1 = (0 + \frac{\pi}{4} - 1 + \frac{1}{2}) - (0) = \frac{\pi}{4} - \frac{1}{2} = \frac{\pi-2}{4}$. $(C)$ is true.Thus,the correct options are $(B)$ and $(C)$.

Column $I$	Column $II$
$(A)$ Interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y^{\prime}+y=0$	$(p)$ $(-\frac{\pi}{2}, \frac{\pi}{2})$
$(B)$ Interval containing the value of the integral $\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) dx$	$(q)$ $(0, \frac{\pi}{2})$
$(C)$ Interval in which at least one of the points of local maximum of $\cos^2 x+\sin x$ lies	$(r)$ $(\frac{\pi}{8}, \frac{5\pi}{4})$
$(D)$ Interval in which $\tan^{-1}(\sin x+\cos x)$ is increasing	$(s)$ $(0, \frac{\pi}{8})$
	$(t)$ $(-\pi, \pi)$

Similar Questions

The solution of the differential equation $y dx - x dy + 3x^2 y^2 e^{x^3} dx = 0$ satisfying $y = 1$ when $x = 1$ is:

The solution of $(1+xy)y \, dx + (1-xy)x \, dy = 0$ is:

Suppose $f(x) = e^{ax} + e^{bx},$ where $a \neq b,$ and that $f''(x) - 2f'(x) - 15f(x) = 0$ for all $x.$ Then the product $ab$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module