Match the statements/expressions in Column I | vedclass

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

Question

(A) The differential equation is $(x-3)^2 \frac{dy}{dx} + y = 0$. Separating variables,we get $\int \frac{dy}{y} = -\int \frac{dx}{(x-3)^2}$. Integrat

Vedclass · Accepted Answer

$(A) \rightarrow p, q, s; (B) \rightarrow q, t; (C) \rightarrow p, q, r, t; (D) \rightarrow s$

Vedclass · Answer

(A) The differential equation is $(x-3)^2 \frac{dy}{dx} + y = 0$. Separating variables,we get $\int \frac{dy}{y} = -\int \frac{dx}{(x-3)^2}$. Integrating,$\ln|y| = \frac{1}{x-3} + C$. The solution is defined for $x 
eq 3$. Thus,the intervals $(-\frac{\pi}{2}, \frac{\pi}{2})$,$(0, \frac{\pi}{2})$,and $(0, \frac{\pi}{8})$ are contained in the domain $R - \{3\}$.$(B)$ Let $I = \int_1^5 (x-1)(x-2)(x-3)(x-4)(x-5) dx$. Let $x-3 = t$,then $dx = dt$. The limits change from $x=1, 5$ to $t=-2, 2$. $I = \int_{-2}^2 (t+2)(t+1)t(t-1)(t-2) dt = \int_{-2}^2 t(t^2-1)(t^2-4) dt$. Since the integrand is an odd function,$I = 0$. The value $0$ is contained in $(0, \frac{\pi}{2})$ and $(-\pi, \pi)$.$(C)$ Let $f(x) = \cos^2 x + \sin x = 1 - \sin^2 x + \sin x = \frac{5}{4} - (\sin x - \frac{1}{2})^2$. Maxima occur when $\sin x = \frac{1}{2}$,i.e.,$x = \frac{\pi}{6}, \frac{5\pi}{6}$. These points lie in $(-\frac{\pi}{2}, \frac{\pi}{2})$,$(0, \frac{\pi}{2})$,$(\frac{\pi}{8}, \frac{5\pi}{4})$,and $(-\pi, \pi)$.$(D)$ Let $g(x) = 	an^{-1}(\sin x + \cos x)$. $g'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2}$. $g(x)$ is increasing when $\cos x - \sin x > 0$,i.e.,$\cos x > \sin x$,which holds for $x \in (0, \frac{\pi}{4})$. This interval is contained in $(0, \frac{\pi}{8})$.

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

$A$ function $y = f(x)$ satisfies the condition $f'(x) \sin x + f(x) \cos x = 1$,where $f(x)$ is bounded as $x \rightarrow 0$. If $I = \int_{0}^{\frac{\pi}{2}} f(x) \, dx$,then:

If $y$ is a function of $x$,then $\frac{d^2y}{dx^2} + y \frac{dy}{dx} = 0$. If $x$ is a function of $y$,then the equation becomes:

The general solution of the differential equation $\frac{dy}{dx} = \frac{2x-3y+4}{3x+2y-7}$ is

The curve that satisfies the differential equation $x y \, dy - (1 + y^2) \, dx = 0$ passes through $(1, 0)$ and intersects the curve $x^2 + 3y^2 = 3$ at an angle $\theta$. Then $\frac{2\theta}{\pi} =$

The solution of the differential equation $\cos^2 x \frac{d^2y}{dx^2} = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module