Mathematics - Definite Integration Question with Solution | TestHub

(a) Given, $(x-3{)}^2\cdot y^{\prime }+y=0$ $$\begin{array}{rlrl}\Rightarrow & \frac{dy}{dx}& =-\frac{y}{(x-3{)}^2}& \\ \Rightarrow & & \int \frac{dy}{y}& =-\int \frac{dx}{(x-3{)}^3}\\ \Rightarrow & & \ln y& =\frac{1}{(x-3)}+\ln C\\ \Rightarrow & & y& =C{e}^{\frac{1}{x-3}},C\ne 0\end{array}$$ $\therefore$ Domain of $y$ is $x\in R-\{3\}$ Aliter Given differential equation is homogeneous linear differential equation and has $x=3$ as a singular point, hence $x=3$ cannot be in domain of solution. (b) Let $I={\int }_1^5(x-1)(x-2)(x-3)(x-4)$ $(x-5)dx$ Let $x-3=t\Rightarrow dx=dt$ $\therefore I={\int }_{-2}^2(t+2)(t+1)t(t-1)(t-2)dt$ $\because$ Integrand is an odd function. $\therefore I=0$ Aliter Let $I={\int }_1^5(x-1)(x-2)(x-3)(x-4)$ $(x-5)dx$ Using, ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$ $I={\int }_1^5(5-x)(4-x)(3-x)(2-x)$ $(1-x)dx$ On adding Eqs. (i) and (ii) we get 2I $=0\Rightarrow I=0$ (c) Let $f(x)={\cos }^{2}x+\sin x$ $\Rightarrow f^{\prime }(x)=-2\cos x\sin x+\cos x$ $=\cos x(1-2\sin x)=0$ (say) Sign scheme for first derivative
Points of local maxima are $\frac{\pi }{6},\frac{5\pi }{6}$. Aliter $y={\cos }^{2}x+\sin x$ $$\Rightarrow y=\frac{5}{4}-{\left(\sin x-\frac{1}{2}\right)}^2$$ For $y$ to be maximum, $$\begin{array}{rl}& {\left(\sin x-\frac{1}{2}\right)}^2=0\Rightarrow \sin x=\frac{1}{2}\\ \Rightarrow & x=n\pi +(-1{)}^n\frac{\pi }{6},n\in I\end{array}$$ (d) Let $y={\tan }^{-1}(\sin x+\cos x)$ $$\Rightarrow \frac{dy}{dx}=\frac{\cos x-\sin x}{1+(\sin x+\cos x{)}^2}$$
Clearly, by graph, $\cos x>\sin x$ is true for option(s).