GATE ME 2010

The volume of a solid generated by revolution about the x-axis, of the area bounded by curve y=f(x),the x-axis and the ordinates
\text{x=a, y=b is}
\text { Volume }=\int_{a}^{b} \pi y^{2} d x
Here, a=1, b=2 and y=\sqrt{x} \Rightarrow y^{2}=x
\begin{aligned} \therefore Volume &=\int_{1}^{2} \pi \cdot x \cdot d x \\ &=\pi \cdot\left[\frac{x^{2}}{2}\right]_{1}^{2}=\frac{\pi}{2}\left[x^{2}\right]_{1}^{2} \\ &=\frac{\pi}{2}\left[2^{2}-1^{2}\right]=\frac{3}{2} \pi \end{aligned}

\frac{d^{3} f}{d n^{3}}+\frac{f}{2} \frac{d^{2} f}{d n^{2}}=0 is third order \left(\frac{d^{3} f}{d n^{3}}\right) and it is non linear, since the product f \times \frac{d^{2} f}{d n^{2}} is not allowed in linear differential equation.

\begin{aligned} \int_{-\infty}^{\infty} \frac{d x}{1+x^{2}} &=\left[\tan ^{-1} x\right]_{-\infty}^{\infty} \\ &=\tan ^{-1}(\infty)-\tan ^{-1}(-\infty) \\ &=\frac{\pi}{2}-\left[\frac{-\pi}{2}\right]=\pi \end{aligned}

\begin{aligned} Z &=\frac{3+4 i}{1-2 i}=\frac{(3+4 i)(1+2 i)}{(1-2 i)(1+2 i)} \\ &=\frac{-5+10 i}{5}=-1+2 i \\ |Z| &=\sqrt{(-1)^{2}+(2)^{2}}=\sqrt{5} \end{aligned}

\begin{aligned} y=|2-3 x| &=2-3 x \quad 2-3 x \geq 0 \\ &=3 x-2 \quad 2-3 x \lt 0 \end{aligned}
Therefore
y=2-3 x \quad x \leq \frac{2}{3}
=3 x-2 \quad x gt \frac{2}{3}
since 2-3 x and 3 x-2 are polynomials, these are continuous at all points. The only concern is at
x=\frac{2}{3}
Left limit at x=\frac{2}{3} is 2-3 \times \frac{2}{3}=0
Right limit at x=\frac{2}{3} is 3 \times \frac{2}{3}-2=0
f\left(\frac{2}{3}\right)=2-3 \times \frac{2}{3}=0
since, Left limit = Right limit =f\left(\frac{2}{3}\right)
Function is continuous at \frac{2}{3}
y is therefore continuous \forall x \in R
Now since 2-3 x and 3 x-2 are polynomials,they are differentiable.
only concern is at x =\frac{2}{3}
Now, at x=\frac{2}{3}, L D= Left derivative =-3
R D= Right derivative =+3
LD \neq RD
\therefore The function y is not differentiable at
x=\frac{2}{3}
So, we can say that y is differential \forall \bar{x} \in R , except at x=\frac{2}{3}