Question 1 |
The parabolic arc y=\sqrt{x}, 1\leq x\leq 2
is revolved around the x-axis. The volume of the solid of revolution is
\frac{\pi }{4} | |
\frac{\pi }{2} | |
\frac{3\pi }{4} | |
\frac{3\pi }{2} |
Question 1 Explanation:
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}
\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}
Question 2 |
The Blasius equation, \frac{\mathrm{d}^{3}f }{\mathrm{d} \eta ^{3}}+\frac{f}{2}\frac{\mathrm{d} ^{2}f }{\mathrm{d} \eta ^{2}}=0 is a
Second order nonlinear ordinary differential equation | |
Third order nonlinear ordinary differential equation | |
Third order linear ordinary differential equation | |
Mixed order nonlinear ordinary differential equation |
Question 2 Explanation:
\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.
Question 3 |
The value of the integral \int_{-\infty }^{\infty }\frac{\mathrm{d} x}{1+x^{2}}
is
-\pi | |
\frac{-\pi }{2} | |
\frac{\pi }{2} | |
\pi |
Question 3 Explanation:
\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}
Question 4 |
The modulus of the complex number \left ( \frac{3+4i}{1-2i} \right )
is
5 | |
\sqrt{5} | |
\frac{1}{\sqrt{5}} | |
\frac{1}{5} |
Question 4 Explanation:
\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}
Question 5 |
The function y=\left | 2-3x \right |
is continuous \forall x\in R
and differentiable \forall x\in R
| |
is continuous \forall x\in R
and differentiable \forall x\in R
except at x=\frac{3}{2}
| |
is continuous \forall x\in R
and differentiable \forall x\in R
except at x=\frac{2}{3}
| |
is continuous \forall x\in R except at x=3
and differentiable \forall x\in R
|
Question 5 Explanation:
\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}
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}
There are 5 questions to complete.