If y = x sinx, find \(\frac{dy}{dx}\)
sin x - x cosx
sinx + x cosx
sinx - cosx
sinx + cosx
Correct answer is B
If y = x sinx, then
Let u = x and v = sinx
\(\frac{du}{dx}\) = 1 and \(\frac{dv}{dx}\) = cosx
Hence by the product rule,
\(\frac{dy}{dx}\) = v \(\frac{du}{dx}\) + u\(\frac{dv}{dx}\)
= (sin x) x 1 + x cosx
= sinx + x cosx
If cot\(\theta\) = \(\frac{8}{15}\), where \(\theta\) is acute, find sin\(\theta\)
\(\frac{8}{17}\)
\(\frac{15}{17}\)
\(\frac{16}{17}\)
\(\frac{13}{17}\)
Correct answer is B
cot\(\theta\) = \(\frac{1}{\cos \theta}\)
= \(\frac{8}{15}\)(given)
tan\(\theta\) = \(\frac{15}{18}\)
By Pythagoras theorem,
x2 = 152 + 82
x2 = 225 + 64 = 289
x = \(\sqrt{289}\)
= 17
Hence sin\(\theta\) = \(\frac{15}{x}\)
= \(\frac{15}{17}\)
Find the gradient of the line passing through the points P(1, 1) and Q(2, 5).
3
2
5
4
Correct answer is D
Let (x1, y1) = (1, 1) and (x2, y2)= (2, 5)
then gradient m of \(\bar{PQ}\) is
m = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{5 - 1}{2 - 1}\)
= \(\frac{4}{1}\)
= 4
1
0
√3
√2
Correct answer is D
Let D denote the distance between (\(\frac{1}{2}\), -\(\frac{1}{2}\)) then using
D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
= \(\sqrt{(-{\frac{1}{2} - \frac{1}{2}})^2 + (-{\frac{1}{2} - \frac{1}{2}})^2}\)
= \(\sqrt{(-1)^2 + (-1)^2}\)
= \(\sqrt{1 + 1}\)
= √2
749\(\pi\)m2
700\(\pi\)m2
350\(\pi\)m2
98\(\pi\)m2
Correct answer is A
Total surface area of the cylindrical pipe = area of circular base + curved surface area
= \(\pi\)r\(^2\) + 2\(\pi\)rh
= \(\pi\) x 7\(^2\) + 2\(\pi\) x 7 x 50
= 49\(\pi\) + 700\(\pi\)
= 749\(\pi\)m\(^2\)