If \(y = x(x^4 + x + 1)\), evaluate \(\int \limits_{0} ^{1} y \mathrm d x\).
\(\frac{11}{12}\)
1
\(\frac{5}{6}\)
zero
Correct answer is B
\(y = x(x^{4} + x + 1) = x^{5} + x^{2} + x\)
\(\int \limits_{0} ^{1} (x^{5} + x^{2} + x) \mathrm d x = \frac{x^{6}}{6} + \frac{x^{3}}{3} + \frac{x^{2}}{2}\)
= \([\frac{x^{6}}{6} + \frac{x^{3}}{3} + \frac{x^{2}}{2}]_{0} ^{1}\)
= \(\frac{1}{6} + \frac{1}{3} + \frac{1}{2}\)
= \(1\)
Integrate \(\frac{1}{x}\) + cos x with respect to x
-\(\frac{1}{x^2}\) + sin x + k
x + sin x - k
x - sin x + k
-\(\frac{1}{x^2}\) - sin x + k
Correct answer is C
\(\int \frac{1}{x} + \cos x = ln x - \sin x + k\)
\(\frac{d}{dx}\) cos(3x\(^2\) - 2x) is equal to
-sin(6x - 2)
-sin(3x2 - 2x)dx
(6x - 2) sin(3x2 - 2x)
-(6x - 2)sin(3x2 - 2x)
Correct answer is D
Let \(3x^{2} - 2x = u\)
\(y = \cos u \implies \frac{\mathrm d y}{\mathrm d u} = - \sin u\)
\(\frac{\mathrm d u}{\mathrm d x} = 6x - 2\)
\(\therefore \frac{\mathrm d y}{\mathrm d x} = (6x - 2) . - \sin u\)
= \(- (6x - 2) \sin (3x^{2} - 2x)\)
Differentiate \(\frac{6x^3 - 5x^2 + 1}{3x^2}\) with respect to x
\(\frac{2 + 2}{3x^3}\)
2 + \(\frac{1}{6x}\)
2 - \(\frac{2}{3x^3}\)
\(\frac{1}{5}\)
Correct answer is C
\(\frac{6x^3 - 5x^2 + 1}{3x^2}\)
let y = 3x2
y = \(\frac{6x^3}{3x^2}\) - \(\frac{6x^2}{3x^2}\) + \(\frac{1}{3x^2}\)
Y = 2x - \(\frac{5}{3}\) + \(\frac{1}{3x^2}\)
\(\frac{dy}{dx}\) = 2 + \(\frac{1}{3}\)(-2)x-3
= 2 - \(\frac{2}{3x^3}\)
In a triangle XYZ, if < ZYZ is 60, XY = 3cm and YZ = 4cm, calculate the length of the sides XZ.
√23cm
√13cm
2√5cm
2√3cm
Correct answer is B
(XZ)2 = 32 + 42 - 2 x 3 x 4 cos60o
= 25 - 24\(\frac{1}{2}\)
XZ = √13cm