\(\frac{d}{dx}\) 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
\(\frac{20}{√3}\)m
\(\frac{40}{√3}\)m
20√3m
40√3m
Correct answer is D
\(\frac{40}{x}\) = tan 30°
x = \(\frac{40}{tan 30}\)
= \(\frac{40}{1\sqrt{3}}\)
= 40√3m
3√10
3√5
√26
√13
Correct answer is D
2x - y .....(i)
x + y.....(ii)
from (i) y = 2x - 4
from (ii) y = -x + 2
2x - 4 = -x + 2
x = 2
y = -x + 2
= -2 + 2
= 0
\(x_1\) = 2
\(y_1\) = 0
\(x_2\) = 4
\(y_2\) = 3
Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\)
= \(\sqrt{(3 - 0)^2 + (4 - 2)^2}\)
= \(\sqrt{3^2 + 2^2}\)
= \(\sqrt{13}\)