Evaluate ∫\(^{\pi}_{2}\)(sec2 x - tan2x)dx
\(\frac{\pi}{2}\)
\(\pi\) - 2
\(\frac{\pi}{3}\)
\(\pi\) + 2
Correct answer is B
∫\(^{\pi}_{2}\)(sec2 x - tan2x)dx
∫\(^{\pi}_{2}\) dx = [X]\(^{\pi}_{2}\)
= \(\pi\) - 2 + c
when c is an arbitrary constant of integration
Differentiate \(\frac{x}{cosx}\) with respect to x
1 + x sec x tan x
1 + sec2 x
cos x + x tan x
x sec x tan x + secx
Correct answer is D
let y = \(\frac{x}{cosx}\) = x sec x
y = u(x) v (x0
\(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\)
dy x [secx tanx] + secx
x = x secx tanx + secx
If y = 243(4x + 5)-2, find \(\frac{dy}{dx}\) when x = 1
\(\frac{-8}{3}\)
\(\frac{3}{8}\)
\(\frac{9}{8}\)
-\(\frac{8}{9}\)
Correct answer is A
y = 243(4x + 5)-2, find \(\frac{dy}{dx}\)
= -1944(4x + 5)-3
= 1944(9)-3
\(\frac{dy}{dx}\) when x = 1
= -\(\frac{1944}{9^3}\)
= -\(\frac{1944}{729}\)
= \(\frac{-8}{3}\)
150(1 + \(\sqrt{3}\))m
50( \(\sqrt{3}\) - \(\sqrt{3}\))m
150 \(\sqrt{3}\)m
\(\frac{50}{\sqrt{3}}\)m
Correct answer is B
\(\frac{150}{Z}\) = tan 60o,
Z = \(\frac{150}{tan 60^o}\)
= \(\frac{150}{3}\)
= 50\(\sqrt{3}\)cm
\(\frac{150}{X x Z}\) = tan45o = 1
X + Z = 150
X = 150 - Z
= 150 - 50\(\sqrt{3}\)
= 50( \(\sqrt{3}\) - \(\sqrt{3}\))m
\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)
\(\frac{\pi}{3}\), \(\frac{2\pi}{3}\)
0, \(\frac{\pi}{3}\)
0, \(\pi\)
Correct answer is D
cos x + sin x \(\frac{1}{cos x - sinx}\)
= (cosx + sinx)(cosx - sinx) = 1
= cos2x + sin2x = 1
cos2x - (1 - cos2x) = 1
= 2cos2x = 2
cos2x = 1
= cosx = \(\pm\)1 = x
= cos-1x (\(\pm\), 1)
= 0, \(\pi\) \(\frac{3}{2}\pi\), 2\(\pi\)
(possible solution)