Simplify the given expression \(\sqrt{\frac{1 - cos x}{1 + cos x}}\)
\(\frac{1 - cos x}{sin x}\)
1 - cos x
sin x
1 + cos x
\(\frac{1 + cos x}{sin x}\)
Correct answer is A
\(\sqrt{\frac{1 - cos x}{1 + cos x}}\) = a
a2 = \(\frac{1 - cosx}{1 + cosx}\)
\(\frac{1 - cosx}{1 + cosx}\) = \(\frac{1 - cosx}{1 - cosx}\)
= \(\frac{(1 - cosx)^2}{cos^2 x}\)
a2 = \(\frac{(1 - cos x)^2}{sin^2 x}\)
a = \(\frac{1 - cos x}{sin x}\)
Solve the system of equation 2x + y = 32, 33y - x = 27
(3, 2)
(-3, 2)
(3, -2)
(-3, -2)
(2, 2)
Correct answer is A
2x + y = 32, 33y - x = 27
2x + y = 25
33y + x = 33
x + y = 5
\(\frac{3y - x = 3}{4y = 8}\)
y = 2
If it is given that 5x + 1 + 5x = 150, then the value of x is equal to
3
4
1
2
\(\frac{1}{2}\)
Correct answer is D
5x + 1 + 5x = 150
5(5x) + 5x = 150
6(5x) = 150
5x = \(\frac{150}{6}\)
= 25
= 52
= 2
Find the solution of the equation x + 2\(\sqrt{x} - 8\) = 0
(4, 16)
(2, 4)
(4, 1)
(1, 16)
(16, 16)
Correct answer is A
x + 2\(\sqrt{x} - 8 = 0, Let \sqrt{x} = y\)
x = \(y^2\)
\(y^2 + 2y\) - 8 = 0
(y + 4)(x - 2) = 0
y = -4 or 2
x = 16 or 4
What will be the value of k so that the quadratic equation kx2 - 4x + 1 = 0 has two equal roots?
2
3
4
8
\(\frac{1}{4}\)
Correct answer is C
kx2 - 4x + 1 = 0, comparing with ax2 + bx + c = 0
a = k, b = -4, c = 1 for equal root b2 = 4ac
(-4)2 = 4k
k = \(\frac{16}{4}\)
= 4