Find the value of y, if log (y + 8) + log (y - 8) = 2log 3 + 2log 5
y = ±5
y = ±10
y = ±17
y = ±13
Correct answer is C
log (y + 8) + log (y - 8) = 2log 3 + 2log 5
⇒ log (y + 8)(y - 8) = 2log 3 + 2log 5 (log ab = log a + log b)
⇒ log (y2 -8y + 8y - 64) = log 32 + log 52
⇒ log (y2 - 64) = log 32 + log 52
⇒ log (y2 - 64) = log 9 + log 25
⇒ log (y2 - 64) = log (9 × 25)
⇒ log(y2 - 64) = log 225
⇒ y2 - 64 = 225
⇒ y2 = 225 + 64
⇒ y2 = 289
⇒ y = ±√289
∴ y = ±17
Find the value of y if 402y=102ten
4
2
5
3
Correct answer is C
402y=102ten
⇒ 4 x y2 + 0 x y1 + 2 x y0 = 102
⇒ 4y2 + 0 + 2 x 1 = 102
⇒ 4y2 + 2 = 102
⇒ 4y2 = 102 - 2
⇒ 4y2 = 100
⇒ y2=1004
⇒ y2 = 25
∴ y = √25 = 5
Determine the area of the region bounded by y = 2x2 + 10 and Y = 4x+16.
18
−103
443
643
Correct answer is D
To find the point of intersection, equate the two equations
⇒ 2x2 + 10 = 4x + 16
⇒ 2x2 + 10 - 4x - 16 = 0
⇒ 2x2 - 4x - 6 = 0
Factorize
⇒ 2(x + 1)(x - 3) = 0
∴ x = -1 or 3
The curves will intersect at -1 and 3
Area = ∫ba [upper function] - [lower function] dx
= ∫314x+16−(2x2+10)dx
= ∫31−2x2+4x+6dx
= (−2x33+2x2+6x)31
= (−2(3)33+2(3)2+6(3))−(−2(−1)33+2(−1)2+6(−1))
= 18 - (103)
= 18 + 103
= 643
Calculate the mean deviation of the first five prime numbers.
2.72
5.6
5.25
13.6
Correct answer is A
The first five prime numbers are 2, 3, 5, 7 and 11
M.D = ∑(x−x̄)n
n = 5
x̄=∑xn=2+3+5+7+115=285=5.6
M.D = (2−5.6)+(3−5.6)+(5−5.6)+(7−5.6)+(11−5.6)5
= (−3.6)+(−2.6)+(−0.6)+(1.4)+(5.4)5=13.65
Take the absolute value of all numbers in bracket
= 3.6+2.6+0.6+1.4+5.45=13.65
∴
Differentiate the function y = \sqrt[3]{x^2}(2x - x^2)
\frac {dy}{dx} = \frac {10x^{5/3}}{3} - \frac {8x^{2/3}}{3}
\frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{5/3}}{3}
\frac {dy}{dx} = \frac {10x^{5/3}}{3} - \frac {8x^{5/3}}{3}
\frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{2/3}}{3}
Correct answer is B
y = \sqrt[3]{x^2(2x - x^2)} = x^{2/3} (2x - x^2)
= 2x^{5/3} - x^{8/3}
Now, we can differentiate the function
\therefore \frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{5/3}}{3}