Mathematics Questions and Answers

16\(x^4 - y^4\)

= 2\(^4x^4 - y^4\)

= \((2x)^4 - y^4\)

= \(((2x)^2)^2 - (y^2)^2\)

Using a\(^2 - b^2\) = (a - b)(a + b) identity

= ((2x)\(^2 - y^2)((2x)^2 + y^2)\)

Using the identity one more time

= \((2x - y)(2x + y)((2x)^2 + y^2)\)

∴ \((2x - y)(2x + y)(4x^2 + y^2)\)

Let F be the set of people who can speak French and E be the set of people who can speak English. Then,

n(F) = 400

n(E) = 350

n(F ∪ E) = 500

We have to find n(F ∩ E).

Now, n(F ∪ E) = n(F) + n(E) – n(F ∩ E)

⇒ 500 = 400 + 350 – n(F ∩ E)

⇒ n(F ∩ E) = 750 – 500 = 250.

∴ 250 people can speak both languages.

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 (y\(^2\) -8y + 8y - 64) = log 3\(^2\) + log 5\(^2\)

⇒ log (y\(^2\) - 64) = log 3\(^2\) + log 5\(^2\)

⇒ log (y\(^2\) - 64) = log 9 + log 25

⇒ log (y\(^2\) - 64) = log (9 × 25)

⇒ log(y\(^2\) - 64) = log 225

⇒ y\(^2\) - 64 = 225

⇒ y\(^2\) = 225 + 64

⇒ y\(^2\) = 289

⇒ y = ±√289

∴ y = ±17

\(402_y = 102_ten\)

⇒ 4 x y\(^2\) + 0 x y\(^1\) + 2 x y\(^0\) = 102

⇒ 4y\(^2\) + 0 + 2 x 1 = 102

⇒ 4y\(^2\) + 2 = 102

⇒ 4y\(^2\) = 102 - 2

⇒ 4y\(^2\) = 100

⇒ y\(^2 = \frac {100}{4}\)

⇒ y\(^2\) = 25

∴ y = √25 = 5

To find the point of intersection, equate the two equations

⇒ 2x\(^2\) + 10 = 4x + 16

⇒ 2x\(^2\) + 10 - 4x - 16 = 0

⇒ 2x\(^2\) - 4x - 6 = 0

Factorize

⇒ 2(x + 1)(x - 3) = 0

∴ x = -1 or 3

The curves will intersect at -1 and 3

Area = \(\int^b_a\) [upper function] - [lower function] dx

= \(\int^3_1 4x + 16 - (2x^2 + 10) dx\)

= \(\int^3_1 -2x^2 + 4x + 6dx\)

= \((\frac {-2x^3}{3} + 2x^2 + 6x)^3_1\)

= \((\frac {-2(3)^3}{3} + 2(3)^2 + 6(3)) - (\frac {-2(-1)^3}{3} + 2(-1)^2 + 6(-1))\)

= 18 - \((\frac {10}{3})\)

= 18 + \(\frac {10}{3}\)

= \(\frac {64}{3}\)