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MATHMAA

SECTION - C

15)

Proof:

Let us assume that √2 is an rational number.

√2 = $$\large \frac{p}{q}$$      where HCF(p,q) = 1

Squaring on both sides,

2 = $$\large (\frac{p}{q})^{2}$$

2 = $$\large \frac{p^{2}}{q^{2}}$$

2$$q^{2}$$ = $$p^{2}$$

So, 'p' is a even number.

Then, p = 2m     where 'm' is any integer.

Squaring on both sides,

$$p^{2}$$ = $$2m^{2}$$

$$p^{2}$$ = 4$$m^{2}$$

$$2q^{2}$$ = 4$$m^{2}$$

$$q^{2}$$ = 2$$m^{2}$$

So 'q' is also an even number,

Therefore 2 is a common factor of 'p' and 'q', this contradicts our assumption that √2 is a rational number.

Hence √2 is an irrational number, so addition and multiplication with an irrational number is an irrational number.

$$\therefore$$ 3 + 5√2 is an irrational number.

Hence proved.

16)

So, 3x³ + 10x² - 14x + 9 = 3x-2(x² + 4x - 2) + 5

Here the remainder is 5.

Therefore 5 should be subtracted from 3x³ + 10x² - 14x + 9 so that 3x - 2 divides it exactly.

17)

5x + $$\large \frac{4}{y}$$ = 9     ----------(1)
7x - $$\large \frac{2}{y} = 5 ----------(2) (2)*2, we get 14x - \(\large \frac{4}{y}$$ = 10  ----------(3)

(3)+(1)

14x - $$\large \frac{4}{y}$$ + 5x - $$\large \frac{4}{y}$$  = 10 + 9

19x = 19

x = 1

Plug x = 1 in (2),

7 - $$\large \frac{2}{y}$$ = 5

$$\large \frac{2}{y}$$ = 2

y = 1

Therefore x = 1 and y = 1.

18)

Let the fraction be $$\large \frac{x}{y}$$.

Then,

x + y = 8          ----------(1)

$$\large \frac{x+3}{y+3}$$ = $$\large \frac{3}{4}$$
4(x+3) = 3(y+3)
4x + 12 = 3y + 9
4x - 3y = -3      ----------(2)

(1)*3

3x + 3y = 24     ----------(3)

(2) + (3)

4x - 3y + 3x + 3y = -3 + 24

7x = 21

x = 3

Plug x=3 in (1), we get

y = 5

Therefore the fraction becomes $$\large \frac{3}{5}$$.
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