Question

Solve the following questions.

Answer 1

13. Given  $\frac{ 3 - \sqrt{5} }{3 + 2\sqrt{5} }$ = $a\sqrt{5} - b$

⇒ $\frac{ (3 - \sqrt{5}) (3 - 2\sqrt{5}) }{ (3 + 2\sqrt{5}) (3 - 2\sqrt{5}) }$ = $a\sqrt{5} - b$

⇒$\frac{[9 + (2 x 5)]}{(9 -20)}$ = $a\sqrt{5} - b$

⇒$\frac{- 19}{11}$ = $a\sqrt{5} - b$

⇒$0 \sqrt{5} - \frac{19}{11} = a \sqrt{5} - b$

Substituting the values, we get ,

a = 0

b = $\frac{19}{11}$


14. p$p(x) = x^{4} - 2 x^{3} + 3 x^{2} - ax + 3a -7 \\ \\ Given = p(-1) = 19 \\ \\ But, p(-1) = (-1)^{4} - 2 (-1)^{3} + 3 (-1)^{2} - a(-1) + 3a - 7 = 19 \\ \\ Therefore, \\ \\ 1 - (-2) + 3 + a +3a - 7 = 19 \\ \\ Therefore, \\ \\ 4a-1 =19 \\ \\ 4a = 20 \\ \\ a =5$

15. (a) Total distance in km = $x$

Total fare =$y$

Fare = 20 + (Total distance in km - 1) x 9

⇒y = 9(x-1) + 20

 y = 9x - 9 + 20

 y =9x +11

(b) Total distance = $x$ = 15

Fare = 9 (15 ) + 11

= 135 + 11

= 146

As the man gave him 150 Rs. The driver would have returned him 4 Rs. The driver shows moral fiber and an will to do his work honestly.