Directions: In the following questions, two equations numbered I and II are given. You have to solve both the equations and give an answer.
I. a2 + 16a + 55 = 0
II. b2 + 7b + 12 = 0I. a2 + 16a + 55 = 0
⇒ a2 + 11a + 5a + 55 = 0
⇒ a(a + 11) + 5(a + 11) = 0
⇒ (a + 11)(a + 5) = 0
Then, a = -11 or a = -5
II. b2 + 7b + 12 = 0
⇒ b2 + 4b + 3b + 12 = 0
⇒ b(b + 4) + 3(b + 4) = 0
⇒ (b + 4)(b + 3) = 0
Then, b = -4 or b = -3
So, when a = -11, a < b for b = -4 and a < b for b = -3
And when a = -5, a < b for b = -4 and a < b for b = -3
∴ We can observe that a < b.3. Divide 150 into two parts such that the sum of their reciprocals is 3/112. Calculate both the parts.
4. If √($cosec^{2}A$ - 1) = x, then the value of x is
6. Onechord of a circle is known to be 10.1 cm. The radius of this circle must be:
7. The 3rd and 7th term of an arithmetic progression are -9 and 11 respectively. What is the 15th term?
8. If $cot^{2}A$ - $cos^{2}A$ = x, then value of x is
9. Simplify: \(\frac{{{{\left( {a + b} \right)}^2}}}{{\left( {{a^2} - {b^2}} \right)}} = ?\)