Find the domain & range of the following functions y = 2 x 1 + x 2

y = 2 x 1 + x 2 In order to define y , the denominator should not be zero and denominator is always positive. So, domain of y is R . Range: y = 2 x 1 + x 2 ⇒ x 2 y + y = 2 x ⇒ x 2 y - 2 x + y = 0 ⇒ x = 2 ± 4 - 4 y 2 2 y Here, x to be real, 4 - 4 y 2 ≥ 0 ⇒ 4 ( y 2 - 1 ) ≤ 0 ⇒ ( y - 1 ) ( y + 1 ) ≤ 0 Therefore y ∈ - 1 , 1 . Domain is R ; Range is − 1 , 1 .

Find the domain & range of the following functions y = 2 − x + 1 + x

Domain of y = 2 − x + 1 + x For 2 - x , 2 - x ≥ 0 ⇒ x ≤ 2 . For 1 + x , 1 + x ≥ 0 ⇒ x ≥ - 1 Therefore, domain of y is ( x ≤ 2 ) ∩ ( x ≥ - 1 ) ⇒ - 1 ≤ x ≤ 2 . Range of y = 2 − x + 1 + x After squaring on both sides, y 2 = 2 - x + 1 + x + 2 ( 2 - x ) ( 1 + x ) ⇒ y 2 = 3 + 2 2 + x - x 2 ⇒ y 2 - 3 2 2 = 2 + x - x 2 ⇒ x 2 - x + y 2 - 3 2 2 - 2 = 0 This is a quadratic equation in the form of a x 2 + b x + c = 0 . To get real roots, b 2 - 4 a c ≥ 0 ⇒ 1 - 4 y 2 - 3 2 2 - 2 ≥ 0 ⇒ y 4 - 6 y 2 + 1 ≤ 1 ⇒ y 2 ( y 2 - 6 ) ≤ 0 ⇒ y 2 ∈ 0 , 6 ⇒ y ∈ 0 , 6 But, the domain is [ - 1 , 2 ] and it is an increasing function. f ( - 1 ) = 2 - - 1 + 1 + ( - 1 ) = 3 Therefore, the range is y ∈ 3 , 6 .

The equation x 2 - a x + b = 0 & x 3 - p x 2 + q x = 0 , where b ≠ 0 , q ≠ 0 , have one common root & the second equation has two equal roots. Prove that 2 q + b = a p .

x 2 - a x + b = 0 … 1 x 3 - p x 2 + q x = 0 … 2 ⇒ x x 2 - p x + q = 0 2 has two equal roots, but q ≠ 0 . ⇒ x = 0 is one root which is not equal root. ⇒ x 2 - p x + q is a perfect square. ⇒ p 2 = 4 q , common root = p 2 . 1 has one common root with 2 & since b ≠ 0 , so product of roots cannot be zero, then x = 0 is not a root of 1 ⇒ Common root = p 2 Root of 1 is a ± a 2 - 4 b 2 = p 2 ⇒ a ± a 2 - 4 b = p ⇒ a - p = ± a 2 - 4 b (squaring on both sides) a - p 2 = a 2 - 4 b ⇒ a 2 + p 2 - 2 a p = a 2 - 4 b ⇒ p 2 + 4 b = 2 a p ⇒ 4 q + 4 b = 2 a p ⇒ 2 q + b = a p