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**Example text**

Y = log F 5x − x I > 0 ⇔ 5x – GH 4 JK 2 Solution: y is defined when x2 > 0 ⇔ x (5 – x) > 0 ⇔ (x – 0) (x – 5) < 0 ⇔ 0 < x a f ∴ D a y f = a0 , 5f F xI 2. : In the above two examples the functions in denominator are positive. This is why considerable function to be greater than zero is only the function in numerator. 3. y = log Fx GH x 2 2 I J + 4 x + 6K − 5x + 6 2 2 I >0 ⇔ J + 4x + 6K − 5x + 6 a f between 2 and 3 ⇔ x < 2 or x > 3 ⇔ x ∈ −∞ , 2 ∪ a3, + ∞f . : In the above example (3), the discriminant D = 16 – 4 × 1 × 6 = 16 – 24 = –ve for the function x2 + 4x + 6 in denominator which ⇒ x2 + 4x + 6 > 0.

The method adopted in the above example is called “if method”. 3. A perfect square is always positive which is greater than any negative number. Method 2. This method consists of showing that 2 a x + b x + c > 0 , ∀ x if a > 0 and discriminant = b2 – 4ac < 0 here 3 > 0, and discriminant = 16 – 60 = – 34 <0 ∴ y is defined ∀ x ∈ R af a f Therefore, D y = R = −∞ , + ∞ 6. y = log (x3 – x) Solution: y is defined when (x3 – x) > 0 ⇔ x (x2 – 1) > 0 ⇔ x (x + 1) (x – 1) > 0 ⇔ (x – 0) (x + 1) (x – 1) > 0 ⇔ (x – (1)) (x – 0) (x – 1) > 0 Now let f (x) = (x – (–1)) (x – 0) (x – 1) If x < –1, then f (x) < 0 as all the three factors are < 0.

In such cases, it is required to be found out the domain and the range of the given function. Already, how to find out the domains of different types of functions has been discussed. Now the methods of finding the range of a given function will be explained. Firstly, domains and range sets of standard functions will be put in a tabular form. 1. y = kx, k ≠ 0 2. y = kx + l ≤1 Now −1 ≤ 2 − 3 x 2 45 6. y = x 7. (i) Domain Range (–∞, ∞) (–∞, ∞) (–∞, ∞) (–∞, ∞) R – {0} R – {0} (–∞, ∞) (–∞, ∞) (0, ∞) (–∞, ∞) [0, ∞) [0, ∞) (–∞, ∞) LM− D , + ∞IJ , N 4a K D = b2 – 4ac a>0 8.