Question

Let f(x) = 121 - $x^2$, g(x) = |x - 8| + |x + 8| and h(x) = min {f(x), g(x)}. What is the number of integer values of x for which h(x) is equal to a positive integral value?

• A. 17
• B. 19
• C. 21
• D. 23

Answer 1

The correct option is C 21

In order to understand this question, you first need to develop your thought process about what the value of h(x) is in various cases. A little bit of trial and error would show you that the value of h(x) since it depends on the minimum of f(x) and g(x), would definitely be dependant on the value of f(x) once x becomes greater than 11 or less than - 11. Also, the value of g(x) is fixed as an integer at 16. It can be observed that when f(x) > g(x) i.e. f(x) > 16, the values of h(x) would also be 16 and hence would be a positive integer.
With this thought when you look at the expression of f(x) = 121 - $x^2$, you realise that the value of x can be - 10, -9, -8, -7, ...0, 1, 2, 3,....8, 9, 10, i.e., 21 values of x when h(x) = g(x) = 16. When we use x = 11 or x = -11, the value of f(x) = min (0, 16) = 0 and is not a positive integral value.
Hence, the correct answer is Option (c).