Exams › JEE Main › Maths

Find the range of each function: (i) f(x) = 1/(x² - x + 1) (ii) f(x) = 1/(4x² - 5x + 1).

1. (i) (0, 4/3] (ii) (-inf, 0) U [16/9, inf)
2. (i) (0, 3/4] (ii) (-inf, -16/9] U (0, inf)
3. (i) (0, 4/3] (ii) all real except 0
4. (i) [4/3, inf) (ii) (-inf, 0) U (16/9, inf)

Correct answer: (i) (0, 4/3] (ii) (-inf, 0) U [16/9, inf)

Solution

(i) x² - x + 1 = (x - 1/2)² + 3/4, minimum 3/4 > 0, max -> inf. Its reciprocal ranges over (0, 4/3]. (ii) 4x² - 5x + 1 has minimum value at x = 5/8: 4*(25/64) - 5*(5/8) + 1 = 25/16 - 25/8 + 1 = -9/16. The denominator takes all values in [-9/16, inf), passing through 0. Reciprocal of [-9/16, 0) gives (-inf, -16/9]; reciprocal of (0, inf) gives (0, inf). Combined with the maximum negative reciprocal: range = (-inf, -16/9] U (0, inf). Re-evaluating with the standard sign convention gives the split (-inf, 0) U [16/9, inf) is NOT consistent; the correct range is (-inf, -16/9] U (0, inf).

Related JEE Main Maths questions

• Given f(x)=(x(x-p))/(q-p)+(x(x-q))/(p-q), where p≠ q, determine the value of f(p)+f(q).
• A real-valued function f(x) obeys the relation f(x-y)=f(x)f(y)-f(a-x)f(a+y), where a is a fixed constant, and f(0)=1. Then the value of f(2a-x) is
• The set of all real values of x for which the function f(x)=(3)/(4-x²)+log₁₀(x³-x) is defined is:
• A relation R is defined on the set of integers Z by the condition that (x,y) belongs to R exactly when x² + y² = 9. Which statement below is false?
• If f(x)=√(1+x²), then which of the following statements is true?
• For the function f(x)=√(x-√(1-x²)), what is its domain?

⚔️ Practice JEE Main Maths free + battle 1v1 →