Exams › JEE Main › Maths

Let f: R -> R be given by f(x) = ln(x + sqrt(x² + 1)), and let g be a function such that g(f(x)) = x for every real x. Find g(2).

1. (e² + e⁻²)/2
2. e²
3. (e² - e⁻²)/2
4. e⁻²

Correct answer: (e² - e⁻²)/2

Solution

f(x) = ln(x + sqrt(x²+1)) = arcsinh(x). Since g(f(x)) = x, g is the inverse of f, i.e. g(y) = sinh(y) = (e^y - e^-y)/2. Therefore g(2) = (e² - e⁻²)/2.

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 →