Question

Let \(\mathrm{f}: \mathrm{R} \rightarrow\left[0, \frac{\pi}{2}\right)\) defined by \(f(x)=\tan ^{-1}\left(x^{2}+x+a\right)\), then set of value(s) of 'a' for which \(f\) is onto is (1) \([0, \infty)\) (2) \([2,1]\) (3) \(\left[\frac{1}{4}, \infty\right)\) (4) \(\left\\{\frac{1}{4}\right\\}\) (5) None of these

Short answer

The correct answer is (3) \(\left[\frac{1}{4}, \infty\right)\).

Step-by-step solution

- Understand the function

The function given is \ f(x)=\tan^{-1}(x^2 + x + a). \ This is an arctangent function which means it maps \(\mathbb{R}\) to the range [0, \frac{\pi}{2}). Our goal is to find the values of 'a' such that \(f(x)\) is onto this entire interval.

- Analyze the function inside arctangent

Inside the arctangent, the expression is \(x^2 + x + a\). This is a quadratic function in terms of \(x\). A quadratic function is a parabola, and its minimum value occurs at the vertex because the parabola opens upwards.

- Find the vertex of the quadratic function

The vertex of the quadratic function \(x^2 + x + a\) occurs at \(x = -\frac{b}{2a}\). For the standard quadratic \(ax^2 + bx + c\), in this case, \(a = 1\), \(b = 1\), and \(c = a\). The x-coordinate of the vertex is \(-\frac{1}{2}\). Substitute this back into the quadratic to find the minimum value: \ \left(-\frac{1}{2}\right)^2 - \frac{1}{2} + a = \frac{1}{4} - \frac{1}{2} + a = a - \frac{1}{4}.

- Set the range requirements

For the function \(f(x) = \tan^{-1}(x^2 + x + a)\) to be onto [0, \frac{\pi}{2}), the minimum value of \(x^2 + x + a\) must be 0. This means that: \ a - \frac{1}{4} \ge 0 \ So, \(a \ge \frac{1}{4}\).

- Identify the solution

The value \ a = \frac{1}{4} \ ensures that \(x^2 + x + a\) touches zero and covers every value towards infinity, allowing \(\tan^{-1}(x^2 + x + a)\) to cover the entire interval [0, \frac{\pi}{2}).

- Confirm the correct answer choice

The correct set of values for which \(f(x)\) is onto is therefore \( \left[\frac{1}{4}, \infty\right) \). This corresponds to option (3).

Key concepts

Functions and Mappings

In mathematics, a function is a relationship between a set of inputs and a set of possible outputs, where each input is related to exactly one output. Functions can be thought of as 'rules' that assign each input to one output.

• In this exercise, the function is given by\(f(x) = \tan^{-1}(x^2 + x + a)\). The input is any real number, and the output is constrained to lie within the interval\(\bigg[0, \frac{\pi}{2}\bigg)\).

Mappings are a broader concept where each element of one set is paired with an element of another set. Functions are a specific type of mapping where this pairing rule is strict.

• The function\(f\) in this problem maps every real number\(x\) to a unique value in\(\bigg[0, \frac{\pi}{2}\bigg)\).

Understanding functions and mappings is essential, especially when determining if a function is onto (surjective).
An onto function covers every element in the target set. Here, we need to ensure that the function\(f\) maps to every value in\(\bigg[0, \frac{\pi}{2}\bigg)\).

Quadratic Functions

Quadratic functions are polynomial functions of degree 2. The general form is\(ax^2 + bx + c\), where\(a, b,\) and\(c\) are constants.
In this problem, the expression inside the arctangent function is\(x^2 + x + a\).

• A quadratic function forms a parabola when graphed. The parabola’s minimum or maximum point is called the vertex.

For the quadratic expression\(x^2 + x + a\), it forms an upward-opening parabola because the coefficient of\(x^2\) (which is 1) is positive.

• The vertex of this quadratic is found at\(x = -\frac{b}{2a}\). Substituting\(a = 1\) and\(b = 1\), the vertex is at\(x = -\frac{1}{2}\).

Substituting\(x = -\frac{1}{2}\) back into\(x^2 + x + a\), we get the minimum value as\(a - \frac{1}{4}\).

• The minimum value helps us determine the lower bound of the quadratic expression. Given that our function must be onto, we set this value to ensure it starts from 0 to cover all values within the interval\(\bigg[0, \frac{\pi}{2}\bigg)\).

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverses of the trigonometric functions. They provide the angle that corresponds to a given trigonometric ratio.
\(\tan^{-1}(x)\) represents the angle whose tangent is\(x\). Its range is\(\big[0, \frac{\pi}{2}\big)\) for non-negative values.
In the exercise, the function \(f(x) = \tan^{-1}(x^2 + x + a)\) is an arctangent function.

• To ensure\(f(x)\) is onto, the expression inside the arctangent function\(x^2 + x + a\) must cover all non-negative values including 0.

This is why we analyzed the vertex of the quadratic. For the function to be onto \(\big[0, \frac{\pi}{2}\big)\), it is necessary that the minimum value of\(x^2 + x + a\) is at least 0.

• This condition provided us with the necessary requirement\(a - \frac{1}{4} \geq 0\), leading to\(a \geq \frac{1}{4}\).
Understanding inverse trigonometric functions is crucial for solving problems where the solution lies in precisely determining the output interval.