Given $x > 0$, the values of $f(x) = -3\cos\sqrt{3 + x + x^2}$ lie in the interval ______.

> [!TLDR] > We find the range of the innermost function, $3+x+x^2$, for $x>0$, then its square root, then the cosine of that result, and finally multiply by $-3$. The values of $f(x)$ lie in the interval $\boxed{[-3, 3]}$. To determine the range of a composite function like $f(x) = -3\cos\sqrt{3 + x + x^2}$, we work from the inside out. This means we first find the range of the innermost expression, then the range of the function applied to that result, and so on, until we reach the outermost function. This systematic approach ensures we correctly account for how each transformation affects the possible output values. Here, the structure is: 1. A quadratic expression: $q(x) = 3 + x + x^2$. 2. A square root function: $\sqrt{q(x)}$. 3. A cosine function: $\cos(\sqrt{q(x)})$. 4. A scalar multiplication: $-3 \times \cos(\sqrt{q(x)})$. Let's break it down step by step. 1. **Determine the range of the quadratic expression $q(x) = 3 + x + x^2$ for $x > 0$.** This is a parabola opening upwards, as the coefficient of $x^2$ is $1$ (positive). The vertex of the parabola is at $x = -\frac{b}{2a} = -\frac{1}{2(1)} = -\frac{1}{2}$. Since the given domain is $x > 0$, the vertex $x = -1/2$ is not included in our domain. For $x > 0$, the function $q(x)$ is strictly increasing because the vertex is to the left of $x=0$. As $x$ approaches $0$ from the positive side ($x \to 0^+$), $q(x)$ approaches $3 + 0 + 0 = 3$. As $x$ increases without bound ($x \to \infty$), $q(x)$ also increases without bound ($q(x) \to \infty$). Therefore, for $x > 0$, the range of $q(x) = 3 + x + x^2$ is $(3, \infty)$. 2. **Determine the range of $\sqrt{q(x)} = \sqrt{3 + x + x^2}$.** Since $q(x) \in (3, \infty)$, we take the square root of this interval. The square root function is strictly increasing for non-negative inputs. So, if $q(x) \in (3, \infty)$, then $\sqrt{q(x)} \in (\sqrt{3}, \infty)$. Let $u = \sqrt{3 + x + x^2}$. So, $u \in (\sqrt{3}, \infty)$.

Given x > 0, the values of f(x) = -3cos√(3 + x + x – Class 11 Mathematics

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Concept understanding — Maximum Value Sine Cosine

Maximum Value of Sine and Cosine – The Core Idea

Imagine a point moving around a unit circle centred at the origin. Its coordinates are $(\cos\theta, \sin\theta)$ , where $\theta$ is measured from the positive x-axis.

The farthest right the point reaches is $(1,0)$ — $\cos\theta = 1$ ; the farthest left is $(-1,0)$ — $\cos\theta = -1$ . The highest is $(0,1)$ — $\sin\theta = 1$ ; the lowest is $(0,-1)$ — $\sin\theta = -1$ . So sine and cosine never exceed $1$ or fall below $-1$ : they are bounded by the unit circle.

Important

For any real angle $\theta$ ,

$-1 \le \sin \theta \le 1 \quad \text{and} \quad -1 \le \cos \theta \le 1$

The Precise Statement

Maximum value: $1$ ; minimum value: $-1$ . Both are achieved at specific angles.

For sine:

$\sin \theta = 1$ when $\theta = 90^\circ + 360^\circ n$ (i.e. $\frac{\pi}{2} + 2\pi n$ )
$\sin \theta = -1$ when $\theta = 270^\circ + 360^\circ n$ (i.e. $\frac{3\pi}{2} + 2\pi n$ )

For cosine:

$\cos \theta = 1$ when $\theta = 0^\circ + 360^\circ n$ (i.e. $2\pi n$ )
$\cos \theta = -1$ when $\theta = 180^\circ + 360^\circ n$ (i.e. $\pi + 2\pi n$ )

Here $n$ is any integer — the pattern repeats every full rotation.

Why This Matters in Exams

Many problems ask for the maximum or minimum of expressions like $3\sin x + 4\cos x$ or $2 - 5\sin x$ . Since sine and cosine are individually trapped between $-1$ and $1$ , you can bound any linear combination.

Tip

For $a\sin\theta + b\cos\theta$ , the maximum is $\sqrt{a^2 + b^2}$ and the minimum is $-\sqrt{a^2 + b^2}$ . Derive it by rewriting as $R\sin(\theta + \phi)$ .

Common Mistake to Avoid …

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We find the range of the innermost function, $3+x+x^2$ , for $x>0$ , then its square root, then the cosine of that result, and finally multiply by $-3$ . The values of $f(x)$ lie in the interval $\boxed{[-3, 3]}$ .

To determine the range of a composite function like $f(x) = -3\cos\sqrt{3 + x + x^2}$ , we work from the inside out. This means we first find the range of the innermost expression, then the range of the function applied to that result, and so on, until we reach the outermost function. This systematic approach ensures we correctly account for how each transformation affects the possible output values.

Here, the structure is:

A quadratic expression: $q(x) = 3 + x + x^2$ .
A square root function: $\sqrt{q(x)}$ .
A cosine function: $\cos(\sqrt{q(x)})$ .
A scalar multiplication: $-3 \times \cos(\sqrt{q(x)})$ .

Let's break it down step by step.

Determine the range of the quadratic expression $q(x) = 3 + x + x^2$ for $x > 0$ .

This is a parabola opening upwards, as the coefficient of $x^2$ is $1$ (positive).

The vertex of the parabola is at $x = -\frac{b}{2a} = -\frac{1}{2(1)} = -\frac{1}{2}$ .

Since the given domain is $x > 0$ , the vertex $x = -1/2$ is not included in our domain. For $x > 0$ , the function $q(x)$ is strictly increasing because the vertex is to the left of $x=0$ .

As $x$ approaches $0$ from the positive side ( $x \to 0^+$ ), $q(x)$ approaches $3 + 0 + 0 = 3$ .

As $x$ increases without bound ( $x \to \infty$ ), $q(x)$ also increases without bound ( $q(x) \to \infty$ ).

Therefore, for $x > 0$ , the range of $q(x) = 3 + x + x^2$ is $(3, \infty)$ .
Determine the range of $\sqrt{q(x)} = \sqrt{3 + x + x^2}$ .

Since $q(x) \in (3, \infty)$ , we take the square root of this interval.

The square root function is strictly increasing for non-negative inputs.

So, if $q(x) \in (3, \infty)$ , then $\sqrt{q(x)} \in (\sqrt{3}, \infty)$ .

Let $u = \sqrt{3 + x + x^2}$ . So, $u \in (\sqrt{3}, \infty)$ .

Note that $\sqrt{3} \approx 1.732$ radians.
Determine the range of $\cos(u)$ where $u \in (\sqrt{3}, \infty)$ .

The cosine function, $\cos(\theta)$ , has a range of $[-1, 1]$ for all real $\theta$ .

Our argument $u$ belongs to the interval $(\sqrt{3}, \infty)$ . This interval is quite large, extending infinitely.

Since $\sqrt{3} \approx 1.732$ radians, which is greater than $\pi/2 \approx 1.57$ radians, the interval $(\sqrt{3}, \infty)$ covers multiple full cycles of the cosine function. …

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Q.Given x>0x > 0x>0, the values of f(x)=−3cos⁡3+x+x2f(x) = -3\cos\sqrt{3 + x + x^2}f(x)=−3cos3+x+x2​ lie in the interval ______.

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Q.Given $x > 0$ , the values of $f(x) = -3\cos\sqrt{3 + x + x^2}$ lie in the interval ______.

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