**Question:**

**Find the range of the following functions given by**

**(i) $f(x)=\frac{3}{2-x^{2}}$**

(ii) $f(x)=1-|x-2|$

(iii) $f(x)=|x-3|$

(iv) $f(x)=1+3 \cos 2 x$

**Solution:**

(i)

$f(x)=\frac{3}{2-x^{2}}$

According to the question, Let $f(x)=y$,

Let $f(x)=y$,

$y=\frac{3}{2-x^{2}}$

$\Rightarrow 2-x^{2}=\frac{3}{y}$

$\Rightarrow x^{2}=2-\frac{3}{y}$

But, we know that, $x^{2} \geq 0$

$2-\frac{3}{y} \geq 0$

$\Rightarrow \frac{2 y-3}{y} \geq 0$

⇒ y>0 and 2y–3≥0

⇒ y>0 and 2y≥3

⇒ y>0 and y ≥ 3/2

Or f(x)>0 and f(x) ≥ 3/2

f(x) ∈ ( – ∞, 0) ∪ [ 3/2 , ∞)

$\Rightarrow \mathrm{f}(\mathrm{x}) \in(-\infty, 0) \cup\left[\frac{3}{2}, \infty\right)$

Therefore, the range of f = ( – ∞, 0) ∪ [ 3/2 , ∞)

(ii) f(x) = 1–|x–2|

According to the question,

For real value of f,

|x–2|≥ 0

Adding negative sign, we get

Or –|x–2|≤ 0

Adding 1 we get

⇒ 1–|x–2|≤ 1

Or f(x)≤1

⇒ f(x)∈ (–∞, 1]

Therefore, the range of f = (–∞, 1]

(iii) f(x) = |x–3|

According to the question,

We know |x| are defined for all real values.

And |x–3| will always be greater than or equal to 0.

i.e., f(x) ≥ 0

Therefore, the range of f = [0, ∞)

(iv) f (x) = 1 + 3 cos2x

According to the question,

We know the value of cos 2x lies between –1, 1, so

–1≤ cos 2x≤ 1

Multiplying by 3, we get

–3≤ 3cos 2x≤ 3

Adding 1, we get

–2≤ 1 + 3cos 2x≤ 4

Or, –2≤ f(x)≤ 4

Hence f(x)∈ [–2, 4]

Therefore, the range of f = [–2, 4]

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