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Question: The domain of the function \(f\left( x \right)=\exp \left( \sqrt{5x-3-2{{x}^{2}}} \right)\) is: 1\...

The domain of the function f(x)=exp⁡(5x−3−2x2)f\left( x \right)=\exp \left( \sqrt{5x-3-2{{x}^{2}}} \right) is:
1. [32,∞]\left[ \dfrac{3}{2},\infty \right]
2. \left[ 1,\left. \dfrac{3}{2} \right\\} \right.
3. (−∞,1]\left( -\infty ,1 \right]
4. \left( 1,\left. \dfrac{3}{2} \right\\} \right.

Explanation

Solution

For the domain we have to find the value of xx for which the given function is always defined, in this case we will put the value under root ≥0\ge 0, and then find the values of xx for which it is defined. Now for the range of function we will put the values of xx for which it gives maximum and minimum value and that will be the range.

Complete step-by-step solution:
According to the question, it is asked to find the domain of the function f(x)=exp⁡(5x−3−2x2)f\left( x \right)=\exp \left( \sqrt{5x-3-2{{x}^{2}}} \right). So, the domain of the function f(x)f\left( x \right) is,
f(x)=exp⁡(5x−3−2x2)f\left( x \right)=\exp \left( \sqrt{5x-3-2{{x}^{2}}} \right)
Exp∫xExp\int{x} is defined everywhere or redouble line, then for 5x−3−2x2\sqrt{5x-3-2{{x}^{2}}} to be value 5x−3−2x2≥05x-3-2{{x}^{2}}\ge 0. We can also write it as −2x2−5x−3≥0-2{{x}^{2}}-5x-3\ge 0.
Since we know that roots of a quadratic equation can be expressed as, x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}. So, we can write it here as,
x=−5±25−4(−2)(−3)2x(−2) ⇒x=−5±25−24−4=−5+1−4,−5−1−4 \begin{aligned} & x=\dfrac{-5\pm \sqrt{25-4\left( -2 \right)\left( -3 \right)}}{2x\left( -2 \right)} \\\ & \Rightarrow x=\dfrac{-5\pm \sqrt{25-24}}{-4}=\dfrac{-5+1}{-4},\dfrac{-5-1}{-4} \\\ \end{aligned}
On further simplification we will get the values as follows,
x=−4−4,−6−4 ⇒x=1,1.5 \begin{aligned} & x=\dfrac{-4}{-4},\dfrac{-6}{-4} \\\ & \Rightarrow x=1,1.5 \\\ \end{aligned}
So, 5x−3−2x2≥05x-3-2{{x}^{2}}\ge 0
⇒(x−1)(x−1.5)≥0\Rightarrow \left( x-1 \right)\left( x-1.5 \right)\ge 0
So, the domain is (−∞,1]∪[32,∞)\left( -\infty ,1 \right]\cup \left[ \dfrac{3}{2},\infty \right).
So, the final answer will be (−∞,1]\left( -\infty ,1 \right].
Hence the correct answer is option 3.

Note: While solving this type of questions you have to always take the function as greater than or equal to zero because the value of any function must be positive and this condition is used in finding the domain, thus we can find the domain for any function in this manner.