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# Find the greatest value of $f\left( x \right)=\cos \left( x{{e}^{\left[ x \right]}}+7{{x}^{2}}-3x \right)$, $x\in [-1,\infty )$A. -1B. 1C. 0D. none of these

Last updated date: 09th Sep 2024
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Hint: We first try to define the domain of the given trigonometric function. Then we state the range of the function for any values of t, $\cos t\in \left[ -1,1 \right]$. We also try to get one continuous domain of $2\pi$ distance. At the end we find one-point x for which it attains the maximum value.

Complete step by step answer:
The main function of the given $f\left( x \right)=\cos \left( x{{e}^{\left[ x \right]}}+7{{x}^{2}}-3x \right)$ is $\cos$ function.
Now in a span of $2\pi$, it will take values of $\left[ -1,1 \right]$. Also, for any values of t, $\cos t\in \left[ -1,1 \right]$.
We just have to check that we can get a continuous domain of $2\pi$ distance.
Now we find a range of $x{{e}^{\left[ x \right]}}+7{{x}^{2}}-3x$. Here $\left[ x \right]$ is the greatest integer function which means the output is the greatest integer possible less than x.
As $x\in [-1,\infty )$, we can say $\left[ x \right]\in [-1,\infty )$ with only integer values.
We also know for any value of x; the exponential function always gets only positive value.
So, $x\in [-1,\infty )$, we can say ${{e}^{\left[ x \right]}}\in [\dfrac{1}{e},\infty )$.
So, we can see the function attains at least one continuous domain of $2\pi$ distance as it goes towards infinity.
So, we can say the greatest value of $f\left( x \right)=\cos \left( x{{e}^{\left[ x \right]}}+7{{x}^{2}}-3x \right)$ is 1.
We can also prove it by just showing 1 value of x for which $f\left( x \right)=\cos \left( x{{e}^{\left[ x \right]}}+7{{x}^{2}}-3x \right)=1$.
Let’s take $x=0$.
We find the value of $x{{e}^{\left[ x \right]}}+7{{x}^{2}}-3x$. So, $x{{e}^{\left[ x \right]}}+7{{x}^{2}}-3x=0$.
So, at $x=0$, $\cos 0=1$. We already got a point.

So, the correct answer is “Option B”.

Note: We need to always shoe at least one point which attains the maximum point. As the part $x{{e}^{\left[ x \right]}}$ can have some fixed points due to the factor that ${{e}^{\left[ x \right]}}$ attains only integer value. We have to show that at least one point of x satisfies the equation.