Question

Given: $f(x) = 4{x^3} - 6{x^2}\cos (2a) + 3x \times \sin (2a) \times \sin (6a) + \sqrt {\ln (2a - {a^2})}$ then
$A.)$ $f(x)$ is not defined at $x = \dfrac{1}{2}.$
$B.)$ $f'\left( {\dfrac{1}{2}} \right) < 0.$
$C.)$ $f'(x)$ is not defined at $x = \dfrac{1}{2}.$
$D.)$ $f'\left( {\dfrac{1}{2}} \right) > 0.$

Answer 1

First of all we will consider the value of the function by putting the value of $x$ into the function and then we will try to check the function whether it does hold the value for the function or not. Finally we will do the derivation of the given function.

Formula used: Derivative of $y$ with respect to $x$ is $= \dfrac{{dy}}{{dx}}.$
So, $\dfrac{d}{{dx}}(f(x)) = f'(x).$
Derivatives of any constant term is always $0.$
So, $\dfrac{d}{{dx}}({x^n}) = n.{x^{(n - 1)}}.$
Also, $2\sin (A)\sin (B) = \cos (A - B) - \cos (A + B)$
${(a - 1)^2} = {a^2} - 2a + {b^2}$

Complete step-by-step solution:
It is given that $f(x) = 4{x^3} - 6{x^2}\cos (2a) + 3x \times \sin (2a) \times \sin (6a) + \sqrt {\ln (2a - {a^2})}$
Here $\cos (2a),\sin (2a),\sqrt {ln(2a - {a^2})} ,\sin (6a)$ are constant terms.
So, individual derivatives of all this terms will be $0.$
Now, if we go by the first option, then we can see the following value of the function:
$f(x)$ at $x = \dfrac{1}{2}$ is,
f\left( {\dfrac{1}{2}} \right)$$$$ = 4{x^3} - 6{x^2}\cos (2a) + 3x\sin (2a)\sin (6a) + \sqrt {\ln (2a - {a^2})}
On putting the $x$ values and we get
$\Rightarrow 4 \times \dfrac{1}{{{2^3}}} - 6 \times \dfrac{1}{{{2^2}}} \times \cos (2a) + 3 \times \dfrac{1}{2} \times \sin (2a) \times \sin (6a) + \sqrt {\ln (2a - {a^2})}$
On simply the terms and we get
$\Rightarrow 0.5 - 1.5\cos (2a) + 1.5\sin (2a)\sin (6a) + \sqrt {\ln (2a - {a^2})} ...\left( 1 \right)$
Here we can see that the value of $a$ is not known in the given question.
So, it is not possible to find the value of ${\text{cosine}}$ and ${\text{sine}}$.
So, we need to evaluate the value of $a$.
Now, whatever the value of $a$ has, the value of $\ln (2a - {a^2})$ must be greater than equal to $0$.
So, $\ln (2a - {a^2}) \geqslant 0$
The above equation is true only if the following condition is truth,
$\Rightarrow (2a - {a^2}) \geqslant 1$
Taking constant terms to the R.H.S:
$\Rightarrow ({a^2} - 2a + 1) \leqslant 0$
By using the formula we can write it as,
$\Rightarrow {(a - 1)^2} \leqslant 0$
But any value of $(a - 1)$ does not satisfy the above inequality
So, $(a - 1) = 0$
Or, $a = 1$
So, if we put the value of $a = 1$ in $\left( 1 \right)$ equation, we get:
$\Rightarrow f\left( {\dfrac{1}{2}} \right) = 0.5 - 1.5\cos (2) + 1.5\sin (2)\sin (6) + \sqrt {\ln (2 - {1^2})}$
Since $\ln (1) = 0$
$\Rightarrow f\left( {\dfrac{1}{2}} \right) = 0.5 - 1.5\cos (2) + 1.5\sin (2)\sin (6)$
So, we can see that $f(x)$ is definable at $x = \dfrac{1}{2}.$
So, option A is the wrong choice.
Now, if we take derivative in the both sides of the given equation, we will stand up with the following equation:
$\dfrac{d}{{dx}}(f(x)) = \dfrac{d}{{dx}}(4{x^3} - 6{x^2}\cos (2a) + 3x\sin (2a)\sin (6a) + \sqrt {\ln (2a - {a^2})} )$.
On splitting the differentiation term and we get
$\Rightarrow f'(x) = \dfrac{d}{{dx}}(4{x^3}) - \dfrac{d}{{dx}}(6{x^2}\cos (2a)) + \dfrac{d}{{dx}}(3x\sin (2a)\sin (6a)) + \dfrac{d}{{dx}}(\sqrt {\ln (2a - {a^2})} )$
Let us differentiation by using the formula and we get
$\Rightarrow f'(x) = (4 \times 3{x^{3 - 1}}) - (6 \times 2 \times {x^{2 - 1}}\cos (2a)) + (3 \times 1 \times {x^{1 - 1}} \times \sin (2a) \times \sin (6a)) + 0$
Derivative of the constant terms are always $0$
$\Rightarrow f'(x) = 12{x^2} - 12x\cos (2a) + 3\sin (2a) \times \sin (6a)$
Now, if we put $x = \dfrac{1}{2}$ in the above equation,
$\Rightarrow f'\left( {\dfrac{1}{2}} \right) = 12{\left( {\dfrac{1}{2}} \right)^2} - 12\left( {\dfrac{1}{2}} \right)\cos (2a) + 3\sin (2a)\sin (6a)$
On simply the terms and we get
$\Rightarrow f'\left( {\dfrac{1}{2}} \right) = 12\left( {\dfrac{1}{4}} \right) - 6\cos (2a) + 3\sin (2a)\sin (6a)$
On divide the first term of the RHS we get
$\Rightarrow f'\left( {\dfrac{1}{2}} \right) = 3 - 6\cos (2a) + 3\sin (2a)\sin (6a)$
Now, we got the value $a$ is $1$ and by putting this value to the above equation,
Rewrite the above equation:
$\Rightarrow f'\left( {\dfrac{1}{2}} \right) = 3 - 6\cos (2) + 3\sin (2)\sin (6)$
Putting the value of ${\text{cosine}}$ and ${\text{sine}}$.
$\Rightarrow f'\left( {\dfrac{1}{2}} \right) = 3 - 6 \times ( - 0.416) + 3 \times (0.909) \times ( - 0.279)$
Let us multiply the term and we get
$\Rightarrow f'\left( {\dfrac{1}{2}} \right) = 3 + 2.496 - 0.760$
Let us simplify the RHS and we get,
$\Rightarrow f'\left( {\dfrac{1}{2}} \right) = 4.736.$
So, $f'(x)$ is definable at $x = \dfrac{1}{2}$ and its value is always greater than $0$.
So, option D is the correct choice.

Note: Point to be remembered:
Determine the constant terms.
Before the derivation split up the constant terms from the variables.
Check whether the cosine and sine values are in radians or in degree.