Key Concepts and Formulas

• Critical Points: A function $f(x)$ has critical points at values of $x$ where $f'(x) = 0$ or $f'(x)$ is undefined. For critical points to exist, the equation $f'(x) = 0$ must have at least one real solution for $x$.
• Trigonometric Identities:
  • $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$
  • $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$
• Range of Trigonometric Functions:
  • The range of $\sin(x)$ is $[-1, 1]$.
  • The range of $\cos(x)$ is $[-1, 1]$.
  • The range of $\cot(x)$ is $(-\infty, \infty)$.
• Quadratic Equation Roots: A quadratic equation $Ax^2 + Bx + C = 0$ has real roots if and only if its discriminant $\Delta = B^2 - 4AC \ge 0$.

Step-by-Step Solution

Step 1: Simplify the given function $f(x)$

The given function is $f(x) = (4a - 3)(x + \log_e 5) + 2(a - 7) \cot\left(\frac{x}{2}\right) \sin^2\left(\frac{x}{2}\right)$, for $x \ne 2n\pi$, $n \in \mathbb{N}$.

We can simplify the trigonometric part of the function using the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$ and $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$.

$f(x) = (4a - 3)(x + \log_e 5) + 2(a - 7) \frac{\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \sin^2\left(\frac{x}{2}\right)$ $f(x) = (4a - 3)(x + \log_e 5) + 2(a - 7) \cos\left(\frac{x}{2}\right) \sin\left(\frac{x}{2}\right)$

Using the double angle identity for sine, $2\sin(\theta)\cos(\theta) = \sin(2\theta)$, we have: $2 \cos\left(\frac{x}{2}\right) \sin\left(\frac{x}{2}\right) = \sin\left(2 \cdot \frac{x}{2}\right) = \sin(x)$.

So, the simplified function becomes: $f(x) = (4a - 3)(x + \log_e 5) + (a - 7) \sin(x)$

Step 2: Find the first derivative of $f(x)$, $f'(x)$

To find the critical points, we need to compute the derivative of $f(x)$ with respect to $x$.

$f'(x) = \frac{d}{dx} \left[ (4a - 3)(x + \log_e 5) + (a - 7) \sin(x) \right]$

The derivative of $(4a - 3)(x + \log_e 5)$ with respect to $x$ is $(4a - 3)$, since $(4a - 3)$ and $\log_e 5$ are constants with respect to $x$. The derivative of $(a - 7) \sin(x)$ with respect to $x$ is $(a - 7) \cos(x)$.

Therefore, the first derivative is: $f'(x) = (4a - 3) + (a - 7) \cos(x)$

Step 3: Set $f'(x) = 0$ to find the condition for critical points

For critical points to exist, the equation $f'(x) = 0$ must have at least one real solution for $x$.

$(4a - 3) + (a - 7) \cos(x) = 0$

We need to solve for $\cos(x)$: $(a - 7) \cos(x) = -(4a - 3)$ $(a - 7) \cos(x) = 3 - 4a$

Step 4: Analyze the equation for $\cos(x)$ based on the value of $(a-7)$

We need to consider two cases for the coefficient $(a-7)$:

• Case 1: $a - 7 \ne 0$ (i.e., $a \ne 7$) In this case, we can divide by $(a - 7)$: $\cos(x) = \frac{3 - 4a}{a - 7}$

  For this equation to have a real solution for $x$, the value of $\cos(x)$ must be within its range, which is $[-1, 1]$. So, we must have: $-1 \le \frac{3 - 4a}{a - 7} \le 1$

  This inequality can be split into two separate inequalities: (i) $\frac{3 - 4a}{a - 7} \le 1$ (ii) $\frac{3 - 4a}{a - 7} \ge -1$

  Let's solve (i): $\frac{3 - 4a}{a - 7} - 1 \le 0$ $\frac{3 - 4a - (a - 7)}{a - 7} \le 0$ $\frac{3 - 4a - a + 7}{a - 7} \le 0$ $\frac{10 - 5a}{a - 7} \le 0$ $\frac{5(2 - a)}{a - 7} \le 0$

  To solve this rational inequality, we consider the critical points $a=2$ and $a=7$.

  • If $a < 2$: Numerator is positive, denominator is negative. Ratio is negative. (Satisfied)
  • If $2 < a < 7$: Numerator is negative, denominator is negative. Ratio is positive. (Not satisfied)
  • If $a > 7$: Numerator is negative, denominator is positive. Ratio is negative. (Satisfied) So, from (i), we get $a \in (-\infty, 2] \cup (7, \infty)$.

  Now let's solve (ii): $\frac{3 - 4a}{a - 7} + 1 \ge 0$ $\frac{3 - 4a + (a - 7)}{a - 7} \ge 0$ $\frac{3 - 4a + a - 7}{a - 7} \ge 0$ $\frac{-3a - 4}{a - 7} \ge 0$ $\frac{-(3a + 4)}{a - 7} \ge 0$ $\frac{3a + 4}{a - 7} \le 0$

  To solve this rational inequality, we consider the critical points $a = -\frac{4}{3}$ and $a = 7$.

  • If $a < -\frac{4}{3}$: Numerator is negative, denominator is negative. Ratio is positive. (Not satisfied)
  • If $-\frac{4}{3} < a < 7$: Numerator is positive, denominator is negative. Ratio is negative. (Satisfied)
  • If $a > 7$: Numerator is positive, denominator is positive. Ratio is positive. (Not satisfied) So, from (ii), we get $a \in [-\frac{4}{3}, 7)$.

  For critical points to exist, both conditions (i) and (ii) must be satisfied. We need to find the intersection of $a \in (-\infty, 2] \cup (7, \infty)$ and $a \in [-\frac{4}{3}, 7)$.

  Intersection: $[-\frac{4}{3}, 2]$.

• Case 2: $a - 7 = 0$ (i.e., $a = 7$) If $a = 7$, the equation $(a - 7) \cos(x) = 3 - 4a$ becomes: $0 \cdot \cos(x) = 3 - 4(7)$ $0 = 3 - 28$ $0 = -25$

  This is a contradiction. Therefore, when $a = 7$, there are no solutions for $x$, and thus no critical points. This confirms that $a=7$ is not in the range.

Step 5: Consider the case where $f'(x)$ is undefined

The derivative $f'(x) = (4a - 3) + (a - 7) \cos(x)$ is defined for all real values of $x$ since $\cos(x)$ is defined for all real $x$. Therefore, there are no critical points arising from $f'(x)$ being undefined.

Step 6: Combine the results to find the range of 'a'

From Step 4, the range of 'a' for which $f'(x) = 0$ has solutions is $a \in [-\frac{4}{3}, 2]$.

Let's re-examine the question and the options. The provided correct answer is A, which is $[1, \infty)$. This suggests a potential misunderstanding or a different interpretation of the problem or a typo in the provided correct answer.

Let's re-check the simplification and derivative. $f(x) = (4a - 3)(x + \log_e 5) + (a - 7) \sin(x)$ $f'(x) = (4a - 3) + (a - 7) \cos(x)$

We require $f'(x) = 0$ to have a solution. $(a - 7) \cos(x) = 3 - 4a$

If $a=7$, $0 = 3-28 = -25$, impossible. If $a \ne 7$, $\cos(x) = \frac{3-4a}{a-7}$. We need $-1 \le \frac{3-4a}{a-7} \le 1$.

Inequality 1: $\frac{3-4a}{a-7} \le 1 \implies \frac{3-4a - (a-7)}{a-7} \le 0 \implies \frac{10-5a}{a-7} \le 0 \implies \frac{5(2-a)}{a-7} \le 0$. This holds for $a \in (-\infty, 2] \cup (7, \infty)$.

Inequality 2: $\frac{3-4a}{a-7} \ge -1 \implies \frac{3-4a + (a-7)}{a-7} \ge 0 \implies \frac{-3a-4}{a-7} \ge 0 \implies \frac{3a+4}{a-7} \le 0$. This holds for $a \in [-\frac{4}{3}, 7)$.

The intersection of these two intervals is $[-\frac{4}{3}, 2]$.

Let's consider the phrasing "has critical points". This means $f'(x)=0$ must have at least one real solution. The range of $\cos(x)$ is $[-1, 1]$. So, we require $\frac{3-4a}{a-7}$ to be a value in $[-1, 1]$.

This means the interval $[-\frac{4}{3}, 2]$ is the set of values of $a$ for which critical points exist.

There seems to be a discrepancy between the derived answer and the provided correct answer (A) $[1, \infty)$. Let's re-examine the problem statement and options. It is possible that the question or options have a typo, or a specific interpretation is needed.

Let's assume, for the sake of reaching the provided answer, that the question implies a different condition or there's a context missing. However, based on the standard definition of critical points and the provided function, the derivation leading to $[-\frac{4}{3}, 2]$ is mathematically sound.

Let's check if any part of the problem statement was misinterpreted. $f(x) = (4a - 3)(x + \log_e 5) + 2(a - 7) \cot\left(\frac{x}{2}\right) \sin^2\left(\frac{x}{2}\right)$, $x \ne 2n\pi$, $n \in \mathbb{N}$.

The simplification to $f(x) = (4a - 3)(x + \log_e 5) + (a - 7) \sin(x)$ is correct. The derivative $f'(x) = (4a - 3) + (a - 7) \cos(x)$ is correct. The condition for critical points to exist is that $f'(x) = 0$ has a real solution for $x$. This leads to $\cos(x) = \frac{3-4a}{a-7}$. The condition for this equation to have a solution is $-1 \le \frac{3-4a}{a-7} \le 1$. This yields $a \in [-\frac{4}{3}, 2]$.

Given the provided answer is (A) $[1, \infty)$, let's try to see if there's any way to arrive at this. This would imply that for $a \ge 1$, there are critical points.

Consider the case where the coefficient of $\cos(x)$ is zero, i.e., $a-7=0$, $a=7$. In this case, $f'(x) = 4a-3 = 4(7)-3 = 28-3 = 25$. So $f'(x)=25 \ne 0$, no critical points.

Consider the case where the coefficient of $\cos(x)$ is non-zero. $\cos(x) = \frac{3-4a}{a-7}$.

If $a=1$, $\cos(x) = \frac{3-4}{1-7} = \frac{-1}{-6} = \frac{1}{6}$. This is in $[-1, 1]$, so $a=1$ should be included. If $a=2$, $\cos(x) = \frac{3-8}{2-7} = \frac{-5}{-5} = 1$. This is in $[-1, 1]$, so $a=2$ should be included. If $a=1.5$, $\cos(x) = \frac{3-4(1.5)}{1.5-7} = \frac{3-6}{1.5-7} = \frac{-3}{-5.5} = \frac{3}{5.5} = \frac{6}{11}$. This is in $[-1, 1]$.

Let's assume there's a typo in the question or options and proceed with the derived correct range. If the question intended to ask for the range of 'a' for which $f'(x)$ is always positive or always negative, that would be a different problem.

However, if we are forced to choose from the given options and the correct answer is A, there might be a mistake in our derivation or interpretation. Let's re-read the question carefully. "The range of a$\in$R for which the function f(x) ... has critical points". This means $f'(x)=0$ must have at least one real solution.

Let's consider the possibility of an error in the problem statement or the given correct answer. Based on the standard calculus procedure, the range derived is $[-\frac{4}{3}, 2]$. None of the options match this exactly. Option (C) is $\left[ - {4 \over 3},2 \right]$, which exactly matches our derived range.

Given that the provided correct answer is (A), and our derivation leads to (C), there is a strong indication of an error in the provided "Correct Answer". Assuming our derivation is correct, the answer should be (C).

Let's assume there is a typo in the question and try to reverse-engineer the answer A. For $a \in [1, \infty)$, we need $\cos(x) = \frac{3-4a}{a-7}$ to have solutions.

If $a=1$, $\cos(x) = 1/6$. Solution exists. If $a=2$, $\cos(x) = 1$. Solution exists. If $a=3$, $\cos(x) = \frac{3-12}{3-7} = \frac{-9}{-4} = \frac{9}{4} > 1$. No solution. So, $a \in [1, \infty)$ is not correct.

Let's consider a different interpretation of the problem that might lead to option A. Perhaps the question is asking for the range of 'a' such that $f'(x)$ is not identically zero. $f'(x) = (4a - 3) + (a - 7) \cos(x)$. For $f'(x)$ to be identically zero, we would need $4a-3=0$ and $a-7=0$. This is impossible. So $f'(x)$ is never identically zero.

Let's assume the provided correct answer (A) is indeed correct and try to find a flaw in our reasoning. Our core step is $-1 \le \frac{3-4a}{a-7} \le 1$.

Let's re-verify the inequalities. $\frac{10-5a}{a-7} \le 0 \implies \frac{5(2-a)}{a-7} \le 0$. Sign analysis for $\frac{2-a}{a-7}$:

$\frac{3a+4}{a-7} \le 0$. Sign analysis for $\frac{3a+4}{a-7}$:

The intersection is indeed $[-\frac{4}{3}, 2]$.

Given the strict instruction to match the provided answer, and the strong discrepancy, there is a high probability of an error in the provided "Correct Answer". However, I must provide a solution that leads to the given correct answer. This is proving impossible with standard mathematical interpretation.

Let's consider if there is a condition for the existence of critical points that is different from $f'(x)=0$ having a solution. The definition of critical points is standard.

If we assume that the question is asking for the range of $a$ such that $f'(x)$ is not a constant positive or negative value, that would imply a different approach. But the question is explicitly about critical points.

Let's assume there's a typo in the question and the function was meant to be different, or the options are for a different question.

If we are forced to select an option, and assuming the provided answer (A) is correct, then our derivation is flawed. Let's look at the boundary points of option A: $a=1$. If $a=1$, $f'(x) = (4(1)-3) + (1-7)\cos(x) = 1 - 6\cos(x)$. $f'(x) = 0 \implies 1 - 6\cos(x) = 0 \implies \cos(x) = 1/6$. This has solutions. So $a=1$ should be in the range.

Now consider the upper bound of the interval in option A, which is $\infty$. This means for all $a \ge 1$, there are critical points. Let's test a value greater than 2, say $a=3$. If $a=3$, $f'(x) = (4(3)-3) + (3-7)\cos(x) = 9 - 4\cos(x)$. $f'(x) = 0 \implies 9 - 4\cos(x) = 0 \implies \cos(x) = 9/4$. This has no solution since $9/4 > 1$. So, $a=3$ should not be in the range. This contradicts option A.

Therefore, based on standard mathematical principles, the provided "Correct Answer" (A) is inconsistent with the question. The mathematically derived answer is $[-\frac{4}{3}, 2]$, which corresponds to option (C).

Given the constraints, I cannot "work backwards" to a demonstrably incorrect answer without making arbitrary assumptions or introducing errors. However, if the goal is to provide a solution that justifies option A, it is not possible with the current problem statement and standard definitions.

Let's proceed with the derived correct answer, assuming the provided "Correct Answer" is erroneous.

Step 7: Finalizing the range of 'a'

The equation $f'(x) = 0$ has real solutions for $x$ if and only if $\frac{3 - 4a}{a - 7}$ lies within the range $[-1, 1]$. This condition was found to be satisfied for $a \in [-\frac{4}{3}, 2]$.

This interval represents the range of values for 'a' for which the function $f(x)$ has critical points.

Common Mistakes & Tips

• Algebraic Errors in Inequalities: Solving rational inequalities requires careful consideration of signs and critical points. Ensure correct handling of the denominator's sign.
• Ignoring the Range of Trigonometric Functions: The core of finding the range of 'a' lies in ensuring that the value of $\cos(x)$ derived from the equation falls within its valid range of $[-1, 1]$.
• Case Analysis for Denominators: When solving inequalities involving a variable in the denominator (like $a-7$), it's crucial to consider cases where the denominator is zero or non-zero. In this problem, $a=7$ leads to a contradiction, meaning no critical points exist when $a=7$.

Summary

To find the range of 'a' for which the function $f(x)$ has critical points, we first simplified the function and then computed its derivative $f'(x)$. Critical points exist if and only if the equation $f'(x) = 0$ has at least one real solution for $x$. This led to the condition that $\cos(x)$ must be equal to $\frac{3 - 4a}{a - 7}$. For this equation to have a real solution for $x$, the value of $\frac{3 - 4a}{a - 7}$ must lie within the range $[-1, 1]$. Solving the resulting inequalities for 'a' yielded the range $[-\frac{4}{3}, 2]$.

The final answer is \boxed{\left[ { - {4 \over 3},2} \right]}.