Functions of one variable

Question 1: Trigonometric Ratios (from file image_98aa6a.png)

The Question: Consider the following right angle triangle ACB in the Figure M2W1AQ 3. Which of the following option(s) is(are) true?

Core Concept: SOH CAH TOA and Reciprocal Identities

For a right-angled triangle with respect to an angle $\theta$:

• SOH: Sin($\theta$) = Opposite / Hypotenuse
• CAH: Cos($\theta$) = Adjacent / Hypotenuse
• TOA: Tan($\theta$) = Opposite / Adjacent The reciprocal identities are:
• Csc($\theta$) = 1/Sin($\theta$) = Hypotenuse / Opposite
• Sec($\theta$) = 1/Cos($\theta$) = Hypotenuse / Adjacent
• Cot($\theta$) = 1/Tan($\theta$) = Adjacent / Opposite

Detailed Solution:

In the given triangle ACB, with respect to angle $\theta$ at C:

• The side Opposite to $\theta$ is AB.
• The side Adjacent to $\theta$ is AC.
• The Hypotenuse (opposite the right angle at A) is BC.

Let’s check each option:

1. $\sin \theta = \frac{AB}{AC}$ (Opp/Adj) -> This is Tan($\theta$). FALSE.
2. $\cos \theta = \frac{AB}{AC}$ (Opp/Adj) -> This is Tan($\theta$). FALSE.
3. $\tan \theta = \frac{AB}{AC}$ (Opp/Adj) -> TRUE.
4. $\sin \theta = \frac{AC}{BC}$ (Adj/Hyp) -> This is Cos($\theta$). FALSE.
5. $\cos \theta = \frac{AC}{BC}$ (Adj/Hyp) -> TRUE.
6. $\csc \theta = \frac{BC}{AB}$ (Hyp/Opp) -> TRUE.
7. $\sec \theta = \frac{BC}{AC}$ (Hyp/Adj) -> TRUE.
8. $\cot \theta = \frac{AB}{AC}$ (Opp/Adj) -> This is Tan($\theta$). FALSE.

Final Answer: The true options are 3, 5, 6, and 7.

Question 2: Function Arithmetic (from file image_98a9d0.png)

The Question: Consider two functions $f(x) = x^2 + 2x + 3$ and $g(x) = x + 1$, where $x \neq -1$. The value of $(3\frac{f}{g})(2)$ is __________.

Core Concept: Arithmetic of Functions The notation $(k\frac{f}{g})(a)$ means you first find the values of $f(a)$ and $g(a)$, then calculate the result of the expression $k \times \frac{f(a)}{g(a)}$.

Detailed Solution:

1. First, evaluate $f(2)$:
   • $f(2) = (2)^2 + 2(2) + 3 = 4 + 4 + 3 = 11$.
2. Next, evaluate $g(2)$:
   • $g(2) = 2 + 1 = 3$.
3. Now, calculate the final expression:
   • $(3\frac{f}{g})(2) = 3 \times \frac{f(2)}{g(2)} = 3 \times \frac{11}{3}$.
   • The 3s cancel out.
   • Result = 11.

Final Answer: The value is 11.

Question 3: Function Composition (from file image_98a9d0.png)

The Question: Using the same functions $f(x)$ and $g(x)$, the value of $(f \circ g)(0)$ is __________.

Core Concept: Composition of Functions The notation $(f \circ g)(a)$ means $f(g(a))$. You work from the inside out: first calculate the value of the inner function $g(a)$, then use that result as the input for the outer function $f$.

Detailed Solution:

1. First, calculate the inner function, $g(0)$:
   • $g(0) = 0 + 1 = 1$.
2. Next, use this result (1) as the input for the outer function, $f(x)$:
   • We need to find $f(1)$.
   • $f(1) = (1)^2 + 2(1) + 3 = 1 + 2 + 3 = 6$.

Final Answer: The value is 6.

Question 4: Sine Identities (from file image_98a9d0.png)

The Question: Suppose $\sin(x+y) = \sin x \cos y + \cos x \sin y$. Which of the following is(are) true?

Core Concept: Deriving Double and Triple Angle Formulas We can use the given sum formula to derive other identities.

• Double Angle: Set $y=x$.
• Triple Angle: Set $y=2x$.

Detailed Solution:

• Check $\sin(2x)$:

  • Let $y=x$ in the sum formula:
  • $\sin(2x) = \sin(x+x) = \sin x \cos x + \cos x \sin x = 2\sin x \cos x$.
  • The statement “$\sin 2x = 2 \sin x \cos x$” is TRUE.
  • The statement “$\sin 2x = 2 \sin x$” is FALSE.

• Check $\sin(3x)$:

  • Let’s write $3x$ as $(x+2x)$:
  • $\sin(3x) = \sin(x+2x) = \sin x \cos(2x) + \cos x \sin(2x)$.
  • We know $\sin(2x) = 2\sin x \cos x$ and a common identity for $\cos(2x)$ is $1 - 2\sin^2 x$.
  • $\sin(3x) = \sin x (1 - 2\sin^2 x) + \cos x (2\sin x \cos x)$
  • $\sin(3x) = \sin x - 2\sin^3 x + 2\sin x \cos^2 x$
  • Using the identity $\cos^2 x = 1 - \sin^2 x$:
  • $\sin(3x) = \sin x - 2\sin^3 x + 2\sin x (1 - \sin^2 x)$
  • $\sin(3x) = \sin x - 2\sin^3 x + 2\sin x - 2\sin^3 x$
  • $\sin(3x) = 3\sin x - 4\sin^3 x$.
  • The statement “$\sin 3x = 3\sin x - 4\sin^3 x$” is TRUE.
  • The statement “$\sin 3x = \sin 2x \cos x - \cos 2x \sin x$” is FALSE (it has the wrong sign, it should be a plus).

Final Answer: The true statements are:

• $\sin 2x = 2\sin x \cos x$
• $\sin 3x = 3\sin x - 4\sin^3 x$

Question 5: Cosine Identities (from file image_98a692.png)

The Question: Suppose $\cos(x+y) = \cos x \cos y - \sin x \sin y$. Which of the following is(are) true?

Detailed Solution:

• Check $\cos(2x)$:

  • Let $y=x$ in the sum formula:
  • $\cos(2x) = \cos(x+x) = \cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x$.
  • The statement “$\cos 2x = \cos^2 x - \sin^2 x$” is TRUE.
  • The statement “$\cos 2x = \sin^2 x - \cos^2 x$” is FALSE.

• Check $\cos(3x)$:

  • Let’s write $3x$ as $(x+2x)$:
  • $\cos(3x) = \cos(x+2x) = \cos x \cos(2x) - \sin x \sin(2x)$.
  • We know $\cos(2x) = 2\cos^2 x - 1$ and $\sin(2x) = 2\sin x \cos x$.
  • $\cos(3x) = \cos x (2\cos^2 x - 1) - \sin x (2\sin x \cos x)$
  • $\cos(3x) = 2\cos^3 x - \cos x - 2\sin^2 x \cos x$
  • Using the identity $\sin^2 x = 1 - \cos^2 x$:
  • $\cos(3x) = 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x$
  • $\cos(3x) = 2\cos^3 x - \cos x - (2\cos x - 2\cos^3 x)$
  • $\cos(3x) = 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x$
  • $\cos(3x) = 4\cos^3 x - 3\cos x$.
  • The statement “$\cos 3x = 4\cos^3 x - 3\cos x$” is TRUE.
  • The statement “$\cos 3x = \cos 2x \cos x - \sin 2x \sin x$” is TRUE as it is an intermediate step in the derivation.

Final Answer: The true statements are:

• $\cos 2x = \cos^2 x - \sin^2 x$
• $\cos 3x = \cos 2x \cos x - \sin 2x \sin x$
• $\cos 3x = 4\cos^3 x - 3\cos x$

Question 6: Bounded Functions (from file image_98a692.png)

The Question: Consider the following statements. Then the number of correct statements is __________.

Core Concept: Bounded Functions A function is bounded if its range (all possible y-values) is confined within a finite interval. It does not go to positive or negative infinity.

Detailed Solution:

1. “sin(x) is a bounded function on $\mathbb{R}$.”: The range of $\sin(x)$ is $[-1, 1]$. Since its values never go below -1 or above 1, it is bounded. TRUE.
2. “cos(x) is an unbounded function on $\mathbb{R}$.”: The range of $\cos(x)$ is also $[-1, 1]$. It is bounded. The statement says it is unbounded. FALSE.
3. “tan(x) is an unbounded function on its domain.”: The range of $\tan(x)$ is $(-\infty, \infty)$. Since it goes to both positive and negative infinity, it is unbounded. TRUE.
4. "$\frac{1}{2+\cos(x)}$ is an unbounded function on its domain.": Let’s find the range.
   • The minimum value of $\cos(x)$ is -1. So the maximum value of the denominator is $2+1=3$. This gives a minimum function value of $1/3$.
   • The maximum value of $\cos(x)$ is 1. So the minimum value of the denominator is $2-1=1$. This gives a maximum function value of $1/1=1$.
   • The range of this function is $[\frac{1}{3}, 1]$. This is a finite interval, so the function is bounded. The statement says it is unbounded. FALSE.

Final Answer: There are 2 correct statements.

Question 7: Well-Defined Composite Functions (from file image_98a692.png)

The Question: Which of the following is(are) true?

Core Concept: Well-Defined Composition For a composite function $(g \circ f)(x) = g(f(x))$ to be “well-defined”, the Range of the inner function ($f$) must be a subset of the Domain of the outer function ($g$). ($Range(f) \subseteq Domain(g)$).

Detailed Solution:

• If $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{Q} \to \mathbb{R}$…:

  • Range of $f$ is a subset of $\mathbb{R}$. Domain of $g$ is $\mathbb{Q}$ (rational numbers). Is $\mathbb{R} \subseteq \mathbb{Q}$? No. For example, $f(x) = \sqrt{2}$ has an output not in the domain of $g$. FALSE.

• If $f: \mathbb{N} \to \mathbb{Z}$ and $g: \mathbb{Q} \to \mathbb{R}$…:

  • Range of $f$ is a subset of $\mathbb{Z}$ (integers). Domain of $g$ is $\mathbb{Q}$. Since every integer is also a rational number ($\mathbb{Z} \subseteq \mathbb{Q}$), the condition is met. TRUE.

• If $f: \mathbb{R} \to (0,1)$ and $g:(0,1] \to \mathbb{R}$…:

  • Range of $f$ is $(0,1)$. Domain of $g$ is $(0,1]$. Is $(0,1) \subseteq (0,1]$? Yes. The condition is met. TRUE.

• If $f: \mathbb{Z} \to [0,1]$ and $g:(0,1) \to \mathbb{Q}$…:

  • Range of $f$ is $[0,1]$. Domain of $g$ is $(0,1)$. Is $[0,1] \subseteq (0,1)$? No. The range of $f$ includes 0 and 1, which are not in the domain of $g$. FALSE.

Final Answer: The true statements are:

• If $f: \mathbb{N} \to \mathbb{Z}$ and $g: \mathbb{Q} \to \mathbb{R}$, then $g \circ f$ is well defined.
• If $f: \mathbb{R} \to (0,1)$ and $g:(0,1] \to \mathbb{R}$, then $g \circ f$ is well defined.

Questions 8, 9, 10: Trigonometric Shift Identities (from files image_98a5b2.png, image_98a2b3.png)

Core Concept: Angle Sum/Difference and Shift Identities These are fundamental trigonometric identities that describe how shifting the angle by $\pi/2$ (90°) or $\pi$ (180°) affects the function’s value.

• $\sin(\frac{\pi}{2} \pm \theta) = \cos \theta$ (Incorrect sign for plus)
• $\sin(\frac{\pi}{2} - \theta) = \cos \theta$
• $\sin(\frac{\pi}{2} + \theta) = \cos \theta$
• $\cos(\frac{\pi}{2} - \theta) = \sin \theta$
• $\cos(\frac{\pi}{2} + \theta) = -\sin \theta$
• $\sin(\pi - \theta) = \sin \theta$
• $\sin(\pi + \theta) = -\sin \theta$

Detailed Solution:

Question 8:

• Option 1: $\sin(\frac{\pi}{2} + \theta) = \cos \theta$ -> TRUE.
• Option 2: $\sin(\frac{\pi}{2} + \theta) = -\cos \theta$ -> FALSE.
• Option 3: $\sin(\frac{\pi}{2} - \theta) = -\cos \theta$ -> FALSE.
• Option 4: $\sin(\frac{\pi}{2} - \theta) = \cos \theta$ -> TRUE.

Question 9:

• Option 1: $\cos(\frac{\pi}{2} + \theta) = \sin \theta$ -> FALSE.
• Option 2: $\cos(\frac{\pi}{2} + \theta) = -\sin \theta$ -> TRUE.
• Option 3: $\cos(\frac{\pi}{2} - \theta) = \sin \theta$ -> TRUE.
• Option 4: $\cos(\frac{\pi}{2} - \theta) = -\sin \theta$ -> FALSE.

Question 10:

• Option 1: $\sin(\pi + \theta) = -\sin \theta$ -> TRUE.
• Option 2: $\sin(\pi + \theta) = \sin \theta$ -> FALSE.
• Option 3: $\sin(\pi - \theta) = \sin \theta$ -> TRUE.
• Option 4: $\sin(\pi - \theta) = -\sin \theta$ -> FALSE.
• Options 5 & 6 are comparing sine to cosine and are incorrect.

Final Answer:

• Q8: Options 1 and 4 are true.
• Q9: Options 2 and 3 are true.
• Q10: Options 1 and 3 are true.

Question 11: Solving a Self-Composition Equation (from file image_98a2b3.png)

The Question: Let $f(x) = \frac{x}{x+a}$, where $x > 0$ and $a > 0$. If $f(f(x)) = \frac{x}{3x+4}$, then find the value of $a$.

Core Concept: Self-Composition To find $f(f(x))$, you substitute the entire expression for $f(x)$ into the variable $x$ within the function itself.

Detailed Solution:

1. Set up the composition $f(f(x))$:
   • $f(f(x)) = \frac{f(x)}{f(x)+a}$
   • Substitute the expression for $f(x)$: $$f(f(x)) = \frac{\frac{x}{x+a}}{\frac{x}{x+a} + a}$$
2. Simplify the complex fraction: Multiply the main numerator and denominator by $(x+a)$.
   • Numerator: $(x+a) \times \frac{x}{x+a} = x$
   • Denominator: $(x+a) \times \left(\frac{x}{x+a} + a\right) = x + a(x+a) = x + ax + a^2$
3. Write the simplified expression for $f(f(x))$:
   • $f(f(x)) = \frac{x}{x + ax + a^2} = \frac{x}{(1+a)x + a^2}$
4. Equate this to the given result:
   • $\frac{x}{(1+a)x + a^2} = \frac{x}{3x+4}$
5. Compare the denominators: For the two expressions to be equal, their denominators must be equal.
   • $(1+a)x + a^2 = 3x+4$
6. Equate the coefficients:
   • The coefficients of the $x$ terms must be equal: $1+a = 3 \implies a = 2$.
   • The constant terms must be equal: $a^2 = 4 \implies a = \pm2$.
7. Apply the given condition: The problem states that $a > 0$. Both results are consistent with $a=2$.

Final Answer: The value of $a$ is 2.