Answer :
Let's solve for [tex]\( A \)[/tex] and [tex]\( B \)[/tex] given the curve [tex]\( y = A \sin x + B \sin 2x \)[/tex] passes through the point [tex]\( P\left(\frac{\pi}{2}, 3\right) \)[/tex] and has a gradient of [tex]\( \frac{3 \sqrt{2}}{2} \)[/tex] when [tex]\( x = \frac{\pi}{4} \)[/tex].
### Step 1: Applying the Point Condition
First, use the condition that the curve passes through the point [tex]\( P\left(\frac{\pi}{2}, 3\right) \)[/tex].
At [tex]\( x = \frac{\pi}{2} \)[/tex]:
[tex]\[ y = A \sin\left(\frac{\pi}{2}\right) + B \sin\left(2 \cdot \frac{\pi}{2}\right) \][/tex]
Since [tex]\(\sin\left(\frac{\pi}{2}\right) = 1\)[/tex] and [tex]\(\sin(\pi) = 0\)[/tex]:
[tex]\[ 3 = A \cdot 1 + B \cdot 0 \][/tex]
[tex]\[ 3 = A \][/tex]
Thus,
[tex]\[ A = 3 \][/tex]
### Step 2: Applying the Gradient Condition
Next, use the gradient condition that the derivative of [tex]\( y \)[/tex] equals [tex]\( \frac{3 \sqrt{2}}{2} \)[/tex] when [tex]\( x = \frac{\pi}{4} \)[/tex].
First, compute the derivative [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ y = A \sin x + B \sin 2x \][/tex]
The derivative is:
[tex]\[ \frac{dy}{dx} = A \cos x + B \cdot 2 \cos 2x \][/tex]
[tex]\[ \frac{dy}{dx} = A \cos x + 2B \cos 2x \][/tex]
Since [tex]\( A = 3 \)[/tex], substitute and simplify:
[tex]\[ \frac{dy}{dx} = 3 \cos x + 2B \cos 2x \][/tex]
Evaluate this expression at [tex]\( x = \frac{\pi}{4} \)[/tex]:
[tex]\[ \left.\frac{dy}{dx}\right|_{x=\frac{\pi}{4}} = 3 \cos\left(\frac{\pi}{4}\right) + 2B \cos\left(2 \cdot \frac{\pi}{4}\right) \][/tex]
Using [tex]\(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)[/tex] and [tex]\(\cos\left(\frac{\pi}{2}\right) = 0\)[/tex]:
[tex]\[ \frac{3 \sqrt{2}}{2} = 3 \cdot \frac{1}{\sqrt{2}} + 2B \cdot 0 \][/tex]
[tex]\[ \frac{3 \sqrt{2}}{2} = \frac{3 \sqrt{2}}{2} \][/tex]
Since the equation holds true, this step confirms our obtained value for [tex]\( A \)[/tex]. To find [tex]\( B \)[/tex], there was no further dependency, so validate the conditions were correctly used.
Thus, the final values are:
[tex]\[ A = 3 \][/tex]
[tex]\[ B = \text{(any value as there's no constraint from gradient condition that affects B independently after A is fixed)} \][/tex]
### Step 1: Applying the Point Condition
First, use the condition that the curve passes through the point [tex]\( P\left(\frac{\pi}{2}, 3\right) \)[/tex].
At [tex]\( x = \frac{\pi}{2} \)[/tex]:
[tex]\[ y = A \sin\left(\frac{\pi}{2}\right) + B \sin\left(2 \cdot \frac{\pi}{2}\right) \][/tex]
Since [tex]\(\sin\left(\frac{\pi}{2}\right) = 1\)[/tex] and [tex]\(\sin(\pi) = 0\)[/tex]:
[tex]\[ 3 = A \cdot 1 + B \cdot 0 \][/tex]
[tex]\[ 3 = A \][/tex]
Thus,
[tex]\[ A = 3 \][/tex]
### Step 2: Applying the Gradient Condition
Next, use the gradient condition that the derivative of [tex]\( y \)[/tex] equals [tex]\( \frac{3 \sqrt{2}}{2} \)[/tex] when [tex]\( x = \frac{\pi}{4} \)[/tex].
First, compute the derivative [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ y = A \sin x + B \sin 2x \][/tex]
The derivative is:
[tex]\[ \frac{dy}{dx} = A \cos x + B \cdot 2 \cos 2x \][/tex]
[tex]\[ \frac{dy}{dx} = A \cos x + 2B \cos 2x \][/tex]
Since [tex]\( A = 3 \)[/tex], substitute and simplify:
[tex]\[ \frac{dy}{dx} = 3 \cos x + 2B \cos 2x \][/tex]
Evaluate this expression at [tex]\( x = \frac{\pi}{4} \)[/tex]:
[tex]\[ \left.\frac{dy}{dx}\right|_{x=\frac{\pi}{4}} = 3 \cos\left(\frac{\pi}{4}\right) + 2B \cos\left(2 \cdot \frac{\pi}{4}\right) \][/tex]
Using [tex]\(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)[/tex] and [tex]\(\cos\left(\frac{\pi}{2}\right) = 0\)[/tex]:
[tex]\[ \frac{3 \sqrt{2}}{2} = 3 \cdot \frac{1}{\sqrt{2}} + 2B \cdot 0 \][/tex]
[tex]\[ \frac{3 \sqrt{2}}{2} = \frac{3 \sqrt{2}}{2} \][/tex]
Since the equation holds true, this step confirms our obtained value for [tex]\( A \)[/tex]. To find [tex]\( B \)[/tex], there was no further dependency, so validate the conditions were correctly used.
Thus, the final values are:
[tex]\[ A = 3 \][/tex]
[tex]\[ B = \text{(any value as there's no constraint from gradient condition that affects B independently after A is fixed)} \][/tex]
