Q.14. If and are two vectors such that and , then the maximum value of is :

Q: What is the correct answer for this exam Mathematics Q14?

The correct answer is C. Sol. Let the expression be $E = 3|(3\mathbf{a}+2\mathbf{b}) + 4(3\mathbf{a}-2\mathbf{b})|$. First, simplify the vector expression inside the modulus: $(3\mathbf{a}+2\mathbf{b}) + 4(3\mathbf{a}-2\mathbf{b}) = 3\mathbf{a}+2\mathbf{b} + 12\mathbf{a}-8\mathbf{b} = 15\mathbf{a}-6\mathbf{b}$. So, $E = 3|15\mathbf{a}-6\mathbf{b}| = 3 \times 3 |5\mathbf{a}-2\mathbf{b}| = 9 |5\mathbf{a}-2\mathbf{b}|$. We w

Q: Q.14. If and are two vectors such that and , then the maximum value of is :

The correct answer is C. Sol. Let the expression be $E = 3|(3\mathbf{a}+2\mathbf{b}) + 4(3\mathbf{a}-2\mathbf{b})|$. First, simplify the vector expression inside the modulus: $(3\mathbf{a}+2\mathbf{b}) + 4(3\mathbf{a}-2\mathbf{b}) = 3\mathbf{a}+2\mathbf{b} + 12\mathbf{a}-8\mathbf{b} = 15\mathbf{a}-6\mathbf{b}$. So, $E = 3|15\mathbf{a}-6\mathbf{b}| = 3 \times 3 |5\mathbf{a}-2\mathbf{b}| = 9 |5\mathbf{a}-2\mathbf{b}|$. We w

Q: What are the options for this this exam Mathematics question?

A. 30 · B. 36 · C. 60 · D. 72

Question

Q.14. If  and  are two vectors such that  and , then the maximum value of  is :

FirstInTest · Accepted Answer

Sol. Let the expression be $E = 3|(3\mathbf{a}+2\mathbf{b}) + 4(3\mathbf{a}-2\mathbf{b})|$.
First, simplify the vector expression inside the modulus:
$(3\mathbf{a}+2\mathbf{b}) + 4(3\mathbf{a}-2\mathbf{b}) = 3\mathbf{a}+2\mathbf{b} + 12\mathbf{a}-8\mathbf{b} = 15\mathbf{a}-6\mathbf{b}$.
So, $E = 3|15\mathbf{a}-6\mathbf{b}| = 3 	imes 3 |5\mathbf{a}-2\mathbf{b}| = 9 |5\mathbf{a}-2\mathbf{b}|$.
We want to maximize $|5\mathbf{a}-2\mathbf{b}|$.
We know that $|\mathbf{u} - \mathbf{v}|^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2 - 2|\mathbf{u}||\mathbf{v}|\cos	heta$, where $	heta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$.
$|5\mathbf{a}-2\mathbf{b}|^2 = |5\mathbf{a}|^2 + |2\mathbf{b}|^2 - 2(5\mathbf{a}) \cdot (2\mathbf{b})$
$= 25|\mathbf{a}|^2 + 4|\mathbf{b}|^2 - 20(\mathbf{a} \cdot \mathbf{b})$
Given $|\mathbf{a}|=2$ and $|\mathbf{b}|=3$.
$= 25(2^2) + 4(3^2) - 20|\mathbf{a}||\mathbf{b}|\cos	heta$
$= 25(4) + 4(9) - 20(2)(3)\cos	heta$
$= 100 + 36 - 120\cos	heta = 136 - 120\cos	heta$.
To maximize $|5\mathbf{a}-2\mathbf{b}|$, we need to maximize $|5\mathbf{a}-2\mathbf{b}|^2$. This occurs when $\cos	heta$ is minimized, i.e., $\cos	heta = -1$.
Maximum value of $|5\mathbf{a}-2\mathbf{b}|^2 = 136 - 120(-1) = 136 + 120 = 256$.
Maximum value of $|5\mathbf{a}-2\mathbf{b}| = \sqrt{256} = 16$.
Therefore, the maximum value of $E = 9 |5\mathbf{a}-2\mathbf{b}| = 9 	imes 16 = 144$.

Let's recheck the solution in the image. It seems to be using a different approach or has a typo.
The image solution: $E = 3\sqrt{9a^2+4b^2+12a \cdot b} + 4\sqrt{9a^2+4b^2-12a \cdot b}$
This looks like $3|3\mathbf{a}+2\mathbf{b}| + 4|3\mathbf{a}-2\mathbf{b}|$, which is different from the question $3|(3\mathbf{a}+2\mathbf{b}) + 4(3\mathbf{a}-2\mathbf{b})|$.
Assuming the question is $3|(3\mathbf{a}+2\mathbf{b}) + 4(3\mathbf{a}-2\mathbf{b})|$, my calculation of 144 is correct.
If the question meant $3|3\mathbf{a}+2\mathbf{b}| + 4|3\mathbf{a}-2\mathbf{b}|$, then:
$|3\mathbf{a}+2\mathbf{b}|^2 = 9|\mathbf{a}|^2+4|\mathbf{b}|^2+12\mathbf{a}\cdot\mathbf{b} = 9(4)+4(9)+12(2)(3)\cos	heta = 36+36+72\cos	heta = 72+72\cos	heta$
$|3\mathbf{a}-2\mathbf{b}|^2 = 9|\mathbf{a}|^2+4|\mathbf{b}|^2-12\mathbf{a}\cdot\mathbf{b} = 36+36-72\cos	heta = 72-72\cos	heta$
So, $3\sqrt{72+72\cos	heta} + 4\sqrt{72-72\cos	heta}$
$= 3\sqrt{72(1+\cos	heta)} + 4\sqrt{72(1-\cos	heta)}$
$= 3 	imes 6\sqrt{2} \sqrt{2\cos^2(	heta/2)} + 4 	imes 6\sqrt{2} \sqrt{2\sin^2(	heta/2)}$
$= 36|\cos(	heta/2)| + 48|\sin(	heta/2)|$
To maximize this, we can use Cauchy-Schwarz inequality or treat it as $A\cos X + B\sin X$.
The maximum value of $A\cos X + B\sin X$ is $\sqrt{A^2+B^2}$.
So, max value is $\sqrt{36^2+48^2} = \sqrt{1296+2304} = \sqrt{3600} = 60$.
This matches the answer (3) 60.
It seems the question text was interpreted differently by the OCR/my initial read vs. the provided solution. I will assume the solution's interpretation of the question, which is $3|3\mathbf{a}+2\mathbf{b}| + 4|3\mathbf{a}-2\mathbf{b}|$. The original text $3|(3\mathbf{a}+2\mathbf{b}) + 4(3\mathbf{a}-2\mathbf{b})|$ is unambiguous and leads to 144. Given the answer is 60, the question must have been $3|3\mathbf{a}+2\mathbf{b}| + 4|3\mathbf{a}-2\mathbf{b}|$. I will use the explanation from the solution.

Revised explanation based on solution:
Sol. Let the expression be $E = 3|3\mathbf{a}+2\mathbf{b}| + 4|3\mathbf{a}-2\mathbf{b}|$.
We have $|3\mathbf{a}+2\mathbf{b}|^2 = (3\mathbf{a}+2\mathbf{b}) \cdot (3\mathbf{a}+2\mathbf{b}) = 9|\mathbf{a}|^2 + 4|\mathbf{b}|^2 + 12\mathbf{a}\cdot\mathbf{b}$.
And $|3\mathbf{a}-2\mathbf{b}|^2 = (3\mathbf{a}-2\mathbf{b}) \cdot (3\mathbf{a}-2\mathbf{b}) = 9|\mathbf{a}|^2 + 4|\mathbf{b}|^2 - 12\mathbf{a}\cdot\mathbf{b}$.
Given $|\mathbf{a}|=2$ and $|\mathbf{b}|=3$.
$|3\mathbf{a}+2\mathbf{b}|^2 = 9(2^2) + 4(3^2) + 12(2)(3)\cos	heta = 36 + 36 + 72\cos	heta = 72(1+\cos	heta)$.
$|3\mathbf{a}-2\mathbf{b}|^2 = 9(2^2) + 4(3^2) - 12(2)(3)\cos	heta = 36 + 36 - 72\cos	heta = 72(1-\cos	heta)$.
Using $1+\cos	heta = 2\cos^2(	heta/2)$ and $1-\cos	heta = 2\sin^2(	heta/2)$:
$|3\mathbf{a}+2\mathbf{b}| = \sqrt{72 \cdot 2\cos^2(	heta/2)} = \sqrt{144\cos^2(	heta/2)} = 12|\cos(	heta/2)|$.
$|3\mathbf{a}-2\mathbf{b}| = \sqrt{72 \cdot 2\sin^2(	heta/2)} = \sqrt{144\sin^2(	heta/2)} = 12|\sin(	heta/2)|$.
So, $E = 3(12|\cos(	heta/2)|) + 4(12|\sin(	heta/2)|) = 36|\cos(	heta/2)| + 48|\sin(	heta/2)|$.
To find the maximum value of $A|\cos X| + B|\sin X|$, we can use the property that the maximum value of $A\cos X + B\sin X$ is $\sqrt{A^2+B^2}$. Since $|\cos X|$ and $|\sin X|$ are always non-negative, the maximum occurs when $\cos(	heta/2)$ and $\sin(	heta/2)$ are positive.
Maximum value of $E = \sqrt{36^2 + 48^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60$.

Answer

30

Answer

36

Answer

60

Answer

72

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