Question

Let \(x_0\) be the point of local maxima of \(f(x) = \overset{̄}{a}\cdot (\overset{̄}{b}\times \overset{̄}{c})\) where \(\overset{̄}{a} = x\hat{i}-2\hat{j}+3\hat{k},\overset{̄}{b} = -2\hat{i}+x\hat{j}-\hat{k}\) and \(\overset{̄}{c} = 7\hat{i}-2\hat{j}+x\hat{k}\) then the value of \(\overset{̄}{a}\cdot \overset{̄}{c}\) at \(x = x_0\) is

• A. \(26\)
• B. \(0\)
• C. \(-15\)
• D. \(-26\)

Hint:

Expand the determinant to get a cubic in x, find its local maximum, then compute a dot c.

Answer 1

The Correct Option is D

Step 1: Use the derivative test in a different order:
Expanding gives $f(x)=x^3-27x+26$. The cubic has a positive leading coefficient, so it rises, falls, then rises again. The local maximum is the smaller critical point.

Step 2: Critical points:
$f'(x)=3(x^2-9)$ gives $x=-3$ and $x=3$. The smaller one, $x=-3$, is the maximum.

Step 3: Dot product:
$\vec a\cdot\vec c=7x+4+3x$ evaluated properly: $\vec a\cdot\vec c=7x+(-2)(-2)+3x=10x+4$. At $x=-3$, this is $-30+4=-26$. Option D.

Final Answer:
The local maximum is at x = -3, which gives a . c = -26. \[ \boxed{\text{(D) }-26} \]