Let x0 be the point of local maxima of f(x)=a→·(b→×c→), where a→=xi^-2j^+3k^,b→=-2i^+xj^-k^ and c→=7i^-2j^+xk^ . Then the value of a→·b→+b→·c→+c→·a→ at x=x0 is

fx=x−23−2x−17−2x=xx2-2+2(-2x+7)+3(4-7x)=x3-2x-4x+14+12-21x=x3-27x+26⇒f'(x)=3x2-27⇒f'(x)=0⇒x=±3f"x=6x ⇒f"-3=-18<0 ⇒ fx has local maximum at x=-3⇒ x0=-3a¯=-3i^-2j^+3k^;b¯=-2i^-3j^-k^;c¯=7i^-2j^-3k^a¯·b¯=6+6-3=9b¯·c¯=-14+6+3=-5∣c¯.a¯=-21+4-9=-26a¯.b¯+b¯.c¯+c¯.a¯=-22

Let x0 be the point of local maxima of f(x)=a→·(b→×c→), where a→=xi^-2j^+3k^,b→=-2i^+xj^-k^ and c→=7i^-2j^+xk^ . Then the value of a→·b→+b→·c→+c→·a→ at x=x0 is

fx=x−23−2x−17−2x=xx2-2+2(-2x+7)+3(4-7x)=x3-2x-4x+14+12-21x=x3-27x+26⇒f'(x)=3x2-27⇒f'(x)=0⇒x=±3f"x=6x ⇒f"-3=-18<0 ⇒ fx has local maximum at x=-3⇒ x0=-3a¯=-3i^-2j^+3k^;b¯=-2i^-3j^-k^;c¯=7i^-2j^-3k^a¯·b¯=6+6-3=9b¯·c¯=-14+6+3=-5∣c¯.a¯=-21+4-9=-26a¯.b¯+b¯.c¯+c¯.a¯=-22

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$Let x_{0} be the point of local maxima of f (x) = \vec{a} \cdot (\vec{b} \times \vec{c}), where$
$\vec{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = - 2 \hat{i} + x \hat{j} - \hat{k} and \vec{c} = 7 \hat{i} - 2 \hat{j} + x \hat{k} . Then the value of \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ $at x = x_{0} is$

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Detailed Solution

$f (x) = |\begin{matrix} x & - 2 & 3 \\ - 2 & x & - 1 \\ 7 & - 2 & x \end{matrix}|$

$\begin{array}{l} = x (x^{2} - 2) + 2 (- 2 x + 7) + 3 (4 - 7 x) \\ = x^{3} - 2 x - 4 x + 14 + 12 - 21 x = x^{3} - 27 x + 26 \\ \Rightarrow f^{'} (x) = 3 x^{2} - 27 \Rightarrow f^{'} (x) = 0 \Rightarrow x = \pm 3 \end{array}$

$f " (x) = 6 x \Rightarrow f " (- 3) = - 18 < 0 \Rightarrow f (x) h a s l o c a l m a x i m u m a t x = - 3 \Rightarrow x_{0} = - 3$

$\begin{array}{l} \bar{a} = - 3 \hat{i} - 2 \hat{j} + 3 \hat{k}; \bar{b} = - 2 \hat{i} - 3 \hat{j} - \hat{k}; \bar{c} = 7 \hat{i} - 2 \hat{j} - 3 \hat{k} \\ \bar{a} \cdot \bar{b} = 6 + 6 - 3 = 9 \\ \bar{b} \cdot \bar{c} = - 14 + 6 + 3 = - 5 ∣ \end{array}$