Let x_0 be the point of local minima of f(x) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The function $f(x)$ is defined as the scalar triple product of vectors $\overline{a}, \overline{b}, 	ext{ and } \overline{c}$:$f(x) = \begin{vmat

Vedclass · Accepted Answer

$-15$

Vedclass · Answer

(B) The function $f(x)$ is defined as the scalar triple product of vectors $\overline{a}, \overline{b}, 	ext{ and } \overline{c}$:$f(x) = \begin{vmatrix} x & -2 & 3 \ -2 & x & -1 \ 7 & -2 & x \end{vmatrix}$Expanding the determinant along the first row:$f(x) = x(x^2 - 2) - (-2)(-2x + 7) + 3(4 - 7x)$$f(x) = x^3 - 2x + 4x - 14 + 12 - 21x$$f(x) = x^3 - 19x - 2$ (Note: Correcting the expansion: $x(x^2-2) + 2(-2x+7) + 3(4-7x) = x^3 - 2x - 4x + 14 + 12 - 21x = x^3 - 27x + 26$)To find the local minima,we find the critical points by setting $f'(x) = 0$:$f'(x) = 3x^2 - 27 = 0$$3(x^2 - 9) = 0 \Rightarrow x = \pm 3$Using the second derivative test,$f''(x) = 6x$:At $x = 3$,$f''(3) = 18 > 0$,so $x_0 = 3$ is the point of local minima.At $x = 3$,$\overline{a} = 3 \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\overline{b} = -2 \hat{i} + 3 \hat{j} - \hat{k}$.Calculating the dot product $\overline{a} \cdot \overline{b}$:$\overline{a} \cdot \overline{b} = (3)(-2) + (-2)(3) + (3)(-1) = -6 - 6 - 3 = -15$.

Similar Questions

The maximum area of a parallelogram inscribed in a triangle $ABC$ (given that $A$ is also one of the vertices of the parallelogram) is equal to (where symbols have their usual meaning in $\Delta ABC$):

The number of points at which the function $f(x) = \int\limits_0^x {{e^{t - 3}}} \left( {{t^2} + 2} \right)\left( {t - 3} \right){\left( {t + 4} \right)^2}dt$ has a local minimum is:

Let $P(x)$ be a real polynomial of degree $3$ which vanishes at $x = -3$. Let $P(x)$ have local minima at $x = 1$,local maxima at $x = -1$,and $\int_{-1}^{1} P(x) dx = 18$. Then the sum of all the coefficients of the polynomial $P(x)$ is equal to ....... .

The maximum value of the function $f(x)=e^x+x \ln x$ on the interval $1 \leq x \leq 2$ is

Maximum value of the function $f(x)=\frac{x}{8}+\frac{2}{x}$ on the interval $[1,6]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module