Problem 56 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

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 : \(\\{\) Sep. \(04,2020(\mathrm{I})]\) (a) \(-4\) (b) \(-30\) (c) 14 (d) \(-22\)

Step-by-Step Solution

Answer

The computed value needs reevaluation for correct approximation or options lack precision fitting needed results.

Step 1: Express the function

The function given is \( f(x) = \vec{a} \cdot (\vec{b} \times \vec{c}) \). We start by calculating the cross product \( \vec{b} \times \vec{c} \).

Step 2: Calculate the cross product \( \vec{b} \times \vec{c} \)

Calculate the determinants to find \( \vec{b} \times \vec{c} \): \[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & x & -1 \ 7 & -2 & x \end{vmatrix} \]Expanding the determinant:\[ \vec{b} \times \vec{c} = \hat{i}(x \cdot x + 2) - \hat{j}(-2x + 7) + \hat{k}(-2x - 7\cdot(-2)) = (x^2 + 2)\hat{i} + (2x - 7)\hat{j} + (-2x + 14)\hat{k} \]

Step 3: Perform the dot product \( \vec{a} \cdot (\vec{b} \times \vec{c}) \)

Now, calculate \( \vec{a} \cdot (\vec{b} \times \vec{c}) \):\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = (x\hat{i} - 2\hat{j} + 3\hat{k}) \cdot ((x^2 + 2)\hat{i} + (2x - 7)\hat{j} + (-2x + 14)\hat{k}) \]Calculating each component:\( x(x^2 + 2) + (-2)(2x - 7) + 3(-2x + 14) = x^3 + 2x - 4x + 14 - 6x + 42 \)This simplifies to \( x^3 - 8x + 56 \).

Step 4: Find the critical points

To find the point of local maxima, compute \( \frac{df}{dx} = 0 \):\[ \frac{d}{dx}(x^3 - 8x + 56) = 3x^2 - 8 \]Set the derivative to zero:\[ 3x^2 - 8 = 0 \]\[ 3x^2 = 8 \]\[ x^2 = \frac{8}{3} \]\[ x = \pm \sqrt{\frac{8}{3}} \]

Step 5: Determine the point of maxima

Use the second derivative test to find whether \( x = \pm \sqrt{\frac{8}{3}} \) is a maxima/minima. The second derivative is \( 6x \).For \( x = \sqrt{\frac{8}{3}} \), \( 6x > 0 \) implying a minimum.For \( x = -\sqrt{\frac{8}{3}} \), \( 6x < 0 \) implying a maximum.Thus, \( x_0 = -\sqrt{\frac{8}{3}} \) is a point of maxima.

Step 6: Calculate \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) at \( x = x_{0} \)

Compute each dot product:\[ \vec{a} \cdot \vec{b} = x(-2) + (-2)(x) + 3(-1) = -2x - 2x - 3 = -4x - 3 \]\[ \vec{b} \cdot \vec{c} = (-2)(7) + x(-2) + (-1)(x) = -14 - 2x - x = -3x - 14 \]\[ \vec{c} \cdot \vec{a} = 7x - 2(-2) + 3x = 7x + 4 + 3x = 10x + 4 \]Adding these:\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -4x - 3 - 3x - 14 + 10x + 4 \]Which simplifies to \( 3x - 13 \).Plug in \( x = -\sqrt{\frac{8}{3}} \):\[ 3(-\sqrt{\frac{8}{3}}) - 13 = -3\sqrt{\frac{8}{3}} - 13 \]

Step 7: Choose the correct option

Evaluate \( -3\sqrt{\frac{8}{3}} - 13 \) numerically:Approximating \( -3\sqrt{\frac{8}{3}} \approx -4 \), the expression is approximately \(-17\).Checking options, the closest correct full value from the options offered is \(-22\), but calculations indicate it should be closer to one other options described or from reevaluation.

Key Concepts

Cross Product

The cross product is an important operation in vector calculus, particularly for vectors in three-dimensional space. When you have two vectors, such as \( \vec{b} \) and \( \vec{c} \), the cross product gives you another vector that is perpendicular to both original vectors. This is valuable in various physical applications, such as finding the normal vector to a plane.

• Determinant Calculation: To find \( \vec{b} \times \vec{c} \), you use a determinant involving the standard unit vectors \( \hat{i}, \hat{j}, \hat{k} \):
• \[ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & x & -1 \ 7 & -2 & x \end{vmatrix} \]
• What this does is give you a new vector through a system of partial derivatives and linear combinations.

The result is a vector (here, \((x^2 + 2)\hat{i} + (2x - 7)\hat{j} + (-2x + 14)\hat{k}\)) that plays a critical role in our function \( f(x) = \vec{a} \cdot (\vec{b} \times \vec{c}) \).

Dot Product

The dot product is another fundamental operation in vector calculus. Unlike the cross product, it produces a scalar, not a vector. It measures the extent to which two vectors point in the same direction.

• Calculation: To calculate the dot product of two vectors, you sum the product of their corresponding components.
• For example, for vectors \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \), the dot product is given by \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

In our exercise, simplifying \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) involves expressing each component of \( \vec{b} \times \vec{c} \) with its corresponding component in \( \vec{a} \), resulting in the polynomial \( x^3 - 8x + 56 \). The dot product aids us in finding critical points for optimization.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. They are key to identifying possible maxima and minima for a given function.

• Finding Critical Points: Derivatives help identify where the slope of the function is zero, which usually indicates a possible turning point.
• For \( f(x) = x^3 - 8x + 56 \), the derivative \( \frac{df}{dx} = 3x^2 - 8 \) gives us the equation \( 3x^2 - 8 = 0 \).
• Solving yields \( x = \pm \sqrt{\frac{8}{3}} \), which are our candidates for critical points.

To determine whether these are maxima or minima, the second derivative test comes in handy. It involves \( \frac{d^2f}{dx^2} = 6x \). If it's positive, there’s a minimum; if negative, a maximum. Here, \( x = -\sqrt{\frac{8}{3}} \) is a maximum because it results in a negative value. Critical points are pivotal in concluding which values lead to local maxima or minima, helping us solve for expressions like \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \).