Question

Let x be a real number and a be any non-zero vector such \(|(4-x)a|<|3a|\).Then which of the following options is correct?

• A. \(0<x<6 \)
• B. \(0<x<7 \)
• C. \(1<x<7 \)
• D. \(1≤x≤7\)
• E. \(0≤x≤6\)

Answer 1

The Correct Option is C

Given: \(|4 - x| × |a| < |3a|\)

Here, \(|a|\) represents the magnitude of vector \(a\).

Since \(a\) is a non-zero vector, its magnitude \(|a|\) is also non-zero.

So, we can now divide both sides of the inequality by \(|a|\)

\(|4 - x| < |3|\)

Now, we have an absolute value inequality with a constant on the right side.

 To solve this, we can take two cases:

Case 1:

\(4 - x\) is positive \((4 - x > 0) |4 - x| = 4 - x\)

In this case, our inequality becomes: \(4 - x < 3\)

Now, solve for \(x\) : \(x > 4 - 3 x > 1\)

Case 2: 4 - x is negative \((4 - x < 0) |4 - x| = -(4 - x) = x - 4\)

In this case, our inequality becomes: \(x - 4 < 3\)

Now, solve for \(x: x < 3 + 4 x < 7\)

So, the valid values of x in this case are \(1 < x < 7.\)

Combining both cases, we find that the valid values of \(x\) are \(1 < x < 7.\) 

Answer 2

Given the inequality \( |(4 - x)\mathbf{a}| < |3\mathbf{a}| \), and that \( \mathbf{a} \) is a non-zero vector, we can simplify this. The magnitude of a scalar multiple of a vector is the absolute value of the scalar times the magnitude of the vector:

\[ |4 - x| \|\mathbf{a}\| < 3 \|\mathbf{a}\| \]

Since \( \mathbf{a} \) is non-zero, \( \|\mathbf{a}\| > 0 \), so we can divide both sides by \( \|\mathbf{a}\| \):

\[ |4 - x| < 3 \]

This inequality can be rewritten as:

\[ -3 < 4 - x < 3 \]

Subtracting 4 from all parts of the inequality:

\[ -7 < -x < -1 \]

Multiplying by \(-1\) (and reversing the inequality signs):

\[ 7 > x > 1 \]

Therefore, the solution is \( 1 < x < 7 \).