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

Answer: (C)

$1<x<7$

Answer 2

_Given that _

$|4 - x| × |a| < |3a|$

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 $1 < x < 7.$ (_Ans)$$