\(0<x<6 \)
\(0<x<7 \)
\(1<x<7 \)
\(1≤x≤7\)
\(0≤x≤6\)
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)\(\)
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
The minimum value of the function \( f(x) = x^4 - 4x - 5 \), where \( x \in \mathbb{R} \), is:
The critical points of the function \( f(x) = (x-3)^3(x+2)^2 \) are:
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: