The solution set for the inequality $ 13x - 5 \leq 15x + 4<7x + 12; x \in W $
We are given the inequality: \[ 13x - 5 \leq 15x + 4<7x + 12 \] Let's first solve each part of the inequality separately.
Step 1: Solving the first part \( 13x - 5 \leq 15x + 4 \):
\[ 13x - 15x \leq 4 + 5 \] \[ -2x \leq 9 \] \[ x \geq -\frac{9}{2} \]
Step 2:
Solving the second part \( 15x + 4 < 7x + 12 \): \[ 15x - 7x<12 - 4 \] \[ 8x<8 \] \[ x<1 \]
Step 3: Combining both inequalities:
We have the solution to the system of inequalities as: \[ -\frac{9}{2} \leq x<1 \] However, since \( x \in W \) (where \( W \) is the set of whole numbers), the only valid value for \( x \) in this range is \( x = 0 \). Thus, the solution set is \( \{ 0 \} \).
Let \( \alpha, \beta, \) and \( \gamma \) be real numbers. Consider the following system of linear equations:
\( x + 2y + z = 7 \)
\( x + \alpha z = 11 \)
\( 2x - 3y + \beta z = \gamma \)
Match each entry in List I to the correct entries in List II
List I | List II | ||
(P) | If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma = 28 \), then the system has | (1) | a unique solution |
(Q) | If \( \beta = \frac{1}{2}(7\alpha - 3) \) and \( \gamma \neq 28 \), then the system has | (2) | no solution |
(R) | If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma \neq 28 \), then the system has | (3) | infinitely many solutions |
(S) | If \( \beta \neq \frac{1}{2}(7\alpha - 3) \) where \( \alpha = 1 \) and \( \gamma = 28 \), then the system has | (4) | \( x = 11, y = -2 \) and \( z = 0 \) as a solution |
(5) | \( x = -15, y = 4 \) and \( z = 0 \) as a solution |
Solution of \( 2^x + 2^{|x|} \geq 2\sqrt{2} \) is:
If the ratio of lengths, radii and Young's Moduli of steel and brass wires in the figure are $ a $, $ b $, and $ c $ respectively, then the corresponding ratio of increase in their lengths would be: