CIE 2023 March 0606/22 Question 3/4 and solution

Question 3

Solve the inequality $|5 x+4| \leqslant|2 x-3|$. [4]

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Solution

Step 1: Analyze the inequality

The given inequality is:

\[ |5x + 4| \leqslant |2x - 3|. \]

To solve, we need to determine where the absolute values are equal, i.e., solve:

\[ |5x + 4| = |2x - 3|. \]

Step 2: Solve $ |5x + 4| = |2x - 3| $

This equality breaks into two cases:

\begin{align*} 5x + 4 &= 2x - 3, \\ 5x + 4 &= -(2x - 3). \end{align*}

Case 1: $ 5x + 4 = 2x - 3 $

Simplify:

\[ 5x - 2x = -3 - 4. \]

\[ 3x = -7. \]

\[ x = -\frac{7}{3}. \]

Case 2: $ 5x + 4 = -(2x - 3) $

Simplify:

\[ 5x + 4 = -2x + 3. \]

Rearrange:

\[ 5x + 2x = 3 - 4. \]

\[ 7x = -1. \]

\[ x = -\frac{1}{7}. \]

Thus, the critical values are:

\[ x = -\frac{7}{3} \quad \text{and} \quad x = -\frac{1}{7}. \]

Step 3: Divide the real line into intervals

The critical values divide the real line into three intervals:

\[ (-\infty, -\frac{7}{3}), \quad \left(-\frac{7}{3}, -\frac{1}{7}\right), \quad \left(-\frac{1}{7}, \infty\right). \]

We analyze $ |5x + 4| \leqslant |2x - 3| $ on each interval.

Step 4: Solve on each interval

Interval 1: $ (-\infty, -\frac{7}{3}) $

On this interval:

\[ 5x + 4 < 0, \quad 2x - 3 < 0. \]

The inequality becomes:

\[ -(5x + 4) \leqslant -(2x - 3). \]

Simplify:

\[ -5x - 4 \leqslant -2x + 3. \]

Rearrange:

\[ -5x + 2x \leqslant 3 + 4. \]

\[ -3x \leqslant 7. \]

Divide by $-3$ (reverse the inequality):

\[ x \geqslant -\frac{7}{3}. \]

On this interval, $ x \geqslant -\frac{7}{3} $ conflicts with $ x < -\frac{7}{3} $. Hence, no solution exists on this interval.

Interval 2: $ \left(-\frac{7}{3}, -\frac{1}{7}\right) $

On this interval:

\[ 5x + 4 > 0, \quad 2x - 3 \lt 0. \]

The inequality becomes:

\[ 5x + 4 \leqslant -(2x - 3). \]

Simplify:

\[ 5x + 4 \leqslant -2x + 3. \]

Rearrange:

\[ 5x + 2x \leqslant 3 - 4. \]

\[ 7x \leqslant -1. \]

Divide by 7:

\[ x \leqslant -\frac{1}{7}. \]

Thus, the solution on this interval is:

\[ -\frac{7}{3} \lt x \leqslant -\frac{1}{7}. \]

Interval 3: $ \left(-\frac{1}{7}, \infty\right) $

On this interval:

\[ 5x + 4 > 0, \quad 2x - 3 > 0. \]

The inequality becomes:

\[ 5x + 4 \leqslant 2x - 3. \]

Simplify:

\[ 5x - 2x \leqslant -3 - 4. \]

\[ 3x \leqslant -7. \]

Divide by 3:

\[ x \leqslant -\frac{7}{3}. \]

On this interval, $ x > -\frac{1}{7} $ conflicts with $ x \leqslant -\frac{7}{3} $. Hence, no solution exists on this interval.

Step 5: Combine the results

The only valid solution is from Interval 2:

\[ -\frac{7}{3} \lt x \leqslant -\frac{1}{7}. \]

Thus, the solution to the inequality is:

\[ \boxed{-\frac{7}{3} \lt x \leqslant -\frac{1}{7}}. \]

Question 4

\[ y=\frac{\sec ^2 5 x-\tan ^2 5 x}{\operatorname{cosec} 5 x} \]

Show that $ y=a \sin b x $, where $ a $ and $ b $ are integers, and hence find the value of \[ \int_0^{\frac{\pi}{5}} y \, \mathrm{d}x. \] [4]

Solution

Step 1: Simplify the expression for $ y $

We are given the expression:

\[ y = \frac{\sec^2(5x) - \tan^2(5x)}{\csc(5x)} \]

Using the identity $ \sec^2 \theta - \tan^2 \theta = 1 $, we can simplify the numerator:

\[ \sec^2(5x) - \tan^2(5x) = 1. \]

Thus, the expression for $ y $ becomes:

\[ y = \frac{1}{\csc(5x)}. \]

Since $ \csc \theta = \frac{1}{\sin \theta} $, we can rewrite the denominator as:

\[ y = \sin(5x). \]

Step 2: Express $ y $ in the form $ a \sin(bx) $

We now have:

\[ y = \sin(5x). \]

This is already in the required form $ a \sin(bx) $, where $ a = 1 $ and $ b = 5 $.

Step 3: Compute the integral

We are tasked with finding the value of the integral:

\[ \int_0^{\frac{\pi}{5}} y \, dx = \int_0^{\frac{\pi}{5}} \sin(5x) \, dx. \]

The antiderivative of $ \sin(5x) $ is $ -\frac{1}{5} \cos(5x) $. Thus, we evaluate the integral:

\[ \int_0^{\frac{\pi}{5}} \sin(5x) \, dx = \left[ -\frac{1}{5}\cos(5x) \right]_0^{\frac{\pi}{5}}. \]

Substituting the limits of integration:

\[ = -\frac{1}{5} \left[ \cos\left(5 \times \frac{\pi}{5}\right) - \cos(0) \right] \]

\[ = -\frac{1}{5} \left[ \cos(\pi) - \cos(0) \right]. \]

Since $ \cos(\pi) = -1 $ and $ \cos(0) = 1 $, we get:

\[ = -\frac{1}{5}(-1 - 1) \]

\[ = -\frac{1}{5}(-2) \]

\[ = \frac{2}{5}. \]

Conclusion

\[ \int_0^{\frac{\pi}{5}} y \, dx = \frac{2}{5}. \]