Question 2, Exercise 9.1

Solutions of Question 2 of Exercise 9.1 of Unit 09: Trigonometric Functions. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the maximum and minimum values of the reciprocal trigonometric function: $y=\dfrac{1}{4+3 \operatorname{Sin} \theta}$

Solution.

We know $$\begin{align*} -1 \leq \operatorname{Sin} \theta \leq 1 \end{align*}$$ Multiplying with $3$ $$\begin{align*} -3 \leq 3 \operatorname{Sin} \theta \leq 3 \end{align*}$$ Adding $4$ $$\begin{align*} & 1 \leq 4+3 \operatorname{Sin} \theta \leq 7 \\ \implies & 1 \geq \frac{1}{4+3 \operatorname{Sin}} \theta \geq \frac{1}{7} \\ \implies & \frac{1}{7} \leq \frac{1}{4+3 \operatorname{Sin}} \theta\leq 1 \\ \end{align*}$$ Hence maximum value $(M) = 1$
and minimum value $(m) = \dfrac{1}{7}$.

Find the maximum and minimum values of the reciprocal trigonometric function: $y=\dfrac{1}{\frac{1}{2}-5 \operatorname{Cos} \theta}$

Solution.

From the graph, we see that given $y$ is not bounded and hence its maximum and minimum value doesn't exist.

As $y=\dfrac{1}{\frac{1}{2}-5 \operatorname{Cos} \theta}$ don't have any maximum or minimum value so asking to find out the maximum and minimum values is not appropriate.

Find the maximum and minimum values of the reciprocal trigonometric function: $y=\dfrac{1}{\frac{1}{3}-4 \sin (2 \theta-5)}$

Solution.

Same as Question 2(ii), we see that given $\dfrac{1}{\frac{1}{3}-4 \sin (2 \theta-5)}$ is not bounded and hence its maximum and minimum value doesn't exist.

As $\dfrac{1}{\frac{1}{3}-4 \sin (2 \theta-5)}$ don't have any finite maximum or minimum value so asking to find out the maximum and minimum values is not appropriate.

Find the maximum and minimum values of the reciprocal trigonometric function: $y=\dfrac{1}{3+\frac{2}{5} \sin (5 \theta-7)}$

Solution.

We know $$\begin{align*} -1 \leq \operatorname{Sin(5 \theta-7)} \theta \leq 1 \end{align*}$$ Multiplying by $\frac{2}{5}$:
$$\begin{align*} &-\frac{2}{5} \leq \frac{2}{5} \operatorname{Sin(5 \theta-7)} \theta \leq \frac{2}{5}\end{align*}$$ Adding 3:
$$\begin{align*} & 3 - \frac{2}{5} \leq 3 + \frac{2}{5} \operatorname{Sin(5 \theta-7)} \theta \leq 3 + \frac{2}{5}\\ \implies & \dfrac{13}{5} \geq \frac{1}{4+3 \operatorname{Sin(5 \theta-7)}} \theta \geq \dfrac{17}{5} \\ \implies & \dfrac{5}{13} \geq \frac{1}{4+3 \operatorname{Sin(5 \theta-7)}} \theta \geq \dfrac{5}{17} \\ \end{align*}$$ Hence maximum value $(M) = \dfrac{5}{13}$
and minimum value $(m) = \dfrac{5}{17}$.

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