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- Question 1, Exercise 9.1
- ame{Cos} \theta$ ** Solution. ** We know \begin{align*} -1 \leq \operatorname{Cos} \theta \leq 1 \end{align*} Multiplying with $-2$ \begin{align*} & 2 \geq -2 \operatorname{Cos} \theta \geq -2 \end{align*} Adding $2$ \begin{align*} & 4 \geq 2-2 \operat
- Question 2, Exercise 9.1
- me{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 \operat
- Question 3, Exercise 9.1
- range: $y=7 \cos 4x$ ** Solution. ** AS \begin{align*} & -1\leq \cos 4x \leq 1 \,\, \forall \,\, x\in... {R} \\ \implies & -7\leq 7 \cos 4x \leq 7 \\ \end{align*} Thus domain $= ]-\infty, \infty[ = \mathbb{R}$ ... $y=\cos \frac{x}{3}$ ** Solution. ** AS \begin{align*} & -1\leq \cos \frac{x}{3} \leq 1 \,\, \forall \,\, x\in \mathbb{R} \\ \end{align*} Thus domain $= ]-\infty, \infty[ = \mathbb{R}$
- Question 4(i-iv), Exercise 9.1
- sider $f(x)=\sin x+x \cdot \cos x$. Take \begin{align*} f(-x) = \sin (-x) + (-x)\cdot \cos (-x) \end{align*} As we know $\sin(-x)=-\sin x$ and $\cos (-x) = \cos x$, so \begin{align*} f(x) & = -\sin x - x \cdot \cos x \\ & = -(\sin x + x \cdot \cos x) \\ & = -f(x) \end{align*} Thus the given function is odd. =====Question
- Question 4(v-viii), Exercise 9.1
- \[y = \frac{\sin^2 x}{x + \tan x}\] Take \begin{align*} y(-x) &= \frac{\big(-\sin x\big)^2}{-x - \tan x... \\ & = -\frac{\sin^2 x}{x + \tan x}\\ &=-y(x)\end{align*} Thus, the given function is odd. =====Question... = \frac{\tan x - \sin x}{\sin^3 x}\] Take \begin{align*} y(-x) &= \frac{-\tan x - (-\sin x)}{(-\sin x)^3... \frac{\tan x - \sin x}{\sin^3 x}\\ & = -y(x) \end{align*} Thus, the given function is odd. =====Question
- Question 6, Exercise 9.1
- period of the $\sec$ is $2\pi$, therefore \begin{align*} 6 \sec(2 x-3) & = 6 \sec(2 x-3+2\pi) \\ & = 6 \sec(2(x+\pi)-3) \end{align*} Hence period of $6 \sec(2 x-3)$ is $\pi$. GOOD ... period of the $\cos$ is $2\pi$, therefore \begin{align*} \cos (5 x+4) & = 6 \cos(5x+4+2\pi) \\ & = \cos\... left(5\left(x+\frac{2\pi}{5}\right)+4\right) \end{align*} Hence period of $\cos (5 x+4)$ is $\dfrac{2\pi}
- Question 2 and 3, Review Exercise
- in \theta=\sqrt{2}\sin \theta$$ This gives \begin{align*} & \cos \theta=\sqrt{2}\sin \theta + \sin \theta... heta=\frac{1}{\sqrt{2}+1}\cos \theta ... (1) \end{align*} Now \begin{align*} LHS & = \cos \theta+ \sin \theta \\ & = \cos \theta+ \frac{1}{\sqrt{2}+1}\cos \the... theta \\ & = \sqrt{2} \cos \theta \\ & = RHS \end{align*} =====Question 3(i)===== Verify: $\dfrac{\tan x
- Question 5(i-v), Exercise 9.1
- x}}{2}$ ** Solution. ** ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit... 9-1-p5|< Question 4(v-viii) ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:uni
- Question 5(vi-x), Exercise 9.1
- {x}{2}$ ** Solution. ** ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit... :ex9-1-p6|< Question 5(i-v) ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:uni
- Question 7 & 8, Exercise 9.1
- cale. ** Solution. ** ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit09:ex9-1-p8|< Question 6 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:uni
- Question 9, Exercise 9.1
- n x=x$ ** Solution. ** ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit... 9:ex9-1-p9|< Question 7 & 8 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:uni
- Question 2 and 3,Review Exercise
- rd, Islamabad, Pakistan. ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit09:Re-ex-p1|< Question 1 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:uni
- Question 4, Review Exercise
- rd, Islamabad, Pakistan. ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit... 9:Re-ex-p2|< Question 2 & 3 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:uni
- Question 4, Review Exercise
- 4===== ** Solution. ** ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit... 9:Re-ex-p2|< Question 2 & 3 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:uni
- Question 5 and 6, Review Exercise
- ===== ** Solution. ** ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit09:Re-ex-p3|< Question 4 ]]</btn></text> <text align="right"><btn type="success">[[math-11-nbf:sol:uni