Search
You can find the results of your search below.
Fulltext results:
- Question 1, Exercise 9.1
- 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 \... & 0 \leq 2-2 \operatorname{Cos} \theta \leq 4 \\ \end{align*} Hence maximum value $(M) = 4$ \\ and mini... value (M)} & = a+|b|\\ & = 2+|-2| \\ & = 2+2 = 4 \end{align*} \begin{align*} \text{Minimum value (m)} &
- Question 2, Exercise 9.1
- 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 \... \frac{1}{4+3 \operatorname{Sin}} \theta\leq 1 \\ \end{align*} Hence maximum value $(M) = 1$ \\ and mini... leq \operatorname{Sin(5 \theta-7)} \theta \leq 1 \end{align*} Multiplying by \(\frac{2}{5}\): \\ \begi
- Question 3, Exercise 9.1
- thbb{R} \\ \implies & -7\leq 7 \cos 4x \leq 7 \\ \end{align*} Thus domain $= ]-\infty, \infty[ = \mathb... }{3} \leq 1 \,\, \forall \,\, x\in \mathbb{R} \\ \end{align*} Thus domain $= ]-\infty, \infty[ = \mathb... 2x}{3} \leq 1 \,\, \forall \,\, x \in \mathbb{R} \end{align*} Thus domain $= ]-\infty, \infty[ = \mathb... frac{\pi}{2} x \neq n\pi \\ \implies & x \neq 2n \end{align*} Hence domain of $y=\left\{x: x\in \math
- Question 4(i-iv), Exercise 9.1
- align*} f(-x) = \sin (-x) + (-x)\cdot \cos (-x) \end{align*} As we know $\sin(-x)=-\sin x$ and $\cos (... x \\ & = -(\sin x + x \cdot \cos x) \\ & = -f(x) \end{align*} Thus the given function is odd. =====Que... (-x) = (-x)^{3} \cdot \sin (-x) \cdot \cos (-x) \end{align*} As we know $\sin(-x)=-\sin x$ and $\cos (... & = x^{3} \cdot \sin x \cdot \cos x \\ & = f(x) \end{align*} Thus the given function is even. =====Qu
- Question 4(v-viii), Exercise 9.1
- x)}\\ & = -\frac{\sin^2 x}{x + \tan x}\\ &=-y(x)\end{align*} Thus, the given function is odd. =====Qu... &= -\frac{\tan x - \sin x}{\sin^3 x}\\ & = -y(x) \end{align*} Thus, the given function is odd. =====Qu... \\ &=-\frac{\tan x - \sin x}{\sin^3 x}\\ &= -y(x)\end{align*} Thus, the given function is odd. =====Qu... \cot x \\ &= -x^2 \cdot \sin x + \cot x\\ &=-y(x)\end{align*} Thus, the given function is odd. ====Go
- Question 6, Exercise 9.1
- & = 6 \sec(2 x-3+2\pi) \\ & = 6 \sec(2(x+\pi)-3) \end{align*} Hence period of $6 \sec(2 x-3)$ is $\pi$.... cos\left(5\left(x+\frac{2\pi}{5}\right)+4\right) \end{align*} Hence period of $\cos (5 x+4)$ is $\dfrac... eft(3\left(x + \frac{2\pi}{3}\right) + 3\right). \end{align*} Hence, the period of \( 7 \sin(3x + 3) \)... &= 5 \sin\left(2\left(x + \pi\right) + 3\right). \end{align*} Hence, the period of $ 5 \sin(2x + 3)$ is
- Question 2 and 3, Review Exercise
- n \theta=\frac{1}{\sqrt{2}+1}\cos \theta ... (1) \end{align*} Now \begin{align*} LHS & = \cos \theta+ ... os \theta \\ & = \sqrt{2} \cos \theta \\ & = RHS \end{align*} =====Question 3(i)===== Verify: $\dfrac{... {\sin^2 x} \\ & = \sec^2 x - \csc^2 x \\ & = RHS \end{align*} GOOD =====Question 3(ii)===== Verify: $\... tan^2 x} \\ & = \sec^2 x - \tan^2 x \\ & = RHS \end{align*} GOOD =====Question 3(iii)===== Verify: $