Question

Plz Any class 11 Moderator please help. if you don't know share with other moderator​

Answer 1

SOLUTION

TO DETERMINE

1.

TO PROVE

$\displaystyle \sf{ \lim_{x \to 0} \frac{ \sin x}{x} = 1}$

TO EVALUATE

$\displaystyle \sf{ \lim_{x \to 0}\frac{ \sin ax}{bx} }$

2. To find the Geometric Progression

EVALUATION

Answer to Question : 1

First Part

Let us consider a circle with center at O and radius OA = 1 unit

Where ∠ AOP = x radian

Now area of the triangle OAP

$\displaystyle \sf{ = \frac{1}{2}. OA. PB}$

$\displaystyle \sf{ = \frac{1}{2}.1. \sin x \: \: \: sq.unit}$

$\displaystyle \sf{ = \frac{1}{2} \sin x \: \: \: sq.unit}$

Area of the sector OAP

$\displaystyle \sf{ = \frac{\pi. {1}^{2} }{2\pi} x \: \: \: sq.unit}$

$\displaystyle \sf{ = \frac{x}{2} \: \: \: sq.unit}$

Area of the triangle OAQ

$\displaystyle \sf{ = \frac{1}{2}.1. \tan x \: \: \: sq.unit}$

$\displaystyle \sf{ = \frac{1}{2} \frac{ \sin x}{ \cos x} \: \: \: sq.unit}$

Now we have

Δ OAP ⊆ Area of Sector OAP ⊆ Δ OAQ

Thus we get

$\displaystyle \sf{ \frac{1}{2} \sin \: x < \frac{x}{2} < \frac{1}{2} \frac{ \sin x}{ \cos x} \: \: }$

$\displaystyle \sf{ \implies \: 1 < \frac{x}{ \sin x} < \frac{1}{ \cos x} \: \: }$

$\displaystyle \sf{ \implies \: \cos x < \frac{\sin x}{ x} < 1 }$

This inequality is true for all x > 0

Also

$\sf{\cos( - x) = \cos x}$

$\displaystyle \sf{ \frac{\sin ( - x)}{ ( - x)} = \frac{ - \sin x}{ - x} = \frac{\sin x}{ x} }$

So the inequality is also true for x < 0

Now

$\displaystyle \sf{ \lim_{x \to 0} \: 1 = 1 \: \: \: \: and \: \: \lim_{x \to 0} \: \cos x= 1}$

So by the Pinching Theorem

$\displaystyle \sf{ \lim_{x \to 0}\frac{ \sin x}{x} = 1}$

Hence proved

Second Part

$\displaystyle \sf{ \lim_{x \to 0}\frac{ \sin ax}{bx} }$

$\displaystyle \sf{ = \lim_{x \to 0}\frac{ \sin ax}{ax} . \lim_{x \to 0} \frac{ax}{bx} }$

$\displaystyle \sf{ = \lim_{x \to 0}\frac{ \sin ax}{ax} . \frac{a}{b} }$

$\displaystyle \sf{ = \frac{a}{b} \: \lim_{x \to 0}\frac{ \sin ax}{ax} }$

Let y = ax

Then y → 0 as x → 0

Thus we get

$\displaystyle \sf{ \lim_{x \to 0}\frac{ \sin ax}{bx} }$

$\displaystyle \sf{ = \frac{a}{b} \: \lim_{x \to 0}\frac{ \sin ax}{ax} }$

$\displaystyle \sf{ = \frac{a}{b} \: \lim_{y \to 0}\frac{ \sin y}{y} }$

$\displaystyle \sf{ = \frac{a}{b} \times 1}$

$\displaystyle \sf{ = \frac{a}{b} }$

Answer to Question : 2

Let first term = a and common ratio = r

Then

$\sf{ \: n \: \: th \: term = a \times {r}^{n - 1} }$

By the first condition

a + ar = - 4 - - - - (1)

By the second condition

$\sf{a {r}^{4} = 4 \times a {r}^{2} }$

$\sf{ \implies \: {r}^{2} = 4 }$

$\sf{ \implies \: r = \pm \: 2 }$

When r = 2 , from Equation 1 we get

$\displaystyle \sf{a + 2a = - 4}$

$\displaystyle \sf{ \implies \: a = - \frac{ 4}{3} }$

In that case the progression is

$\displaystyle \sf{ - \frac{4}{3} \:, \: - \frac{8}{3} \: , \: - \frac{16}{3} \: , \: ... }$

When r = - 2 , from Equation 1 we get

$\displaystyle \sf{ \: a - 2a = - 4}$

$\displaystyle \sf{ \implies \: - a = - 4}$

$\displaystyle \sf{ \implies \: a = 4}$

In that case the progression is

4 , - 8 , 16 , - 32 , 64 , - 128 ,...

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