If y 1 - x 2 + x 1 - y 2 = 1 , then prove that d y d x = - 1 - y 2 1 - x 2

We have, y 1 - x 2 + x 1 - y 2 = 1 Let, x = sin A , y = sin B ⇒ sin B 1 - sin 2 A + sin A 1 - sin 2 B = 1 ⇒ sin B cos A + sin A cos B = 1 ∵ sin x + y = sin x cos y + cos x sin y ⇒ sin A + B = 1 ⇒ A + B = sin - 1 1 ⇒ sin - 1 x + sin - 1 y = π 2 ∵ x = sin A , y = sin B Differentiate with respect to x , ⇒ d d x sin - 1 x + d d x sin - 1 y = d d x π 2 ⇒ 1 1 - x 2 + 1 1 - y 2 d y d x = 0 ⇒ d y d x = - 1 - y 2 1 - x 2

Find x → if x → - a → . x → + a → = 12 , where a → is a unit vector.

Given, a → is a unit vector, ∴ a → = 1 ............... ( i ) Consider, x → - a → ⋅ x → + a → = 12 [ given ] ⇒ x → · x → + a → - a → · x → + a → = 12 [Using distributivity of scalar product] ⇒ x → ⋅ x → + x → ⋅ a → - a → ⋅ x → - a → ⋅ a → = 12 ⇒ x → ⋅ x → + x → ⋅ a → - x → ⋅ a → - a → ⋅ a → = 12 [ Using x → · a → = a → · x → ] ⇒ x → 2 - a → 2 = 12 [Using a ⇀ · a → = a → 2 ] ⇒ x → 2 - 1 = 12 [ Using i ] ⇒ x → 2 = 13 ∴ x → = 13

Find ∫ d x 3 - 2 x - x 2

Given, ∫ d x 3 - 2 x - x 2 = ∫ d x 4 - x 2 + 2 x + 1 = ∫ d x ( 2 ) 2 - ( x + 1 ) 2 We know that ∫ d x a 2 - x 2 = sin - 1 x a + c = sin - 1 x + 1 2 + c

Evaluate: ∫ π 6 π 3 d x 1 + tan x .

Let I = ∫ π 6 π 3 d x 1 + tan x → 1 Applying the property: ∫ a b f x d x = ∫ a b f a + b - x d x Therefore, I = ∫ π 6 π 3 d x 1 + tan π 3 + π 6 - x ⇒ I = ∫ π 6 π 3 d x 1 + tan π 2 - x ⇒ I = ∫ π 6 π 3 d x 1 + cot x ⇒ I = ∫ π 6 π 3 tan x d x 1 + tan x → 2 Now, adding 1 and 2 , we get: 2 I = ∫ π 6 π 3 d x 1 + tan x + ∫ π 6 π 3 tan x d x 1 + tan x ⇒ 2 I = ∫ π 6 π 3 1 + tan x d x 1 + tan x ⇒ 2 I = ∫ π 6 π 3 1 · d x ⇒ 2 I = x π 6 π 3 ⇒ 2 I = π 3 - π 6 ⇒ 2 I = π 6 ⇒ I = π 12 Hence, ∫ π 6 π 3 d x 1 + tan x = π 12 .

Evaluate : ∫ 0 4 x - 1 d x

We need to find ∫ 0 4 x - 1 d x We know that x = x if x ≥ 0 - x if x < 0 ⇒ x - 1 = - x - 1 if 0 ≤ x < 1 1 - x if 1 ≤ x ≤ 4 ⇒ ∫ 0 4 x - 1 d x = ∫ 0 1 - x - 1 d x + ∫ 1 4 x - 1 d x = - x 2 2 - x 0 1 + x 2 2 - x 1 4 = - 1 2 - 1 - 0 2 - 0 + 4 2 2 - 4 - 1 2 2 - 1 = - - 1 2 + 4 - - 1 2 = 5

Evaluate: ∫ π 6 π 3 d x 1 + tan x .

Let I = ∫ π 6 π 3 d x 1 + tan x → 1 Applying the property: ∫ a b f x d x = ∫ a b f a + b - x d x Therefore, I = ∫ π 6 π 3 d x 1 + tan π 3 + π 6 - x ⇒ I = ∫ π 6 π 3 d x 1 + tan π 2 - x ⇒ I = ∫ π 6 π 3 d x 1 + cot x ⇒ I = ∫ π 6 π 3 tan x d x 1 + tan x → 2 Now, adding 1 and 2 , we get: 2 I = ∫ π 6 π 3 d x 1 + tan x + ∫ π 6 π 3 tan x d x 1 + tan x ⇒ 2 I = ∫ π 6 π 3 1 + tan x d x 1 + tan x ⇒ 2 I = ∫ π 6 π 3 1 · d x ⇒ 2 I = x π 6 π 3 ⇒ 2 I = π 3 - π 6 ⇒ 2 I = π 6 ⇒ I = π 12 Hence, ∫ π 6 π 3 d x 1 + tan x = π 12 .

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Sample Paper 1>Exercise>Q 25

EASY

12th Uttar Pradesh Board

IMPORTANT

Earn 100

Find the direction ratio and direction cosines of a line parallel to the line whose equations are

$6 x - 12 = 3 y + 9 = 2 z - 2$

Important Questions on Sample Paper 1

EASY

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Sample Paper 1>Exercise>Q 26

If

y \sqrt{1 - x^{2}} + x \sqrt{1 - y^{2}} = 1,

then prove that

\frac{d y}{d x} = - \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}

EASY

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Sample Paper 1>Exercise>Q 27

Find

|\vec{x}|

if

(\vec{x} - \vec{a}) . (\vec{x} + \vec{a}) = 12,

where

\vec{a}

is a unit vector.

EASY

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Sample Paper 1>Exercise>Q 28

Find

\int \frac{d x}{\sqrt{3 - 2 x - x^{2}}}

EASY

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Sample Paper 1>Exercise>Q 29

Three friends go for coffee. They decide who will pay the bill, by each tossing a coin and then letting the “odd person” pay. There is no odd person if all three tosses produce the same result. If there is no odd person in the first round, they make a second round of tosses and they continue to do so until there is an odd person. What is the probability that exactly three rounds of tosses are made?

EASY

12th Uttar Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Sample Paper 1>Exercise>Q 30

Find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing