Integrate with respect to x : x + 3 x 2 + 2 x + 2

Given ∫ x + 3 x 2 + 2 x + 2 Let I = ∫ x + 3 x 2 + 2 x + 2 d x Consider, x + 3 = A d d x x 2 + 2 x + 2 + B ⇒ x + 3 = A 2 x + 2 + B ⇒ x + 3 = 2 A x + 2 A + B 2 A = 1 ⇒ A = 1 2 2 A + B = 3 ⇒ 2 × 1 2 + B = 3 ⇒ B = 2 Therefore, I = ∫ 1 2 ( 2 x + 2 ) + 2 x 2 + 2 x + 2 d x = 1 2 ∫ 2 x + 2 x 2 + 2 x + 2 d x + 2 ∫ 1 x 2 + 2 x + 2 d x Put x 2 + 2 x + 2 = t in the first integrand ⇒ ( 2 x + 2 ) · d x = d t Add and subtract by 1 in the denominator of the second integral = 1 2 × 2 t + 2 ∫ 1 x 2 + 2 x + 1 - 1 + 2 d x = t + 2 ∫ 1 ( x + 1 ) 2 + ( 1 ) 2 d x Hence, I = x 2 + 2 x + 2 + 2 log ( x + 1 ) + ( x + 1 ) 2 + ( 1 ) 2 + C

Integrate with respect to x : x + 3 x 2 + 2 x + 2

Given ∫ x + 3 x 2 + 2 x + 2 Let I = ∫ x + 3 x 2 + 2 x + 2 d x Consider, x + 3 = A d d x x 2 + 2 x + 2 + B ⇒ x + 3 = A 2 x + 2 + B ⇒ x + 3 = 2 A x + 2 A + B 2 A = 1 ⇒ A = 1 2 2 A + B = 3 ⇒ 2 × 1 2 + B = 3 ⇒ B = 2 Therefore, I = ∫ 1 2 ( 2 x + 2 ) + 2 x 2 + 2 x + 2 d x = 1 2 ∫ 2 x + 2 x 2 + 2 x + 2 d x + 2 ∫ 1 x 2 + 2 x + 2 d x Put x 2 + 2 x + 2 = t in the first integrand ⇒ ( 2 x + 2 ) · d x = d t Add and subtract by 1 in the denominator of the second integral = 1 2 × 2 t + 2 ∫ 1 x 2 + 2 x + 1 - 1 + 2 d x = t + 2 ∫ 1 ( x + 1 ) 2 + ( 1 ) 2 d x Hence, I = x 2 + 2 x + 2 + 2 log ( x + 1 ) + ( x + 1 ) 2 + ( 1 ) 2 + C

Evaluate the integral: ∫ 0 π 2 sin 2 x cos x + sin x d x .

Let I = ∫ 0 π 2 sin 2 x cos x + sin x d x → 1 ⇒ I = ∫ 0 π 2 sin 2 π 2 - x cos π 2 - x + sin π 2 - x d x ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x ⇒ I = ∫ 0 π 2 cos 2 x d x sin x + cos x → 2 Adding 1 and 2 , we get: 2 I = ∫ 0 π 2 sin 2 x + cos 2 x sin x + cos x d x ⇒ 2 I = ∫ 0 π 2 1 sin x + cos x d x ∵ sin 2 x + cos 2 x = 1 ⇒ I = 1 2 ∫ 0 π 2 1 sin x + cos x d x Now, put cos x = 1 - tan 2 x 2 1 + tan 2 x 2 and sin x = 2 tan x 2 1 + tan 2 x 2 , so we get: I = 1 2 ∫ 0 π 2 1 2 tan x 2 1 + tan 2 x 2 + 1 - tan 2 x 2 1 + tan 2 x 2 d x ⇒ I = 1 2 ∫ 0 π 2 1 + tan 2 x 2 2 tan x 2 + 1 - tan 2 x 2 d x ⇒ I = 1 2 ∫ 0 π 2 sec 2 x 2 2 tan x 2 + 1 - tan 2 x 2 d x Put tan x 2 = t ⇒ 1 2 sec 2 x 2 d x = d t {Differentiating w.r.t x } Also, x = 0 ⇒ t = 0 and x = π 2 ⇒ t = 1 Therefore, I = ∫ 0 1 d t 2 t + 1 - t 2 ⇒ I = ∫ 0 1 d t - t 2 - 2 t - 1 ⇒ I = ∫ 0 1 d t - t - 1 2 - 1 - 1 ⇒ I = ∫ 0 1 d t - t - 1 2 - 2 2 ⇒ I = ∫ 0 1 d t ( 2 ) 2 - ( t - 1 ) 2 ⇒ I = 1 2 2 log 2 + t - 1 2 - t + 1 0 1 ∵ ∫ d x a 2 - x 2 = 1 2 a log a + x a - x + C ⇒ I = 1 2 2 log 2 + 1 - 1 2 - 1 + 1 - log 2 + 0 - 1 2 - 0 + 1 ⇒ I = 1 2 2 - log 2 - 1 2 + 1 ⇒ I = 1 2 2 log 2 + 1 2 - 1 × 2 + 1 2 + 1 ⇒ I = 1 2 2 log ( 2 + 1 ) 2 ⇒ I = 2 2 2 log ( 2 + 1 ) ⇒ I = 1 2 log ( 2 + 1 ) Hence, ∫ 0 π 2 sin 2 x cos x + sin x d x = 1 2 log ( 2 + 1 ) .

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise>Q 40

MEDIUM

12th Meghalaya Board

IMPORTANT

Earn 100

Integrate with respect to $x$ :

$\frac{x + 3}{\sqrt{x^{2} + 2 x + 2}}$

Important Questions on Solved Mathematics Paper-2

EASY

12th Meghalaya Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise-Choice>Q 41

Find the area bounded by the curves

y^{2} = 4 a x

and

x^{2} = 4 a y, a > 0

.

MEDIUM

12th Meghalaya Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise>Q 42

Find the unit vector perpendicular to plane determined by the vectors

\vec{a} = 4 \hat{i} + 3 \hat{j} - \hat{k}

and

\vec{b} = 2 \hat{i} - 6 \hat{j} - 3 \hat{k}

.

HARD

12th Meghalaya Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise>Q 43

Apply corner point method to find the optimal solution of the following linear programming problem. Find also the optimum value of the objective function.

Maximize $z = x + 3 y$

subject to $5 x + 4 y \geq 20$ ,

$7 x + 8 y \leq 56$

where $x \geq 0, y \geq 0$

HARD

12th Meghalaya Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise>Q 44

Solve the following equations by matrix method:

$2 x + y + z = 1$

$x - 2 y - z = \frac{3}{2}$

$3 y - 5 z = 9$ .

MEDIUM

12th Meghalaya Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise-Choice>Q 45

Use elementary row operation to determine whether the following matrix is invertible and hence find the inverse whenever possible.

$(\begin{matrix} 1 & 2 & 1 \\ 2 & 1 & - 1 \\ 1 & 5 & 4 \end{matrix})$

HARD

12th Meghalaya Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise>Q 46

Evaluate the integral: $\int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{\cos x + \sin x} d x$ .

MEDIUM

12th Meghalaya Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise>Q 47

Find the equation of the plane passing through the line of intersection of the planes

r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1 and r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0

and parallel to

X

-axis.

MEDIUM

12th Meghalaya Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise>Q 48

An examinee has to answer a question on a multiple choice type test. The answer to this question is either known to him or else he simply guesses at it. The probability that he knows the answer is