Find the order and the degree of the differential equation: x 2 d 2 y d x 2 = 1 + d y d x 2 4 .

Given differential equation is x 2 d 2 y d x 2 = 1 + d y d x 2 4 ⇒ x 2 d 2 y d x 2 1 = 1 + d y d x 2 4 We know that, The order of a differential equation is the order of the highest derivative occurring in the differential equation. The degree of a differential equation is the power of the highest order derivative when differential coefficients are made free from radicals and fractions. Hence, Order = 2 and Degree = 1 .

Find : ∫ s e c 2 x tan 2 x + 4 d x

We have, ∫ s e c 2 x tan 2 x + 4 d x Put tan x = t ⇒ sec 2 x d x = d t , we get ∫ s e c 2 x tan 2 x + 4 d x = ∫ d t t 2 + 4 = log t + t 2 + 4 + C = log tan x + tan 2 x + 4 + C

Find the shortest distance between the lines r → = ( 4 i ^ - j ^ ) + λ ( i ^ + 2 j ^ - 3 k ^ ) , and r → = ( i ^ - j ^ + 2 k ^ ) + μ ( 2 i ^ + 4 j ^ - 5 k ^ )

We know that the shortest distance between the lines r → = a → 1 + λ b → 1 and r → = a → 2 + λ b → 2 is given by d = a → 2 - a → 1 · b → 1 × b → 2 b → 1 × b → 2 Here, we have, a → 1 = 4 i ^ - j ^ , a → 2 = i ^ - j ^ + 2 k ^ , b → 1 = i + 2 j - 3 k ^ and b → 2 = 2 i ^ + 4 j ^ - 5 k ^ Now, a → 2 - a → 1 = i ^ - j ^ + 2 k ^ - 4 i ^ + j ^ = - 3 i ^ + 0 j ^ + 2 k ^ And b → 1 × b → 2 = i ^ j ^ k ^ 1 2 - 3 2 4 - 5 = - 10 + 12 i ^ - - 5 + 6 j ^ + 4 - 4 k ^ = 2 i ^ - j ^ + 0 k ^ Therefore, a → 2 - a → 1 · b → 1 × b → 2 = ( - 3 i ^ + 0 j ^ + 2 k ^ ) · ( 2 i ^ - j ^ + 0 k ^ ) = - 6 + 0 + 0 = - 6 And b → 1 × b → 2 = 4 + 1 + 0 = 5 Therefore, d = a → 2 - a → 1 · b → 2 × b → 1 b → 2 × b → 1 = - 6 5 = 6 5 5 units. Hence, the shortest distance between the lines is 6 5 5 units.

If y = sin - 1 x 2 , prove that 1 - x 2 d 2 y d x 2 - x d y d x - 2 = 0 .

Given y = sin - 1 x 2 Differentiating w.r.t. x , we get ⇒ d y d x = d sin - 1 x 2 d x ⇒ d y d x = 2 sin - 1 x × 1 1 - x 2 ⇒ d y d x = 2 sin - 1 x 1 - x 2 → 1 Again differentiating w.r.t x , we get ⇒ d 2 y d x 2 = 2 d d x sin - 1 x 1 - x 2 = 2 1 - x 2 × d sin - 1 x d x - sin - 1 x × d 1 - x 2 d x 1 - x 2 (Use Quotient Rule in Differentiation) = 2 1 - x 2 × 1 1 - x 2 - sin - 1 x × 1 2 1 - x 2 × ( - 2 x ) 1 - x 2 ⇒ d 2 y d x 2 = 2 1 + x sin - 1 x 1 - x 2 1 - x 2 ⇒ 1 - x 2 d 2 y d x 2 = 2 + 2 x sin - 1 x 1 - x 2 = 2 + x × 2 sin - 1 x 1 - x 2 ⇒ 1 - x 2 d 2 y d x 2 = 2 + x d y d x from 1 ⇒ 1 - x 2 d 2 y d x 2 - x d y d x - 2 = 0 Hence proved.

Find the vector equation of a line passing through the point ( 2 i ^ - 3 j ^ - 5 k ^ ) and perpendicular to the plane r → · ( 6 i ^ - 3 j ^ + 5 k ^ ) + 2 = 0 . Also, find the point of intersection of this line and the plane.

The vector equation of a line passing through the point ( 2 i ^ - 3 j ^ - 5 k ^ ) is r → = a → + λ b → , λ ∈ R i.e., r → = ( 2 i ^ - 3 j ^ - 5 k ^ ) + λ b → Since it is given that, required line is perpendicular to the plane r → · ( 6 i ^ - 3 j ^ + 5 k ^ ) + 2 = 0 , which means that, the line is parallel to the vector ( - 6 i ^ + 3 j ^ - 5 k ^ ) [Since line is perpendicular to plane, the line will be parallel to the normal of the plane] Now, the Cartesian equation of the plane is ( x i ^ + y j ^ + z k ^ ) · ( 6 i ^ - 3 j ^ + 5 k ^ ) + 2 = 0 ⇒ 6 x - 3 y + 5 z + 2 = 0 Thus, the required vector equation of the line is r → = ( 2 i ^ - 3 j ^ - 5 k ^ ) + λ ( - 6 i ^ + 3 j ^ - 5 k ^ ) Now, r → = 2 - 6 λ i ^ + - 3 + 3 λ j ^ + - 5 - 5 λ k ^ So, any point on the line is ( 2 - 6 λ ) , ( - 3 + 3 λ ) , ( - 5 - 5 λ ) Since this point is also, lie on the plane 6 x - 3 y + 5 z + 2 = 0 i.e., 6 ( 2 - 6 λ ) - 3 - 3 + 3 λ + 5 - 5 - 5 λ + 2 = 0 ⇒ 12 - 36 λ + 9 - 9 λ - 25 - 25 λ + 2 = 0 ⇒ - 70 λ - 2 = 0 ⇒ - 70 λ = 2 ⇒ λ = - 1 35 Hence, the point of intersection of this line and the plane is 2 - 6 - 1 35 , - 3 + 3 - 1 35 , - 5 - 5 - 1 35 i.e., 76 35 , - 108 35 , - 170 35

Maximize z = 6 x - 4 y subject to 2 x + 4 y ≤ 4 4 x + 8 y ≥ 16 where, x , y ≥ 0 .

Given LPP as follows: Objective function: z = 6 x - 4 y Subject to 2 x + 4 y ≤ 4 4 x + 8 y ≥ 16 where x ≥ 0 , y ≥ 0 We first draw the graph of the line 2 x + 4 y = 4 on the x y - plane by joining the points 0 , 1 and 2 , 0 . Now, the line divides the plane into two half. Let us identify the solution region of the inequation 2 x + 4 y ≤ 4 . Since the coordinates of the point 0 , 0 satisfy the inequation 2 x + 4 y ≤ 4 . So, the solution region for this inequation is that half-plane which contains the point 0 , 0 . Also, since the inequality is slack, so all the points on the line are also included in the solution region. Similarly, we can find the solution region for 4 x + 8 y ≥ 16 Also, x ≥ 0 , y ≥ 0 represents the region in the first quadrant. Thus, clearly, we see that, there is no feasible region because there is no such point that satisfy all constraints of the problem. Thus, the given problem has infeasible solution

E M B I B E

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

Embibe Experts Solutions for Chapter: Solved Mathematics Paper-2, Exercise 1: Exercise

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Solved Mathematics Paper-2, Exercise 1: Exercise

Attempt the free practice questions on Chapter 15: Solved Mathematics Paper-2, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Solved Mathematics Paper-2, Exercise 1: Exercise with Hints & Solutions

EASY

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Differential Equations>Exercise>Q 11

Solve: $(x^{2} + x) d x + (y^{2} - y) d y = 0$

EASY

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Differential Equations>Exercise>Q 26

Find the order and the degree of the differential equation: $x^{2} \frac{d^{2} y}{d x^{2}} = {\{1 + {(\frac{d y}{d x})}^{2}\}}^{4}$ .

MEDIUM

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise>Q 25

Find : $\int \frac{s e c^{2} x}{\sqrt{\tan^{2} x + 4}} d x$

HARD

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise-Choice>Q 50

Find the vector and Cartesian equations of the Plane through the points $(2, 2, - 1)$ , $(3, 4, 2) & (7, 0, 6)$ .Also find the vector equation of the plane passing through $(4, 3, 1)$ and parallel to the above obtained plane.

MEDIUM

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise-Choice>Q 51

Find the shortest distance between the lines $\vec{r} = (4 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} - 3 \hat{k})$ , and $\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} + 4 \hat{j} - 5 \hat{k})$

MEDIUM

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise>Q 58

If $y = {(\sin^{- 1} x)}^{2}$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - 2 = 0$ .

HARD

12th Karnataka Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 15 - Solved Mathematics Paper-2>Exercise-Choice>Q 64

Find the vector equation of a line passing through the point $(2 \hat{i} - 3 \hat{j} - 5 \hat{k})$ and perpendicular to the plane $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} + 5 \hat{k}) + 2 = 0$ . Also, find the point of intersection of this line and the plane.