Mathematics Mock Test - 5

Concept:

Direction cosine: It is the cosines of each angles that the directed line makes with the x-axis, y-axis, and z-axis i.e. α, β and γ respectively.

Diagram

Represented as: {l, m, n}

Where, l = cos α, m = cos β and n = cos γ

Also, l² + m² + n² = 1

Calculation:

Given:

Direction cosines of a line {0, 0, 1}

To find: Orientation with coordinate axes.

l = cos α = 0

⇒ α = 90°

m = cos β = 0

⇒ β = 90°

n = cos γ = 1

⇒ γ = 0°

So, Cosine angle along z- axis is zero hence the line is parallel to z-axis.

Concept:

The order of a differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

Given the differential equation is

\(\rm x\left(d^2y\over dx^2\right)^{2\over3} = y^2\left({dy\over dx}\right)^{3\over2}\)

Taking cube on both sides, we get

\(\rm x^3\left(d^2y\over dx^2\right)^{2} = y^6\left({dy\over dx}\right)^{9\over2}\)

Taking square on both sides, we get

\(\rm x^6\left(d^2y\over dx^2\right)^{4} = y^{12}\left({dy\over dx}\right)^{9}\)

Highest derivate is \(\rm d^2y\over dx^2\)

So, the order of the given differential equation = 2

The power of the highest derivate = 4

So, the degree of the given differential equation = 4

Concept:

Chain Rule of Derivatives: For two functions u and v of x, we have: \(\rm \frac{du}{dv}=\frac{du}{dx}\times\frac{dx}{dv}\).

Calculation:

Using the chain rule of derivatives, we have:

\(\rm \frac{dx^2}{dx^3}=\frac{dx^2}{dx}\times\frac{dx}{dx^3}\)

= \(\rm \frac{dx^2}{dx}\times \frac 1 {\frac{dx^3}{dx}}\)

= \(\rm \frac{2x}{3x^2}\)

= \(\rm \frac{2}{3x}\)

\(\begin{array}{l} \frac{{dy}}{{dx}} = 2xy\\ \smallint \frac{1}{y}dy = \smallint 2xdx + c\\ y = {e^{{x^2}}} + c \end{array}\)

I = \(\int\limits_0^{\frac{\pi }{2}} {\log (\tan x)dx} \)  …….(1)

I = \(\int\limits_0^{\frac{\pi }{2}} {\log \tan (\frac{\pi }{2} - x)dx}\)

I = \(\int\limits_0^{\frac{\pi }{2}} {\log (cotx)dx}\) ……(2)

Adding 1 and 2

\(2I = \;\mathop \smallint \limits_0^{\pi /2} {\rm{log}}\{ (\tan x) \times (\cot x)\} dx\)

\(2I = \mathop \smallint \limits_0^{\pi /2} \log 1dx\)

I = 0  {∵ log 1 = 0}

CONCEPT:

In order to find the minor of an element of a determinant, we need to delete the row and column passing through the element aij , thus obtained is called the minor of aij and is usually denoted by Mij .

The co-factor of an element aij is given by (-1)i + j ⋅ Mij, where Mij is the minor of element aij and it is denoted by Cij.

Thus \({C_{ij}} = \;\left\{ {\begin{array}{*{20}{c}} {{M_{ij}},\;when\;i + j\;is\;even}\\ { - \;{M_{ij}},\;when\;i + j\;is\;odd} \end{array}} \right.\)

CALCULATION:

The co-factor of element 4, in he 2^nd row and 3^rd column is,

C23 = (-1)²⁺³ \(\left| {\begin{array}{*{20}{c}} 1&3&1\\ 8&0&1\\ 0&2&1 \end{array}} \right|\) 

⇒ C23 = -{1(-2) - 3 (8 - 0) + 1.16}

⇒ C23 =10.

Concept:

• If f′(x) > 0 then the function is said to be increasing.
• If f′(x) < 0 then the function is said to be decreasing.
• If a function f is increasing so while operating that function on a given relation of two input values of that function, the sign of the inequality will not change.

⇒ If a > b ⇒ f(a) > f(b)

• If a function g is decreasing so while operating that function on a given relation of two input values of that function, the sign of the inequality will change.

​⇒ If a > b ⇒ g(a) < g(b)

Calculation:

Given: f(x) is an increasing function and g(x) is a decreasing function

∴ f’(x) >  0 and g’(x) < 0

Also,

If p > q 

⇒ f(p) > f(q)

and  g(p) < g(q)

So, the correct answer will be option 1.

Concept:

Following steps to finding maxima and minima using derivatives:

Step-1: Find the derivative of the function.

Step-2: Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.

Step-3: Now we have to find the second derivative.

Case-1: If f"(x) is less than 0 then the given function is said to be maxima

Case-2: If f"(x) Is greater than 0 then the function is said to be minima

Calculation:

Given:

f(x) = x² - 4x

Differentiating with respect to x, we get

⇒ f'(x) = 2x - 4

For minimum value, f'(x) = 0

⇒ 2x - 4 = 0

∴ x = 2

Again differentiating with respect to x, we get

⇒ f"(x) = 2 > 0

Hence f(x) attains minimum value at x = 2

Concept:

An identity function is a function that always returns the same value that was used as its argument.

Hence f:R→R, f (x) = x is an identity function

Calculation:

f(g(x)) = x

f'(g(x))g'(x) = 1

Put x = a

⇒ f'(b) g'(a) = 1

f'(b) = \(\frac{1}{5}\).

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

• A new variable is to be chosen, say “t”
• The value of dt is to be determined.
• Substitution is done and the integral function is then integrated.
• Finally, the initial variable t, to be returned.

Calculation:

\(I = \mathop \smallint \limits_{1/\pi }^{2/\pi } \frac{{\cos \left( {1/x} \right)}}{{{x^2}}}dx\)

\(⇒ I = \mathop \smallint \limits_{1/\pi }^{2/\pi } \frac{{\cos \left( {1/x} \right)}}{{{x^2}}}dx\; = \;\mathop \smallint \limits_{2/\pi }^{1/\pi } \frac{{ - \cos \left( {1/x} \right)}}{{{x^2}}}dx\)

\(⇒ I = \;\mathop \smallint \limits_{\frac{\pi }{2}}^\pi {\rm{cosy}}dy\; = \;\left| {siny} \right|_{\pi /2}^\pi\)

⇒ I = 0 - 1 = - 1

Concept:

Let y = f(x) be the equation of a curve, then slope of the tangent at any point say (x1, y1) is given by: 

\(m = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;\;{y_1}} \right)}}\).

Slope of normal at any point say (x1, y1) is given by:

 \(\frac{{ - 1}}{{Slope\;of\;tangent\;at\;point\;\left( {{x_1},\;{y_1}} \right)}} = \; - {\left[ {\frac{{dx}}{{dy}}} \right]_{\left( {{x_1},\;{y_1}} \right)}}\)

Equation of tangent at any point say (x1, y1) is given by:

 \(y - {y_1} = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;{y_1}} \right)}} ⋅ \left( {x - {x_1}} \right)\)

Equation of normal at any point say (x1, y1) is given by: 

\(y - {y_1} = \; - {\left[ {\frac{{dx}}{{dy}}} \right]_{\left( {{x_1},\;{y_1}} \right)}} ⋅ \left( {x - {x_1}} \right)\)

Calculation:

Given: The tangent to the curve y = x + sin y at a point (a, b) is parallel to the line joining \(\left( {0,\frac{3}{2}} \right)\)and \(\left( {\frac{1}{2},2} \right)\),

So, the slope of the tangent at (a, b) and the slope of the line joining \(\left( {0,\frac{3}{2}} \right)\)and \(\left( {\frac{1}{2},2} \right)\) will be same 

y = x + sin y

\(⇒\frac{{dy}}{{dx}} = \frac{1}{{1 - \cos y}} = \frac{{2 - \frac{3}{2}}}{{\frac{1}{2}-0}} = 1\)

\( ⇒ \cos y = 0 ⇒ y = \left( {2n + 1} \right)\frac{\pi }{2}\)

Point lie on the curve,

⇒ b = a + sin y 

⇒ b - a = sin y 

Since, 

-1 ≤ sin y ≤ 1

⇒ |sin y| ≤ 1

⇒ | b - a | = 1

Given:

The cards pack 52

Formula used

P(E) = N(E)/N(S)

P(E) = Probability of an event

N(E) = Number of favorable outcomes

N(S) = total events

Calculation

Total cards = 52

The cards represent = 2, 3, 4, 5, 6, 7, 8, 9, 10

4 set of packs of cards

Then total numbered cards = 9 × 4 = 36

Then P(E) = 36/52

P(E) =  9/13

Hence, the probability of the number cards = 9/13

\(\begin{array}{l} P\left( {\frac{H}{R}} \right) = \frac{{P\left( {H \cap R} \right)}}{{P\left( R \right)}}\\ \Rightarrow 0.2 = \frac{{P\left( {R \cap H} \right)}}{{0.4}} \end{array}\)

∴ P(R∩H) = 0.2 × 0.4 = 0.08

Concept:

The probability mass function,  \(P(X=x)=f(x)\) , of a discrete random variable  X  is a function that satisfies the following properties:

\(P(X=x)=f(x)>0\) for \(x\in Range (X)\)

\( \sum_{x\in Range(X)}^{}P(X=x) =1\)

Calculation:

Given:

\(f(x)=cx^2\)

If f is a probability mass function, then

\( \sum_{x=1}^{3}P(X=x) =f(x)=1\)

\(\Rightarrow f(1)+f(2)+f(3)=1\)

\(\Rightarrow c(1)^2+c(2)^2+c(3)^2=1\)

\(\Rightarrow c+4c+9c=1\)

\(\Rightarrow 14c=1\)

\(\therefore c=\frac{1}{14}\)

Therefore the option (4) correct.

Concept:

Function is a relation between sets that associates to every element of a first set exactly one element of the second set.

Or, A function can only have one output, for each unique input.

Vertical line test: vertical line test is a visual way to determine if a curve is a graph of a function or not.

Calculation:

Given: S = {(x, y): x2 + y2 = 1,  - 1 ≤ x ∈ R ≤ 1 and - 1 ≤ y ∈ R ≤ 1}

x² + y² = 1, It represents the equation of circle.

As we can see through vertical line test it gives two output for unique input.

Concept:

\({\rm{Mean\;}}\left( \mu \right) = \mathop \sum \limits_{i = 1}^n {P_i} \cdot {x_i}\)

Probability of success ‘r’ times out of ‘n’

\( = {}_{}^n{C_r}{p^r}{q^{n - r}}\)

Calculation:

Given: The probabilities of zero, one two, and three successes are 8/27, 4/9, 2/9, and 1/27 respectively.

So, we can define the table as, 

\({\rm{Mean\;}}\left( \mu \right) = \mathop \sum \limits_{i = 0}^3 {x_i}{p_i} = 0 + \frac{4}{9} + 2 \times \frac{2}{9} + 3 \times \frac{1}{{27}}\)

∴ μ = 1

Concept:

One-To-One Function: A function f:A → B is one-to-one if every element of the range B corresponds to exactly one element of the domain A of f.

Important: For a function f:A → B to be one-to-one, the range must have more number of elements than the domain.

Calculation:

Since the function from A to B has to be one-to-one, therefore every element of A must be mapped to exactly one element of B without repetition.

Since A = {a, b, c, d} has 4 elements and B = {1, 2, 3, 4, 5, 6} has 6 elements in it,

The total number of one-to-one functions possible is:

= (6 possibilities for mapping a) × (5 possibilities for mapping b) × (4 possibilities for mapping c) × (3 possibilities for mapping d)

= 360.

Concept:

An operation * on a non-empty set S, is said to be a binary operation if it satisfies the closure property.

Closure Property:

Let S be a non-empty set and a, b ∈ S, if a * b ∈ S for all a, b ∈ S then S is said to be closed with respect to operation *.

Let * be a binary operation on a non-empty set S. Then

• * is associative on S if (a * b) * c = a * (b * c) ∀ a, b, c ∈ S.
• * is commutative on S if a * b = b * a ∀ a, b ∈ S.

Calculation:

Given: A = {1, 2, 3, 4, 5} and let * is an operation on A such that a * b = min {a, b}

First lets draw an operation table for the operation * defined on A

Statement 1: * is a binary operation on A

As we can from the table given above that a * b ∈ A ∀ a, b ∈ A.

Hence, * is a binary operation on A.

Concept: 

There are three main components of linear programming:

• Decision Variables: These variables are the activities that share the available resources while also competing with one another. In other words, these are interrelated in terms of resource utilization and need to be solved simultaneously. They are assumed to be non-negative and continuous.  
• The Objective Function: Each linear programming problem is aimed to have an objective to be measured in quantitative terms such as profits, sales, cost, time, or some other parameter, etc which needs to be maximized or minimized. The objective function will vary depending upon the business requirements or other factors.
• Constraints: These represent real-life limitations such as money, time, labor, or other restrictions and the objective function needs to be maximized or minimized while satisfying the constraints. They are represented as linear equations or inequations in terms of the decision variables.
• For example, suppose x and y are the decision variables. The objective function will be given by:

Z = ax + by ….(1)

Where a and b are constants and Z is the function to be maximized or minimized. There will be conditions x ≥ 0, and y ≥ 0, which indicates the non-negative constraints on the decision variable.

• The equation looks very simple since there are various assumptions involved while forming a linear programming example. These are mentioned below:
• Parameters such as resources available, profit contribution of unit decision variable, and resources used by unit decision variable need to be known.
• Decision variables are continuous. Hence the outputs can be an integer or a fraction.
• The contribution of each decision variable in the objective function is directly proportional to the objective function.

Calculation:

Given:

• From the above problem, Durgesh needs to maximize his daily income. x and y represent the number of packets prepared for dishes A and B. These represent the decision variables for the problem.
• There is a constraint of time mentioned in the problem.
• It has been mentioned that Durgesh works for 12 hours, or 1440 minutes a day in the restaurant, and 1 hour, or 60 minutes in those 1440 minutes is used by Durgesh for his own activities. 
• Dish A takes 20 minutes to prepare while dish B takes 30 minutes. 
• Hence, the time constraint is given by the following equation:

20x + 30y - 60 ≤ 1440,

• Hence, the correct answer will be option (2).

Concept:

Corner point method:

•  As per the method, the coordinates of all corner points of the feasible region are determined and the value of the objective function at these points is computed since the optimal value of the objective function lies at one of the corner points as per the LP theory.
• The first step to find the solution of an LP problem is to determine the feasible region. It is obtained by drawing the X-Y coordinates and plotting all the constraints.
• The common region on the graph obtained that satisfies all the constraints will be the feasible region.
• If no feasible region is obtained via plotting the constraints, then the linear program is infeasible.
• All the possibilities are summarized below: 

Calculation:

Given:

• Objective function Z = 5x + 4y needs to be maximized subject to the constraints,

 x, y ≥ 0 ....(1), x - 2y ≥ 2 ....(2) and 2x - 4y ≤ -3 ....(3). 

• The first thing needed to be done is to plot all the constraints on the X-Y axis and then find the feasible region if it exists. Plotting inequalities (1), (2), and (3) on the X-Y axis gives the following graph:

• To draw the constraints, a line of respective equations has been formed by calculating the points on the line.
• For example, to draw in equation (2), points on line x - 2y = 2 are calculated by putting x = 0 and calculating y as - 1. Hence one point on the line is calculated.
• Similarly, putting y = 0 and calculating x as 2. Hence one other point on the line is calculated.
• Then the two points will be joined via a straight line and extended further on both sides to get a better perspective of the line.
• A similar process will be done for another constraint to form its line.
• Constraint (2) will solution will lie in the lower half-plane to maintain the inequality while constraint (3) will lie in the upper half-plane.
• As it can be seen that the orange area represents the area satisfying constraint (3) and the blue area represents the area satisfying constraint (2).
• However, since no area in the plane is satisfying constraints (2) and (3) due to the different half-planes of their solutions, there is no area on the graph which is satisfying all the constraints. Hence, this problem will have no solution, in other words, the problem has an infeasible solution.     
• So, the correct answer is option 4.

Additional Information 

Examples for all possible results in LP using the Graphical method:

• The LP problem with no region satisfying all the constraints is shown below graphically:          

• This kind of LP problem has no solution.

• Subsequently, a feasible and bounded region (closed) is shown in the figure below:

• The feasible and unbounded region is shown in the graph below. Here the value of y and x has no bound unlike the previous examples:

• When the feasible region is obtained, first the corners of the same are determined.
• Subsequently, the value of the objective function Z is calculated at all the points.
• The point where the objective function attains maximum or minimum value is the optimal solution of the problem.  
• If the feasible bounded region has been formed, then for this case, the objective function has both maximum and minimum value at a corner point of the given feasible region.
• However, if the region is feasible and unbounded, then things become a little complicated. Here also, the objective function will have a maximum and minimum value at a corner point of the feasible region. But to check whether Z has maximum or minimum values in this unbounded region case, we have to draw this region also.

Concept:

Corner point method:

•  As per the method, the coordinates of all corner points of the feasible region are determined and the value of the objective function at these points is computed since the optimal value of the objective function lies at one of the corner points as per the LP theory.
• The first step to find the solution of an LP problem is to determine the feasible region. It is obtained by drawing the X-Y coordinates and plotting all the constraints.
• The common region on the graph obtained that satisfies all the constraints will be the feasible region.
• If no feasible region is obtained via plotting the constraints, then the linear program is infeasible.
• All the possibilities are summarized below: 

Calculation:

Given:

• Objective function Z = 2x + 6y needs to be maximized subject to the constraints,

x, y ≥ 0 ....(1), x + 3y ≤ 8 ....(2) and 2x + y ≤ 6 ....(3). 

• The first thing needed to be done is to plot all the constraints on the X-Y axis and then find the feasible region if it exists. Plotting inequalities (1), (2), and (3) on the X-Y axis gives the following graph:

• The yellow shaded area represents the bounded feasible region. 
• Now, it is needed to determine the corner points of the feasible region. It will be done one by one. We will refer to the figure to find each corner point.
• One point is the intersection of line 4x + 2y = 12 with y-axis where x = 0. At x = 0, y is equal to 6. Hence the corner point is (0, 6).
• One point is the intersection of line 2x + 4y = 6 with a y-axis where x = 0. At x = 0, y is equal to 3/2. Hence the corner point is (0, 3/2).
• Final point is the intersection point of lines 4x + 2y = 12 and 2x + 4y = 6. Solving the two equations. we get x = 3 and y = 0. The final corner point is (3, 0).
• Below is the table showing different corner points along with the value of the objective function at those points:

• Corner Point	Value of the objective function Z = 4x + 3y
  (0, 6)	18
  (0, 3/2)	9/2
  (3, 0)	12

• Clearly, the value of Z is maximum at (0, 6).
• The maximum value is 18.
• So, the correct answer is option 1.  

Additional Information

Examples for all possible results in LP using the Graphical method:

• The LP problem with no region satisfying all the constraints is shown below graphically:          

• Subsequently, a feasible and bounded region (closed) is shown in the figure below:

• The feasible and unbounded region is shown in the graph below. Here the value of y and x has no bound unlike the previous examples:

• When the feasible region is obtained, first the corners of the same are determined.
• Subsequently, the value of the objective function Z is calculated at all the points.
• The point where the objective function attains maximum or minimum value is the optimal solution of the problem.  
• If the feasible bounded region has been formed, then for this case, the objective function has both maximum and minimum value at a corner point of the given feasible region.
• However, if the region is feasible and unbounded, then things become a little complicated. Here also, the objective function will have a maximum and minimum value at a corner point of the feasible region. But to check whether Z has maximum or minimum values in this unbounded region case, we have to draw this region also.

Concept:

Two matrices Am × n and Bp × q 

if AB and BA are defined then p = n and q = m

Calculation:

Given:

O(A) = 2 × 3, O(B) = 3 × 2, O(C) = 3 × 3

⇒ O(B') = 2 × 3 

⇒ O(A + B') = 2 × 3

⇒ C(A + B') is not defined as the number of rows of c ≠ number of columns of (A + B'). 

\(\left( {\mathop \sum \limits_{i = 1}^n {x_i}{p_i}} \right) = 0.2 + 0.8 + \;0.9 + 0.4 = 2.3\)

\(\left( {\mathop \sum \limits_{i = 1}^n x_i^2{p_i}} \right) = 0.2 + 1.6 + \;2.7 + 1.6 = 6.1\)

\(var\left( X \right) = {\sigma ^2} = \;\mathop \sum \limits_{i = 1}^n x_i^2{p_i} - {\left( {\mathop \sum \limits_{i = 1}^n {x_i}{p_i}} \right)^2}\)

var(X) = σ² = 6.1 – 5.29

var(X) = σ² = 0.81

CONCEPT:

In general, the determinant of a square matrix can be evaluated along any row (column) and it is equal to the sum of the product of elements in any row and their corresponding co-factor.

i.e \(\left| A \right| = \;\mathop \sum \limits_{j = 1}^n {a_{ij}}\;{C_{ij}},\;for\;some\;i \in N\), where Cij is the co-factor.

CALCULATION: 

Given:

\(\left| {\begin{array}{*{20}{c}} a&{2b}&{2c}\\ 3&b&c\\ 4&a&b \end{array}} \right| = 0\)

\( ⇒ \left| {\begin{array}{*{20}{c}} {a - 6}&0&0\\ 3&b&c\\ 4&a&b \end{array}} \right| = 0\left| {R{}_1 \to {R_1} - 2{R_2}} \right|\)

\(⇒ \) (a - 6)(b² - ac) = 0

⇒ b² - ac = 0,      [∵ a ≠ 6]

∴ ac = b² 

⇒  abc = b³ .

Concept

Lagrange mean value theorem (LMVT)

Let f(x) be a function defined in [a, b] such that

f (x) is continuous in [a, b] and differentiable in (a, b)

Then there exists at least one point such that c ∈ (a, b) such that

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)

Calculation

Given: f(-7) = - 3 and f'(x) ≤ 2

Using LMVT in [-7, -1]

\(\frac{{f\left( { - 1} \right) - f\left( { - 7} \right)}}{{ - 1 - \left( { - 7} \right)}} \le 2\)

\(f\left( { - 1} \right) - f\left( { - 7} \right) \le 12\)

\(⇒ f\left( { - 1} \right) \le 9\)      ----(i)

Using LMVT in [-7, 0]

\(⇒ \frac{{f\left( 0 \right) - f\left( { - 7} \right)}}{{0 - \left( { - 7} \right)}} \le 2 \)

\(⇒ f\left( 0 \right) - f\left( { - 7} \right) \le 14\)

\(⇒ f\left( 0 \right) \le 11\)        ---(ii)

From (i) and (ii),

⇒ f(0) + f(-1) ≤ 20

Concept:

Let y = f(x) be the equation of a curve, then slope of the tangent at any point say (x1, y1) is given by: 

\(m = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;\;{y_1}} \right)}}\).

Slope of normal at any point say (x1, y1) is given by:

 \(\frac{{ - 1}}{{Slope\;of\;tangent\;at\;point\;\left( {{x_1},\;{y_1}} \right)}} = \; - {\left[ {\frac{{dx}}{{dy}}} \right]_{\left( {{x_1},\;{y_1}} \right)}}\)

Equation of tangent at any point say (x1, y1) is given by:

 \(y - {y_1} = {\left[ {\frac{{dy}}{{dx}}} \right]_{\left( {{x_1},\;{y_1}} \right)}} ⋅ \left( {x - {x_1}} \right)\)

Equation of normal at any point say (x1, y1) is given by: 

\(y - {y_1} = \; - {\left[ {\frac{{dx}}{{dy}}} \right]_{\left( {{x_1},\;{y_1}} \right)}} ⋅ \left( {x - {x_1}} \right)\)

Calculation:

Given: y(x - 2)(x - 3) = x + 6

Intersection with y-axis

⇒ x = 0 in the given curve to get,

⇒ y = 1

⇒  The point of intersection is (0, 1)

Now, \(y = \frac{{x + 6}}{{{x^2} - 5x + 6}}\)

\(y' = \frac{{\left( {{x^2} - 5x + 6} \right) - \left( {x + 6} \right)\left( {2x - 5} \right)}}{{{{\left( {{x^2} - 5x + 6} \right)}^2}}}\)

\(y' = \frac{{6 - \left( { - 30} \right)}}{{36}} = 1\,at\,\left( {0,1} \right)\)

∴ Equation of normal is given by

⇒ (y - 1) = -1(x - 0)

⇒ x + y - 1 = 0.

⇒ Since, \(\left( {\frac{1}{2},\frac{1}{2}} \right)\) is satisfying the equation of the normal so the correct answer is option 2

Concept

Let the system of equations be,

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

\(\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} {{d_1}}\\ {{d_2}}\\ {{d_3}} \end{array}} \right]\)

⇒ AX = B

⇒ X = A-1 B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}\;B\)

⇒ If det (A) ≠ 0, the system is consistent having unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

Calculation:

Given:  C11 = 7, C12 = -7, C13 = 7, C21 = 6, C22 = - 6, C23 = 6, C31 = 5, C32 = -5, C33 = 5

Also, a + b - 2c = 0, 2a - 3b + c = 0 and a -5b + 4c = α

\( \Rightarrow {\rm{A}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{1}}&1&-2\\ 2&{\rm{-3}}&1\\ 1&-5&{\rm{4}} \end{array}} \right],{\rm{B}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{a}}\\ {\rm{b}}\\ {\rm{c}} \end{array}} \right]{\rm{and\;C}} = \left[ {\begin{array}{*{20}{c}} 0\\ {\rm{0}}\\ {{{\rm{α}}}} \end{array}} \right]\)

\(⇒ |A| = \left| {\begin{array}{*{20}{c}} 1&1&{ - 2}\\ 2&{ - 3}&1\\ 1&{ - 5}&4 \end{array}} \right| = 0\)

⇒ If det (A) = 0 and (adj A). B = O, the system is consistent, with infinitely many solutions.

Cofactors are given in the question,

⇒ Adj (A) = \( {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{7}}&-7&7\\ 6&{\rm{-6}}&6\\ 5&-5&{\rm{5}} \end{array}} \right]^{T}\)

⇒ Adj (A).B =  \( {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{7}}&6&5\\ -7&{\rm{-6}}&-5\\ 7&6&{\rm{5}} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} 0\\ {\rm{0}}\\ {{{\rm{α}}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5α \\ {\rm{-5α}}\\ {{{\rm{5α}}}} \end{array}} \right]\)

• For, (adj A). B to be zero, α  = 0 hence, the value of α is zero

Concept:

Here, we have to use the multiplication, addition, and subtraction properties of the matrixes.

Calculation:

Given: 2A + 3B = \(\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&4\\ 3&2&5 \end{array}} \right]\)

and A + 2B = \(\left[ {\begin{array}{*{20}{c}} 5&0&3\\ 1&6&2 \end{array}} \right]\)

⇒ 2A + 4B = \(\left[ {\begin{array}{*{20}{c}} {10}&0&6\\ 2&{12}&4 \end{array}} \right]\)

Solving (i) and (ii), we get

- B = \(\left[ {\begin{array}{*{20}{c}} -8&-1&{ -2}\\ { 1}&{-10}&{ 1} \end{array}} \right]\) 

⇒ B = \(\left[ {\begin{array}{*{20}{c}} 8&1&2\\ { - 1}&{10}&{ - 1} \end{array}} \right]\)

Concept:

Area between Two Curves: Let curves are f(x) and g(x)

\({\rm{Area}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} \left[ {{\rm{f}}\left( {\rm{x}} \right) - {\rm{g}}\left( {\rm{x}} \right)} \right]{\rm{dx}} = \mathop \smallint \nolimits_{\rm{a}}^{\rm{b}} \left[ {{\rm{Top}} - {\rm{bottom}}} \right]{\rm{dx}}\)

Calculation:

\({\rm{y}} = \left| {\rm{x}} \right| = \left\{ {\begin{array}{*{20}{c}} { - x,\;x < 0}\\ {x,\;x \ge 0} \end{array}} \right.\)

y = x² and y = ± x solving these two equation, we get intersection points

⇒ y = x² = ± x

⇒ x² ± x = 0

⇒ x(x ± 1) = 0

∴ x = 0, ± 1

Put the value of x in y = x²

The point of intersection of the given curves are (0, 0), (-1, 1) and (1, 1)

The area bounded by the curves y ≥ x² and y = |x| is represented by the shaded region.

It is clearly observed that the required area is symmetrical about y- axis.

∴ Required area = Area between parabola (y = x²) and line (y = x) between limits x = 0 and x = 1

\({\rm{Area}} = 2 \times \mathop \smallint \nolimits_0^1 \left[ {{\rm{Top}} - {\rm{bottom}}} \right]{\rm{dx}} = 2 \mathop \smallint \nolimits_0^1 \left[ {{\rm{x}} - {{\rm{x}}^2}} \right]{\rm{dx}}\)

\(= 2\left[ {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3}} \right]_0^1 = 2\left[ {\frac{1}{2} - \frac{1}{3}} \right] = \frac{1}{3}{\rm{sq}}.{\rm{\;unit}}\)

Concept:

• There are three main components of linear programming:

• Decision Variables: These variables are the activities that share the available resources while also competing with one another. In other words, these are interrelated in terms of resource utilization and need to be solved simultaneously. They are assumed to be non-negative and continuous.  
• The Objective Function: Each linear programming problem is aimed to have an objective to be measured in quantitative terms such as profits, sales, cost, time, or some other parameter, etc which needs to be maximized or minimized. The objective function will vary depending upon the business requirements or other factors.
• Constraints: These represent real-life limitations such as money, time, labor, or other restrictions and the objective function needs to be maximized or minimized while satisfying the constraints. They are represented as linear equations or inequations in terms of the decision variables.
• For example, suppose x and y are the decision variables. The objective function will be given by:

Z = ax + by ….(1)

• Where a and b are constants and Z is the function to be maximized or minimized. There will be conditions x ≥ 0, and y ≥ 0, which indicates the non-negative constraints on the decision variable.

Explanation: 

• As per the question, the words are clearly mentioning about the decision variable.
• Hence, the correct answer is option (1).  

Concept:

Dot product of two perpendicular lines is zero.

Distance between two points (x1, y1, z1) and (x2, y2, z2) is given by, \(\rm \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

Calculation:

Let M be the foot of perpendicular drawn from the point P(2, 3, 4)

Let, \(\rm \dfrac{x-0}{1}=\dfrac{y-0}{0}=\dfrac{z-0}{0} =k\)

x = k, y  = 0, z = 0

So M = (k, 0, 0)

Now direction ratios of PM = (2 - k, 3 - 0, 4 - 0) = (2- k, 3, 4) and direction ratios of given line are 1, 0, 0

PM is perpedicular to the given line so,

(2 - k) (1) + 3(0) + 4 (0) = 0

∴ k = 2

M = (2, 0, 0)

Perpendicular distance PM =

 \(\rm \sqrt {(2-2)^2+(0-3)^2+(0-4)^2}\\ =\sqrt{9+16}\\ =5\)

Hence, option (2) is correct. 

Concept:

Here, we have to use the multiplication properties of the matrix and the properties of determinants,

| A| = |kA|  = (k)ⁿ||A| 

Where n is the order of determinant.

Also, | A.B| = |A| .|B| 

Calculation:

 A2 + I = 0

⇒ A2 = - I

⇒ | A2| = |- I|  = (-1)²||I| = 1

⇒ | A.A| = |A| .|A| = 1

⇒ |A| = ± 1

Now, check the options,

Only for option 2,

⇒ |A| = i²

⇒ |A| = ± 1

∴ A² + 1 = 0, if A = \(\left[ {\begin{array}{*{20}{c}} {\rm{i}}&{\overline {\rm{v}} }\\ {\rm{0}}&{\rm{i}} \end{array}} \right]\)

CONCEPT:

The inverse of a matrix: The Inverse of an n × n matrix is given by:

\({A^{ - 1}} = \frac{{adj\left( A \right)}}{{\left| A \right|}}\) where adj(A) is called an adjoint matrix.

Adjoint Matrix: If Bn× n is a cofactor matrix of matrix An× n then the adjoint matrix of An× n is denoted by adj(A) and is defined as BT. So, adj(A) = BT.

CALCULATION:

Given: A-1 = x A + y I

\(\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 5}&1 \end{array}} \right]\therefore {A^{ - 1}} = \frac{{adjA}}{{\left| A \right|}} =\frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 5&1 \end{array}} \right]\\ \end{array}\)

\( ⇒ \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} 1&2\\ { - 5}&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} x&{2x}\\ { - 5x}&x \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} y&0\\ 0&y \end{array}} \right]\)

\( ⇒ x + y = \frac{1}{{11}},2x = \frac{{ - 2}}{{11}} \).

\(⇒ x = \frac{{ - 1}}{{11}},y = \frac{2}{{11}}\)

: \(\smallint \frac{{{x^2}}}{{{x^2}\; + \;4}}\) dx = \(\smallint \frac{{{x^2} + \;4\;-\;4}}{{{x^2}\; + \;4}}\) dx = \(\smallint \left( {1 - \frac{4}{{{x^2}\; + \;4}}} \right)\) dx = \(\smallint {\rm{dx}} - 4\smallint \frac{{{\rm{dx}}}}{{{{\rm{x}}^2}{\rm{\;}} + {\rm{\;}}{2^2}}}\)

Concept:

The shortest distance between the lines \(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\)  and \(\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}\) is given by:\(d= \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} & c_{1}\\ a_{2}& b_{2} & c_{2} \end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2}+(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}\)

Calculation:

Here we have to find the shortest distance between the lines ​​\(\frac{x}{-1}=\frac{y-2}{0}=\frac{z}{1}\) and \(\frac{x+2}{1}=\frac{y}{1}=\frac{z}{0}\)

Let line L₁ be represented by the equation \(\frac{x}{-1}=\frac{y-2}{0}=\frac{z}{1}\) and line L₂ be represented by the equation \(\frac{x+2}{1}=\frac{y}{1}=\frac{z}{0}\)

⇒ x1 = 0, y1 = 2, z1 = 0  and a1 = -1, b1 = 0, c1 = 1.

⇒ x2 = -2, y2 = 0, z2 = 0  and a2 = 1, b2 = 1, c2 = 0.

∵ The shortest distance between the lines is given by:  \(d= \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} & c_{1}\\ a_{2}& b_{2} & c_{2} \end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2}+(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}\)

⇒ \(d= \frac{\begin{vmatrix} -2-0 &0-2&0-0 \\ -1& 0 & 1\\ 1& 1 & 0 \end{vmatrix}}{\sqrt{(0-1)^{2}+(0-1)^{2}+(-1-0)^{2}}}\)    

⇒ \(d= \frac{\begin{vmatrix} -2 &-2&0 \\ -1& 0 & 1\\ 1& 1 & 0 \end{vmatrix}}{\sqrt{1+1+1}}\)

⇒ d = 0

Hence, option 4 is correct.

Concept:

The area between the curves y1 = f(x) and y2 = g(x) is given by:

Area enclosed = \(\rm \left|\int_{x_1}^{x_2}(y_1-y_2)dx\right|\)

Where, x1 and x2 are the intersections of curves y1 and y2 

Calculation:

Given

Curve 1: y = sinx = f(x) (say)

Curve 2: Lines \(\rm x = -\frac {π} {3} \) and \(\rm x = \frac {π} {3} \)

According to the figure the sum of area curve OAB and curve OCD

Here OAB and OCD are equal and limit 0 to \(\rm \frac {π} {3} \)

So, Area = 2 × area of OAB

Now the required area (A) is 

Area of OAB = \(\rm \left|\int_{x_1}^{x_2}f(x)dx\right|\)

= \(\rm \left|\int_{0}^{\frac {\pi}{3}}sindx\right|\)

= \(\rm \left|\left[{-cosx}\right]_{0}^{\frac{\pi}{3}}\right|\)

= \(\left| {- \cos \frac{\pi}{3} + \cos 0^{\circ}} \right|\)

= \(\rm |{1 - \frac {1}{2}}| = \frac {1}{2}\)

∴ Shaded  Area = 2 × \(\rm 1\over2\) = 1

Additional Information

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C 

Concept:

For Integration with modulus, first we have to find the point where the sign of the value of the function gets change.

Calculation:

Given:

\(\mathop \smallint \limits_0^1 \left| {5x - 3} \right|dx\)

f(x) = 5x - 3 = 0

x = 3/5

∴ from 0 to 3/5 the function is negative and 3/5 to 1 the function is positive.

\(\mathop \smallint \limits_0^1 \left| {5x - 3} \right|dx = - \mathop \smallint \limits_0^{\frac{3}{5}} \left( {5x - 3} \right)dx + \mathop \smallint \limits_{\frac{3}{5}}^1 \left( {5x - 3} \right)dx\)

\( = \left( { - \frac{5}{2}{x^2} + 3x} \right)_0^{\frac{3}{5}} + \left( {\frac{{5{x^2}}}{2} - 3x} \right)_{\frac{3}{5}}^1\)

\( = \left( { - \frac{9}{{10}} + \frac{9}{5}} \right) + \left[ {\left( {\frac{5}{2} - 3} \right) - \left( {\frac{9}{{10}} - \frac{9}{5}} \right)} \right]\)

\( = \frac{9}{{10}} + \left( {\frac{{ - 1}}{2} + \frac{9}{{10}}} \right) = \frac{{13}}{{10}}\)

Concept:

Probability of an event happening = (Number of ways it can happen) / (Total number of outcomes)

A die is thrown: Sample space S = {1, 2, 3, 4, 5, 6}

Two events are said to be mutually exclusive events when both do not have any common element.

Calculation:

A die is thrown:

S = {1, 2, 3, 4, 5, 6}

Now, A is the event that an even number occurs,

A = {2, 4, 6}

B is the event that an odd number occurs

B = {1, 3, 5}

And C the event that a number greater than 3 occurs

C = {4, 5, 6}

Here, A and B have no element in common so, these events are mutually exclusive.

Hence, option (2) is correct.

Concept:

The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral 

​\(\rm \displaystyle \left|\int_a^b f(x)\ dx\right|\)

This is for curves that are entirely on the same side of the x-axis in the given range.

If the curves are on both sides of the x-axis, then we calculate the areas of both sides separately and add them.

Definite integral: If ∫ f(x) dx = g(x) + C, then \(\rm \displaystyle \int_a^b f(x)\ dx = [ g(x)]_a^b=g(b)-g(a).\)

\(\rm \displaystyle \int x^n\ dx = \dfrac{x^{n+1}}{n+1}+C\).

Calculation:

\(\rm \displaystyle \int x^4\ dx = \dfrac{x^5}{5}+C\).

Using the above concept for area of a curve, we can say that the required area is:

\(\rm I=\displaystyle \int_1^5 x^4\ dx\)

\(\rm = \left [\dfrac{x^5}{5}\right ]_1^5\)

\(\rm =\dfrac{5^5}{5}-\dfrac{1^5}{5}\)

\(\rm =\dfrac{3125-1}{5}\)

\(\rm =\dfrac{3124}{5}\).

Concept: 

\(\rm \frac{ydx-xdy}{y^2} = d\left(\frac{x}{y} \right )\)

ydx + xdy = d(xy)

Calculation:

Given:  ydx + (x - y)dy = 0

⇒ydx + xdy - ydy = 0

⇒ydx + xdy = ydy

⇒ d(xy) = ydy

Integrating both sides, we get

\(\rm ⇒ \int d(xy) =\int ydy\)

\(\rm ⇒ xy = \frac{y^2}{2} + c\)                      .... (1)

Curve passing through the point (1, 1)

⇒ 1 = \(\frac 1 2\) + c

∴ c = \(\frac 1 2\)

Put the value of c in equation 1st we get,

\(\rm ⇒ xy = \frac{y^2}{2} + \frac 1 2\)

⇒ 2xy = y² + 1

Concept: 

If a differential equation in the form of \(\frac{{{\rm{dx}}}}{{{\rm{dy}}}} + {\rm{Px}} = {\rm{Q}}\),

• Then we solve by calculating integrating factor, I.F. = \({{\rm{e}}^{\smallint {\rm{P\;dy}}}}\)
• The solution will be \({\rm{x}}.{\rm{}}\left( {{\rm{I}}.{\rm{F}}.} \right){\rm{}} = {\rm{}}\smallint \left( {{\rm{Q}}.\left( {{\rm{I}}.{\rm{F}}.} \right)} \right){\rm{dy}} + {\rm{c}}\).

Formula used:

e^log_ex = x

\(\frac{d}{dx}log\ x\ =\ \frac{1}{x}\)

\({\int \frac{1}{x}dx}\ =\ log\ x\ +\ C\)

Calculation:

Given that

\(\frac{dy}{dx}(xlogx)+y=2log x\)

⇒ \(\frac{dy}{dx}\ + \ \frac{1}{(xlogx)}y \ =\ \frac{2}{x}\)

Here, P = \(\frac{1}{x\ log\ x}\) and Q = \(\frac{2}{x}\)

Therefore, integrating factor

I.F. = \({{\rm{e}}^{\smallint {\rm{P\;dy}}}}\)

⇒ I.F = \({{\rm{e}}^{\smallint {\rm{\frac{1}{x\ log\ x}dx}}}}\)

log x = t    ---(1)

⇒ (1/x)dx = dt

⇒ I.F = \(e^{\int \frac{1}{t}dt}\)

⇒ I.F = e^log t   [∵ elogex = x]

⇒ I.F = t 

⇒ I.F = log x      [From equation (1)]

We compute the area limited by the x-axis, the ordinates, x = p, x = q and the given curve.

Given curve is, xy = m² which is \(y = \frac{{{m^2}}}{x}\)

\(Area,A = \mathop \smallint \limits_q^p y\;dx\)           Since p > q

\(A = \mathop \smallint \limits_q^p \frac{{{m^2}}}{x}dx\)

\(A = {m^2}\mathop \smallint \limits_q^p \frac{1}{x}dx\)

\(Area = {m^2}\left[ {\log x} \right]_q^p\;\smallint \frac{{dx}}{x} = \log x\)

Applying the lower and upper limits,

\(\begin{array}{l}Area = {m^2}\left[ {\log p - \log q} \right]\\Area = {m^2}log\left( {\frac{p}{q}} \right)\end{array}\)

Concept:

• A critical point x₀ (a point where either f'(x₀) = 0 or f'(x₀) does not exist) is a point of local maxima or local minima in accordance with either f''(x₀) < 0 or  f''(x0) > 0.

Calculation:

Given:

\(f(x)=x^3-3 \log x\)

• Differentiating with respect to x,

\(f'(x)=3x^2-\frac{3}{x}\)

• For maxima or minima,

\(f'(x)=0\)

\(\Rightarrow 3x^2-\frac{3}{x}=0\)

\(\Rightarrow 3x^3-3=0\)

\(\Rightarrow x^3-1=0\)

∴ \(x=1\)

• The second derivative is given by, \(f''(x)=6x+\frac{3}{x^2}\)
• Then \(f''(1)=6(1)+\frac{3}{1^2}=9>0\)
• So x = 1 is a point of local minima.
• Therefore option 4 is correct.

Concept:

• A function f has a critical point at x=x₀ if either f'(x₀) does not exist or  f'(x0) vanishes i.e.  f'(x0) = 0.

Calculation:

Given: f(x) = a tan -1x + 2b log (1 + x) + x + 1

• Differentiating with respect to x,

⇒ \(f'(x)=\frac{a}{1+x^2}+\frac{2b}{1+x}+1\) for all x > - 1

• Since x = 0 and x = 2 critical points.

f'(0) = 0

⇒ a + 2b + 1 = 0      ....(1)

f'(2) = 0

 \(⇒ \frac{a}{5}+\frac{2b}{3}+1=0\)

⇒ 3a + 10b + 15=0      ....(2)

• Now, check by options,

a = 5 and b = -3

• Therefore option 3 is correct.

Alternate Method 

• Let us solve (1) and (2),
• Multiplying (1) by 3 and subtracting it from (2), we get,

(3a +10b +15) - (3a + 6b +3) = 0

⇒ 3a + 10b +15 - 3a -6b -3 =0

⇒ 4 b +12 = 0

⇒ b + 3 =0

∴ b = -3

Put this value in (1),

⇒ a - 6+1=0

∴ a = 5

• Therefore option 3 is correct.

Concept:

Scalar Triple Product: 

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\), 

\(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) and,

\(\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\)

Then their scalar triple product is defined as,

\(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

Calculation:

Given:

We have,

\(\rm \mathop v\limits^ \to = 2\hat i + \hat j - \hat k\)

\(\rm \mathop w\limits^ \to = \hat i + 3\hat k\)

\(\rm \vec v \times \vec w = \begin{bmatrix} \hat i & \hat j & \hat k \\\ 2 & 1 & -1 \\\ 1 & 0 & 3 \end{bmatrix}\)

\(\Rightarrow \vec v \times \vec w = \hat i (3) - \hat j (7) + \hat k (-1)\)

\(\rm \Rightarrow \vec v \times \vec w = 3\hat i - 7\hat j - \hat k \)

Now, \(\rm \vec u . (\vec v \times \vec w) = | \vec v| (3\hat i - 7\hat j - \hat k ) \cos θ \)

When, θ = 0°

Maximum value = \(\sqrt{3^2 +(-7)^2 + (-1)^2 } = \sqrt{59}\)

CONCEPT:

• Projection of a vector \(\vec a\) on other vector \(\vec b\) is given by: \(\vec a \cdot \hat b = \frac{\vec a \cdot \vec b}{|\vec b|}\)

CALCULATION:

Given: \(\vec a = 2\hat i + 3\hat j + 2\hat k\) and \(\vec b = \vec i + 2\vec j + \hat k\)

Here, we have to find the projection of a vector \(\vec a\) on other vector \(\vec b\) is given by: \(\vec a \cdot \hat b = \frac{\vec a \cdot \vec b}{|\vec b|}\)

⇒ \(\vec a \cdot \vec b = 2 + 6 + 2 = 10 \ and \ |\vec b| = \sqrt {6}\)

⇒ \(\vec a \cdot \hat b = \frac{10}{\sqrt {6}} = \frac{5\sqrt6}{3}\)

Hence, option 3 is correct.

Concept:

Let a pair of linear equation in two variable a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

The condition of parallel lines or inconsistent equations.

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{c_1}{c_2}\)

Calculation:

Given equation of lines

3x + 2ky = 2 and 2x + 5y + 1 = 0

a1 = 3; a2 = 2

b1 = 2k; b2 = 5

c1 = -2; c2 = 1

Here, given lines are parallel

So that,

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{c_1}{c_2}\)

\(\therefore \frac{3}{2} = \frac{2k}{{5}}\left( {\because{c_1} \ne {c_2}} \right)\)

∴ \(k=\frac{15}{4}\)

Additional Information

(I) If \(\frac{{{a_1}}}{{{a_1}}} \ne \frac{{{b_1}}}{{{b_2}}}\)

Then the graph will be a pair of lines interesting at a unique point. Which is the solution of the pair of equations.

(II) If 

\(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{c_1}{{{c_2}}}\) then

Then graph will be a pair of coincident lines

CONCEPT:

To find the image of a point in a given plane, following steps should be followed

• The coordinates of the image Q are (x₁ + ar, y₁ + br, z₁ + cr).
• The coordinates of the mid-point R of line PQ is found.
• The value of r is obtained by substituting the coordinates of R in the plane equation.

Finally, the value of r is substituted in the coordinates of Q

CALCULATIONS:

Let Q is image of point P (-2, 1, 1)

Given plane is x + y + z = 0.

So, PQ will be normal to the given plane, so direction ratios of PQ will be proportional to (1, 1, 1).

∵ PQ passes from (-2, 1, 1) and have direction ratios (1, 1, 1).

∴ Equation of line PQ \( = \;\frac{{x + 2}}{1} = \frac{{y - 1}}{1} = \frac{{z - 1}}{1} = r\;\left( {say} \right)\)

So, point Q in the form of r is (r – 2, r + 1, r + 1)

Let R is the middle point of PQ, so by mid-point formula

\(R = \left( {\frac{{r - 4}}{2},\;\frac{{r + 2}}{2},\;\frac{{r + 2}}{2}} \right)\)

Since R lies on the plane x + y + z = 0.

∴ \(1.\left( {\frac{{r - 4}}{2}} \right) + 1.\left( {\frac{{r + 2}}{2}} \right) + 1.\left( {\frac{{r + 2}}{2}} \right) = 0\)

⇒ r = 0

Now substituting the value of r in Q coordinates, i.e. (r - 2, r + 1, r + 1)

Concept:

• The principal value of cos-1 x or cos-1 y or cos-1 z or cos-1t lies in [0, π]

Calculation:

Given:

cos^-1 x + cos^-1 y + cos^-1 z + cos^-1 t = 4 π      ....(1)

• According to the concept used,
• We know that the maximum value of cos-1 x is π so  equation 1 is only possible when

cos^-1 x = cos^-1 y = cos^-1 z = cos^-1 t = π,

∴ x = y = z = t = cos π = -1

∴ x² + y² + z² + t² = 4.  

Concept:

Here, we have to use Inverse trigonometry formulas:

tan-1 (- x) = - tan-1 (x)      [Since tan-1 (x) is a odd function]

Calculation:

According to the concept used,

⇒ cos (2 tan^-1 (-7)) = cos (-2 tan^-1 7) = cos (2 tan ^-1 7)

Let tan^-1 7 = θ

∴ tan θ = 7

⇒ cos θ = \(\frac{1}{{\sqrt {50} }}\)

Now, cos (2 tan^-1 7) = cos 2θ

⇒ cos (2 tan-1 7) = 2 cos² θ - 1 = 2 × \({\left( {\frac{1}{{\sqrt {50} }}} \right)^2} - 1\)

⇒ cos (2 tan-1 7) = \( = \frac{{ - 48}}{{50}} = \frac{{ - 24}}{{25}}\)