Mathematics Mock Test - 4

Concept:

If a matrix has ‘m’ rows and ‘n’ columns, its order is said to be m × n (read as ‘m by n’).

\(\left[ {\begin{array}{*{20}{c}} 1&{ - 5}&{89} \end{array}} \right]\) is of order 1 × 3;

\(\left[ {\begin{array}{*{20}{c}} 2&{54}&{23}\\ {56}&{89}&{123} \end{array}} \right]\) is of order 2 × 3;

Extra concepts:

Matrix is an ordered rectangular arrangement of numbers (real or complex) or functions.

Usually, a matrix is denoted by a capital letter, such as A, B, C, D, M, N, X, Y, Z, etc.

Matrix elements

 Consider the matrix below, in which matrix elements are represented entirely by symbols.

\(\left[ {\begin{array}{*{20}{c}} A_{11}&A_{12}&A_{13}\\ A_{21}&A_{22}&A_{23} \end{array}} \right]\)

The second element in the first row is represented by A₁₂. And so on, until we reach the fourth element in the second row, which is represented by A₂₄.

NOTE:

A matrix of the order m × n has mn elements.

Hence, if the number of elements in a matrix is prime, it must have one row or one column.

Concept:

\(\rm\int \frac{dx}{x}=\ln x+c\)

Logarithm properties: 

\(\rm {\log _a}\left( {mn} \right) = \;{\log _a}m + \;{\log _a}n\)

\(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\)

\(\rm {\log _a}{m^n} = n{\log _a}m\)

Calculation:

xdy - 2ydx = 0

⇒  xdy = 2ydx

\(\rm⇒ \frac{dy}{y}=2\frac{dx}{x}\)

Integrating both sides, we get

\(\rm⇒ \int \frac{dy}{y}=2\int \frac{dx}{x}\)

\(\rm⇒ \ln y=2 \ln x+\ln c\)

\(\rm⇒ \ln y=\ln x^{2}+\ln c\)

\(\rm⇒\ln y-\ln x^{2}=\ln c\)

\(⇒\rm \ln\frac{y}{x^{2}}=\ln c\)

\(\rm \therefore y=x^{2}c\)

\(\therefore\) option 3 is correct

Concept:

Rules for Operations on Inequalities:

• Adding the same number to each side of an inequality does not change the direction of the inequality symbol.
• Subtracting the same number from each side of an inequality does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.
• Dividing each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

Calculation:

Given:

\(\frac{4}{x}<\frac{1}{3}\)

\(\Rightarrow \frac{4}{x}-~\frac{1}{3}<0\)

\(\Rightarrow \frac{12-x}{3x}<0\)

\(\Rightarrow \frac{-~\left( x-12 \right)}{3x}<0\)

Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.

\(\Rightarrow \frac{\left( x-12 \right)}{3x}>0\), Here x ≠ 0

∴ x < 0 OR x > 12

Concept:

Calculation:

\(\rm \vec a \times (\vec b \times \vec a)\)is coplanar with Both \(\rm \vec a\) and \(\rm \vec b\)

Let, \(\rm\vec z = \vec b \times \vec a\), so, z is in perpendicular plane of both a and b

Also assume, \(\rm\vec c = \vec a \times \vec z\), so c will be in perpendicular plane of a and z

∴ Vector c or \(\rm \vec a \times (\vec b \times \vec a)\)is coplanar with both a nor b.

Hence, option (4) is correct.

Concept:

• Order: The order of a differential equation is the order of the highest derivative appearing in it.
• Degree: The degree of a differential equation is the power 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: \(\rm {\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2} + \sin \left( {\frac{{dy}}{{dx}}} \right)} \right]^{3/4}} = \frac{{{d^2}y}}{{d{x^2}}}\)

⇒ \(\rm \left[{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2} + \sin \left( {\frac{{dy}}{{dx}}} \right)} \right]^{3/4}}\right]^4 = \left[\frac{{{d^2}y}}{{d{x^2}}}\right]^4\)

⇒ \(\rm \left[1+(\frac{dy}{dx})^{2} + \sin (\frac{dy}{dx})\right ]^3 = \left[\frac{d^2y}{dx^2} \right ]^4\)

The obtained equation has the highest derivative 2 so the order is two.

The degree is not defined because the equation is not a polynomial.

Concept:

• The standard form of a linear equation of the first order is given by \(\frac{{dy}}{{dx}} + Py = Q\) where P, Q are arbitrary functions of x.
• The integrating factor of the linear equation is given by \(I.F. = {e^{\smallint pdx}}\)
• The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

Calculation:

\(\frac{{dx}}{{dy}} + \frac{x}{{co{s^2}y}} = {e^{ - \tan y}}\)

It is the linear equation of the form of \(\frac{{dx}}{{dy}} + Px = Q\)

\(I.F. = {e^{\smallint pdy}}\)

\(I.F. = {e^{\smallint se{c^2}ydy}} = {e^{\tan y}}\)

• So, the correct answer is option 1.

Concept:

• You can take common k from each and every element of the matrix,

Calculation:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 2&0&0\\ 0&2&0\\ 0&0&2 \end{array}} \right]\)

A = 2 \( \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\)

⇒ A = 2I

⇒ A² = 2I × 2I  = 4 I2 

∵  I2  = I as I is a identity matrix.

⇒ A2 = 4I

⇒ A2A2 = A⁴ = 4I × 4I= 16I 

⇒  A⁵ = 16A

Alternate Method

\(A = \left[ {\begin{array}{*{20}{c}} 2&0&0\\ 0&2&0\\ 0&0&2 \end{array}} \right]\)

\(⇒ {A^5} = \left[ {\begin{array}{*{20}{c}} {{2^5}}&0&0\\ 0&{{2^5}}&0\\ 0&0&{{2^5}} \end{array}} \right] = {2^4}\left[ {\begin{array}{*{20}{c}} 2&0&0\\ 0&2&0\\ 0&0&2 \end{array}} \right] = 16A.\)

Concept:

• Given two positive integers X and Y such that X > Y and Y ≠ 0, there will be two unique integers Q and R such that the following can be written:

X = Y × Q + R ….(1)

Where, R = Remainder, 0 ≤ R < Y

• The same expression can be written as shown below:

X mod Y = R ….(2)

Where mod is the modulo operator which gives the remainder after X is divided by Y.

Calculation:

Given: 15th May is Sunday.

• Number of days in a week = 7 days
• Difference of days between 15th May and 15th August including 15th August = Number of remaining days in July + Days up to 15th August = 16 +15 = 31

∴ (31) mod 7 = 3 (∵ 31 days = 4 weeks and 3 day) ….(3)

∵ 15^th May is Sunday. so from equation (3), 15^th August will be 1 weekday after Sunday, that is, Wednesday.

Given:

A tap can fill the tank in 8 hours

Concept:

Work = 1/Efficiency

Calculation:

⇒ One hour work of one tap = 1/8

⇒ One hour work of 4 taps = 4/8 = 1/2

⇒ Total time to fill the tank = 4 hours to first half + 1 hour to second half

∴ The required result will be 5 hours.

Concept:

• The decrease in the value of the assets such as building machinery and equipment of all kinds is called depreciation. 

Linear method of calculating depreciation:

• The linear method of depreciation is the simplest and the most widely used method to calculate the depreciation for fixed assets.
• According to this method the annual depreciation D is given by,

\(D=\frac{C-S}{n}\)

Where C is the original cost of the asset, S is the estimated scrap value of residual value (the value of a depreciable asset at the end of its useful life ) and n is the useful life in years.

Calculations:

Given: C = Rs 25000, S = Rs 5000, n = 8 Years

• Thus the annual depreciation is given by,

\(\Rightarrow D=\frac{C-S}{n}=\frac{25000-5000}{8}\)

\(\Rightarrow D = \frac{20000}{8}=​​​​2500 \ Rs\)

Therefore the option 2 is correct.

Concept:

\(Var[aX + b] = a^2Var(X)\)

\(V(ax)=a^2V(x)\) where a is constant,

\(V(a)=0\), i.e. variance of constant is zero ; where a is constant

Calculation:

Given:

\(Var(x) = 1\), \(Var(3x + 4)=?\)

Let \(y = 3x + 4\)

\(V(y) = V(3x + 4)\)

\(V(y) = 3^2[V(x)]\)

\(V(y) = 9[1]\)

\(V(y) = V(3x + 4) = 9\).

Concept:

• Convert  both the given inverse in the form of tan^-1,

Calculation:

 \({\sin ^{ - 1}}\frac{1}{{\sqrt {17} }} + {\cos ^{ - 1}}\frac{5}{{\sqrt {34} }} = {\tan ^{ - 1}}\frac{1}{4} + {\tan ^{ - 1}}\frac{3}{5}\)

\(= {\tan ^{ - 1}}\left[ {\frac{{\frac{1}{4} + \frac{3}{5}}}{{1 - \frac{1}{4}.\frac{3}{5}}}} \right] = {\tan ^{ - 1}}(1) = \pi /4.\)

• So, the correct answer is option 1.

Concept:

• Marginal cost is the instantaneous rate of change of cost function with respect to the production output x.

Marginal cost = C'(x) = \( \frac{dC(x)}{dx}\) 

• Marginal revenue is the instantaneous rate of change of revenue function with respect to the production output x.

Marginal revenue = R'(x) = \( \frac{dR(x)}{dx}\) 

• Revenue function R(x) = xp(x)

Calculation:

Given:

Variable cost function C(x) = x(80 - x2)

Price per unit p(x) = 5x 

• Marginal cost = C'(x) = \( \frac{dC(x)}{dx}\) 

 \(⇒ C'(x) = \frac{d}{dx} (x(80 - x^2))\)

 \(⇒ C'(x) = \frac{d}{dx}(80x - x^3)\)

⇒ C'(x) = 80 - 3x²

⇒ Marginal cost at (x = 5) = 80 - 3(5)² = 5

Revenue function R(x) = x p(x)

⇒ R(x) = x(5x)

⇒ R(x) = 5x²

• Marginal revenue,

⇒ R'(x) = \( \frac{dR(x)}{dx}\) 

⇒ R'(x)= \( \frac{d(5x^2)}{dx}=10x\)

⇒ Marginal revenue at (x = 5) = 10 × 5 = 50

• Therefore ratio of marginal cost and marginal revenue = 5 : 50 = 1 : 10

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 *.

Calculation:

Given: a * b = |a -b|

∵ 5 * p ≤ 5 

⇒ 5 * p = |5 -p| ≤ 5

⇒ |5 -p| ≤ 5

⇒ -5 ≤ 5 -p ≤ 5

⇒ -5 ≤ p - 5 ≤ 5

⇒ 0 ≤ p ≤ 10

• The maximum value of p is 10.

Concept:

1. Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is symmetric if for all x, y ∈ A, if xRy, then yRx.

3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.

Calculation:

S = Set of all integers and R = {(a, b), a, b ϵ  S and ab \(≥\) 0}

For reflexive:

aRa = a.a = a² ≥ 0, so it's reflexive.

For symmetric:

aRb = ab ≥ 0 and bRa = ba ≥  0, So relation is symmetric

For transitive:

For all integers, if ab ≥ 0, bc ≥  0, then for all the cases ac ≥ 0 is not true.

For example,

If a = -1, b = 0, and c =1

Then,  ab ≥ 0, bc ≥  0 but for ac = -1 which is not satisfying ac ≥ 0 so R is not transitive.

So, Relation Is reflexive, symmetric but not transitive. 

Hence, option (2) is correct.

Concept:

• The system of n linear equations AX = B in n variables is -

1. Consistent and has a unique solution if |A|≠ 0.
2. Consistent and has infinitely many solutions  if  |A| = 0 and (adj A )B = O
3. inconsistent and has no solution if |A| = 0 and (adj A)B ≠ O

Calculation:

Given:

\(A=\begin{pmatrix} 7 &9 \\ 14&18 \end{pmatrix} \) and \(B=\begin{pmatrix} 15 \\ 30 \end{pmatrix} \)

The determinant of the matrix A is given by,

\(|A|=\begin{vmatrix} 7 &9 \\ 14&18 \end{vmatrix} \)

⇒ |A| = (18) × 7 -(14) × 9

⇒ |A| =126 -126 = 0

• Then the system can not have a unique solution.
• If C_ij  be the cofactor of a_ij of A = (a_ij), then 

​\(A=\begin{pmatrix} 7 &9 \\ 14&18 \end{pmatrix} \)

C₁₁ = 18 , C₁₂ = -14 

C₂₁ = - 9, C₂₂ = 7

Thus \(adj A=\begin{pmatrix} 18 &-14 \\ -9& 7 \end{pmatrix}^T=\begin{pmatrix} 18 &-9 \\ -14& 7 \end{pmatrix} \)

Consider, \((adjA)B=\begin{pmatrix} 18 &-9 \\ -14& 7 \end{pmatrix}\begin{pmatrix} 15\\ 30 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix} \)

Thus the system has infinitely many solutions.

Therefore option 3 is correct.

Concept:

If \(\rm (x_1, y_1), (x_2, y_2)\) and \(\rm (x_3, y_3)\) are the vertices of a triangle then, 

Area = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

Note: Area is always a positive quantity, therefore we always take the absolute value of the determinant for the area. 

Calculation:

Given vertices are A (1, 1) ,B ( 6, 0) and C ( 3, 2).

We know that area of triangle ABC is given by, 

Δ = \(\rm \dfrac 12 \begin{vmatrix} \rm x_1 & \rm y_1 & 1 \\\ \rm x_2 & \rm y_2 & 1 \\\ \rm x_3 & \rm y_3 & 1 \end{vmatrix}\)

⇒ Δ = \(\rm\frac{1}{2}\begin{vmatrix} 1& 1 & 1\\ 6& 0 & 1\\ 3& 2 & 1 \end{vmatrix}\) 

⇒ Δ = \(\frac{1}{2}\left [ 1\left ( 0-2 \right )-1 (6-3)+1 (12-0)\right ]\) 

⇒ Δ =  \(\frac{7}{2}\)  

The correct option is 1.

Concept:

• EMI stands for equated monthly installment. It is a monthly payment that we make towards a loan we opted for at a fixed date every month.
• There are two methods for calculating the EMI -

1. Flat Rate Method
2. Reducing-balance method or Amortization of Loan

• In the Flat Rate Method, each interest charge is calculated based on the original loan amount, even though the loan balance outstanding is gradually being paid down. 
• Let P, I, and n be the principal of the loan, the total interest on the principal, and the number of months in the loan period respectively.
• Then the EMI is given by the formula,

\(EMI=\left(\frac{P+I}{n}\right)\)

Calculations:

Given: P = 10,00,000 Rs, R = 10 %, n = 5 years = 5 × 12 = 60 Months  

I = p × r × n 

• So the Interest,

⇒ I  = 10,00,000 × 0.10 × 5 = 5,00,000 

Therefore the EMI,

 \(EMI=\left(\frac{P+I}{n}\right)\)

\(EMI=\frac{10,00,000+5,00,000}{60}=25000\ Rs\)

Hence option 1 is correct.

Concept:

Property of determinant of a matrix:

Let A be a matrix of order n × n then det(kA) = kn det(A)

Calculation:

Given: \({\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 2&3&1\\ { - 1}&0&2\\ { - 3}&1&2 \end{array}} \right]\)

Here the order of the matrix is 3.

Now,

det(A) = 2(0 - 2) - 3(-2 + 6) + 1(-1 + 0)

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

= - 4 - 12 - 1

= -17

Now using the property the value of det(2²A) is:

det(2²A) = det(4A) = 43 det(A)

= 64 × -17

= -1088

Concept:

• The formula for weighted aggregate,

​\(Index\ number = \frac{Σ P_{1i}Q_i}{Σ P_{0i}Q_i}\space × 100\)

Where,

P1_i = The price of i^th item of the current year

Calculation:

Given:

• Referring to the data given in the table
• The method used for indexing is Simple Weighted Aggregate

⇒ Σ(P_1iQ_i) = p × 60 + 3 × q + 2 × 40

⇒ Σ(P1iQi) = 60p + 3q + 80      ...(1)

⇒ Σ(P₀iQi) = 4 × 60 + 2 × q + 3 × 40

⇒ Σ(P0iQi) = 2q + 360      ...(2)

\(\Rightarrow \frac{\Sigma P_{1i}Q_i}{\Sigma P_{0i}Q_i} \space \times 100=\frac{60p + 3q + 80}{2q + 360}\times 100\))

⇒ 110/100 = (60p + 3q + 80)/(2q + 360)    -----    (A)

∵  60p + 3q = 360, (given)

• Put in equation (A)

⇒ 110/100 = (360 + 80)/(8p + 360) 

• On solving, p = 5

Let speed of boat and current be x km/hr and y km/hr.

⇒ Upstream speed = (x – y) km/hr

⇒ Downstream speed = (x + y) km/hr

As we know,

Speed = Distance/time

⇒ (x – y) = 16/4

⇒ x – y = 4      ----(i)

(x + y) = 16/2

⇒ x + y = 8      ----(ii)

Add equation (i) and equation (ii)

2x = 12

⇒ x = 6

From equation (i)

y = 6 – 4 = 2

Concept:

• If Rolle's theorem holds for the function f(x) for  x ∈ [1, 2] 

Also, f'(c) = 0

Where, c ∈ [1, 2] 

⇒ f(1) = f(2)     

Calculation:

Given: \(f'\left( {\frac{4}{3}} \right) = 0\)

⇒ f(1) = f(2)    

⇒ 1 - a + b - 4 = 8 - 4a + 2b - 4

⇒ 3a - b = 7      ----(i)

Also \(f'\left( {\frac{4}{3}} \right) = 0\) [Given]

\( ⇒ {\left( {3{x^2} - 2ax + b} \right)_{x = \frac{4}{3}}} = 0 ⇒ \frac{{16}}{3} - \frac{{8a}}{3} + b = 0\)

⇒ 8a - 3b -16 = 0

• Solving (i) and (ii), a = 5, b = 8 
• So option 1 is correct.

Given

∑p_oq_o == 160

∑p_oq₁ = 250

∑p₁q_o = 200

∑p₁q₁ = 288

Formula

Fisher’s ideal index = √[(∑(p₁q_o/∑p_oq_o) × (∑p₁q₁/∑p_oq₁] × 100

Here,

P₁ = Price in current year

q₁ = quantity in current year

p_o = price in base year

q₀ = quantity in base year

Calculation

Fisher’s ideal index = √[(∑(p₁q_o/∑p_oq_o) × (∑p₁q₁/∑p_oq₁] × 100

⇒ √[(200/160) × (288/250)] × 100

⇒ (6/5) × 100

∴ Fisher’s index number is 120

The Fisher’s index number obtained as geometric mean  of index obtained by laspeyres’ and paasche’s fpormula

F = √(L × P)

L = Laspeyre’s index = ∑p_nq_o/∑p_oq_o

P = Paasche’s index = ∑p_nq_n/∑p_oq_n

Concept:

The slope of the tangent on the given curve at a point (2, 3),

\( ⇒ m = \frac{{dy}}{{dx}}_{ \ at\ ​​(2,3)}\)

Calculation:

Given: 

The curve y2 = px3 + q      ----(i)

Differentiate with respect to x,

2y.\(\frac{{dy}}{{dx}}\) = 3px²

_{\( ⇒ \frac{{dy}}{{dx}} = \frac{{3p}}{2}\left( {\frac{{{x^2}}}{y}} \right)\)}

_{\(∴ {\left| {\frac{{dy}}{{dx}}} \right|_{2,3}} = \frac{{3p}}{2} × \frac{4}{3} = 2p\)}

For given line, slope of tangent = 4

∴ 2p = 4 

⇒ p = 2

From equation (i),

⇒ 9 = 2 × 8 + q 

⇒ q  = -7

Concept:

The slope of the tangent on the given curve,

\(\frac{{dy}}{{dx}} = 2{x^3} - 15{x^2} + 36x - 19\)

Calculation:

Given: The slope is maximum.

\(⇒ \frac{{{d^2}y}}{{d{x^2}}} = 6{x^2} - 30x + 36 = 0\)

⇒ x² - 5x + 6 = 0

x = 2, 3

At x = 2

\(\frac{{{d^3}y}}{{d{x^3}}} = 12x - 30\)
\(At\ x = 2,\ \frac{{{d^3}y}}{{d{x^3}}} < 0\)

• So, maxima 

\(y = \frac{1}{2} \times 16 - 5 \times 8 + 18 \times 4 - 19 \times 2\)

= 8 - 40 + 72 - 38 = 80 - 78 = 2

Ans = Point (2, 2)

Explanation:

Let the speed be in the ratio 3x : 4x

Total distance is 500 m, therefore the distance to be covered by A is 360 m.

As per the problem:

3x = 360

∴ x = 120.

∴ 4x = 4 × 120 ⇒ 480 m.

Thus when P finishes the race of 500 m, the distance covered by Q is 480 m.

The difference between the two will be 20 m.

Concept:

A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)\) exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a ⇔ \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ + }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{a}}^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\)

Calculation:

Given: \({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} { - 1}&{{\rm{if}}}&{{\rm{x}} \le 0}\\ {{\rm{ax}} + {\rm{b}}}&{{\rm{if}}}&{0 < {\rm{x}} < 1}\\ 1&{{\rm{if}}}&{{\rm{x}} \ge 1} \end{array}} \right.\)

Given function continuous everywhere.

Let’s check the continuity at x tends to zero,

\(\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{f}}\left( {\rm{x}} \right)\)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{f}}\left( {\rm{x}} \right)\)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} - 1{\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{ax}} + {\rm{b}}\)

∴ b = -1       ---(1)

Now check the continuity at x tends to one,

\(\mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {1^ - }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{f}}\left( {\rm{x}} \right)\)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{f}}\left( {\rm{x}} \right)\)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to {1^ + }} 1{\rm{\;}} = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{ax}} + {\rm{b}}\)

⇒ a + b = 1

From equation 1st,

⇒ a - 1 = 1

∴ a = 2 

Concept:

We will use the chain rule for differentiation,

Calculation:

\(\frac{{dx}}{{dt}} = 3{\sec ^2}t\)

\(\frac{{dy}}{{dt}} = 3\sec t\tan t\)

\(\frac{{dy}}{{dx}} = \frac{{\tan t}}{{\sec t}} = \sin t\)

\(\frac{{{d^2}y}}{{d{x^2}}} = \cos t\frac{{dt}}{{dx}} \)

At \(t = \frac{\pi }{4}\),

\(\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{\cos t}}{{3{{\sec }^2}t}} = \frac{{{{\cos }^3t}}}{3} = \frac{1}{{3.2\sqrt 2 }} = \frac{1}{{6\sqrt 2 }}.\)

Concept:

Unbounded Solution: 

• If the feasible region is not bounded, it is possible that the value of the objective function goes on increasing without leaving the feasible region.
• This is known as an unbounded solution.

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• Convert all the constraints to equality and plot on the graph. Put the value of (x1, x2) obtained from the corner points of the feasible region and put it in the objective function.
• Now check the region which we have to choose depending on the sign of inequality.
• To check which region we need to choose put (0,0) in both the inequality. and check whether this inequality is satisfying or not.
• If it is satisfying the inequality then take the region containing t (0,0) else the opposite side of (0,0).

Calculation:

Given:

Maximum Z = 3X1 + 2X2

Subject to

- 2X1 + 3X2 ≤ 9

X1 – 5X2 ≥ -20

X1, X2 ≥ 0

• Plotting the graph for the given constraints as shown in the figure. From the figure, we can see that LPP has an unbounded solution.

• So, the correct answer is option 1.

Explanation:

Lifetime value = Total Revenues - (Fixed Cost + Variable Cost)

• Marketing is closely related to efforts to create and deliver value to customers.
• Customer Value is the incremental benefit that a customer derives from consuming a product after paying in return.
• The term value signifies the benefits that a customer gets from a product.
• It is the difference between the benefits (sum of tangible and intangible benefits) and the cost.
• In another word, Lifetime value = Total Revenues - (Fixed Cost + Variable Cost)
• Customer value is dependent on the three factors – Quality, Service, and Price.
• Hence, these three together form the ‘Customer Value Triad’.
• The value of a product increases with its quality and service, as the benefits increase. On the other hand, the value decreases with an increase in price because of the increase in costs increase in this case.

CONCEPT:

• Fundamental Theorem of Calculus (Newton-Leibniz formula)

\(\rm \frac{d}{dx} \left[ \int_{g(x)}^{h(x)} f(t) dt \right] = f(h(x)) . h'(x) - f(g(x)) g'(x)\)

CONCEPT:

Given: F(x2) = x2(1 + x)

⇒ x²(1 + x) = \(\int\limits_0^{{x^2}} {f\left( t \right)dt.} \)

• Differentiating w.r.t.x,

⇒  2x(1 + x) + x² = f(x²).2x

⇒ f(x²) = 1 + \(\frac{{3x}}{2}\), x > 0

• Putting x = 2,

⇒ f(4) = 1 + 3 × \(\frac{4}{4}\) = 4.

Concepts

• A symmetric matrix and skew-symmetric matrix both are square matrices.
•  The symmetric matrix is equal to its transpose 
• The skew-symmetric matrix is a matrix whose transpose is equal to its negative.

Explanation:

If A is symmetric then A^T = A.

If A is skew-symmetric then AT = – A.

NOTE:

A^T is a transpose matrix

Property:

1) log a^b = b log a

2) \(log~a-log~b=\frac{log~a}{log b}\)

Application:

\(\smallint \frac{x}{{\left( {x - 2} \right)\left( {x - 1} \right)}}dx\)

Applying Partial Fractions, we get:

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

\(= 2\log \left( {{\rm{x}} - 2} \right) - \log \left( {{\rm{x}} - 1} \right) + {\rm{p}}\)

= log(x – 2)² – log(x – 1) + p

= \({\log _e}\frac{{{{\left( {x - 2} \right)}^2}}}{{\left( {x - 1} \right)}} + p\)

Let \(\frac{k}{n}\) = x, \(\frac{1}{n}\) = dx

3\(\int\limits_0^1 {(9{x^2} + 2)} \)dx = \(\int\limits_0^3 {({x^2} + 2)}\)dx

3\(\left[ {\frac{{9{x^3}}}{3} + 2x} \right]_0^1\) = 15

Concept:

For the independent events A and B, 

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• P(A ∩ B) = P(A) × P(B)

Calculation:

Probability of getting head on first toss P(E) = \(1\over2\)

Probability of getting head on first toss P(F) = \(1\over2\)

P(E ∩ F) = \(1\over2\) × \(1\over2\) = \(1\over4\)

P(E ∪ F) = P(E) + P(F) - P(E ∩ F)

P(E ∪ F) = \(1\over2\) + \(1\over2\) - \(1\over4\)

P(E ∪ F) = \(3\over4\)

Concept:

Integration by parts

Integration by parts is used to integrate the product of two or more functions. The two functions

to be integrated f(x) and g(x) are of the form \(\rm \displaystyle\int \)f(x).g(x). Thus, it can be called a product rule of

integration. Among the two functions, the first function f(x) is selected such that its derivative

formula exists, and the second function g(x) is chosen such that an integral of such a function exists.

\(\rm \displaystyle\int \)f(x).g(x) dx = f(x)\(\rm \displaystyle\int \)g(x) dx−\(\rm \displaystyle\int \big[\)f′(x)\(\rm \displaystyle\int \)g(x) dx\(\big]\) dx + C

A useful rule of integral by parts is ILATE.

I: Inverse trigonometric functions : sin-1(x), cos-1(x), tan-1(x)

L: Logarithmic functions : ln(x), log(x)

A: Algebraic functions : x2, x3

T: Trigonometric functions : sin(x), cos(x), tan (x)

E: Exponential functions : ex, 3x

Calculation:

Let I = \(\rm \displaystyle\int e^x \left(√ x + \frac{1}{2 √ x} \right) dx \)

⇒ I = \(\rm \displaystyle\int e^x √ x dx+\rm \displaystyle\int \frac{e^x}{2 √ x} dx \)

Taking √ x is the first function  and ex  is the second function and integrating by parts, we get

I = \(e^x√ x-\rm \displaystyle\int \frac{e^x}{2 √ x} dx+\rm \displaystyle\int \frac{e^x}{2 √ x} dx \)

⇒ I = e^x √x + C

∴ The value of \(\rm \displaystyle\int e^x \left(√ x + \frac{1}{2 √ x} \right) dx \) = ex √x + C

Concept:

• Two curves f(x, y) = 0 and g(x, y) = 0 cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.

• The area under the function y = f(x) from x = a to x = b and the x-axis is given by the definite integral \(\rm \displaystyle \left|\int_a^b f(x)\ dx\right|\), for curves which are entirely on the same side of the x-axis in the given range.
• If the curves are on both the sides of the x-axis, then we calculate the areas of both the 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).
• \(\rm \displaystyle \int x^n\ dx = \dfrac{x^{n+1}}{n+1}+C\).

Calculations:

Let's first find the points of intersection/touching of the two curves y2 = x and y = |x|.

Using the condition for intersecting, we must have:

y2 - x = y - |x| = 0

CASE 1: For, x ≥ 0, |x| = x.

⇒ y2 - x = y - x = 0

⇒ y2 - y = 0

⇒ y(y - 1) = 0

⇒ y = 0 OR y - 1 = 0

⇒ y = 0 OR y = 1.

And, x = 0² = 0 OR x = 1² = 1.

The points of intersection are (0, 0) and (1, 1).

CASE 2: For x < 0, |x| = -x.

⇒ y2 - x = y - (-x) = 0

⇒ y2 - x = - y - x = 0

⇒ y2 + y = 0

⇒ y(y + 1) = 0

⇒ y = 0 OR y + 1 = 0

⇒ y = 0 OR y = -1.

∵ y = |x|, y cannot be negative.

∴ The only solution in this case y = 0 and x = 0² = 0.

Finally, the solutions (points of intersection/touching) of y2 = x and y = |x| are (0, 0) and (1, 1).

The graph of the curves is shown below:

The required (shaded) area is:

(Area under y² = x from x = 0 to x = 1.) - (Area under y = |x| from x = 0 to x = 1.)

= \(\rm \displaystyle \left|\int_0^1 \sqrt x\ dx\right| -\left|\int_0^1 x\ dx\right|\)

= \(\rm \left|\dfrac{2}{3}\left[x^{\tfrac{3}{2}} \right ]_0^1\right|-\left|\dfrac{1}{2}\left[x^{2} \right ]_0^1\right|\)

= \(\rm \left|\dfrac{2}{3}[1^{\tfrac{3}{2}}-0^{\tfrac{3}{2}}]\right|-\left|\dfrac{1}{2}[1^2-0^2]\right|\)

= \(\rm \dfrac 23 - \dfrac 12\)

= \(\dfrac16\) sq. units.

Calculation:

Given: f(1) = 2 and f'(x) = f(x) for all x ∈ R

\(\Rightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = 1\ \forall \ x \epsilon R\)

• Integrate and use f(1) = 2

⇒  f(x) = 2e^x-1 

⇒ f'(x) = 2e^x-1

Now,

⇒ h(x) = f(f(x)) 

⇒ h'(x) = f'(f(x))f'(x)

⇒ h'(1) = f'(f(1))f'(1) = f'(2)f'(1) = 2e.2 = 4e.

Concept:

If the equation of the line is an \(\rm \vec r = \;\vec a + \lambda \;\vec b\) and equation of the plane is \(\rm \overrightarrow{r}. \overrightarrow{n}=d\) . then the angle α  between line and direction parallel to the plane is, 

\(\rm \sin α = \left | \frac{\overrightarrow{b}.\overrightarrow{n}}{\left | \overrightarrow{b} \right |\left | \overrightarrow{n} \right |} \right |\)  

Calculation: 

Equation of line with X-axis is,

\(\rm \overrightarrow{b}= \hat{i}\)

Equation of plane is  2x -3y +6z - = 0 

or in  vector form,  \(\rm \overrightarrow{n}= 2\hat{i}-3\hat{j}+6\hat{k}\)   

∴   \(\rm \sin α = \left | \frac{\overrightarrow{b}.\overrightarrow{n}}{\left | \overrightarrow{b} \right |\left | \overrightarrow{n} \right |} \right |\)

⇒ \(\rm \sin α = \frac{\left ( 2\hat{i}-3\hat{j}+6\hat{k} \right ). \left ( \hat{i}+ 0\hat{j}+0\hat{k} \right )}{\sqrt{2^{2}+(-3)^{2}+6^{2}}. \sqrt{1^{2}}}\) = \(\frac{2}{\sqrt{49}}\)

⇒ sin α = \(\frac{2}{7}\) 

⇒ α = sin^-1 ( \(\frac{2}{7}\) )  

⇒ α = \(\frac{2}{7}\).

The correct option is 2.

We need to find the normal vector to the plane containing \({\rm{\;\hat i}} + {\rm{\hat j}} + {\rm{\hat k\;and\;\hat i}} + 2{\rm{\hat j}} + 3{\rm{\hat k}}\):

So, let’s take the cross product:

\({\rm{\vec n}} = \left( {{\rm{\hat i}} + {\rm{\hat j}} + {\rm{\hat k}}} \right) \times \left( {{\rm{\hat i}} + 2{\rm{\hat j}} + 3{\rm{\hat k}}} \right)\)

\({\rm{\vec n}} = \left| {\begin{array}{*{20}{c}} {{\rm{\hat i}}}&{{\rm{\hat j}}}&{{\rm{\hat k}}}\\ 1&1&1\\ 1&2&3 \end{array}} \right|\)

\({\rm{\vec n}} = {\rm{\hat i}}\left( {3 - 2} \right) - {\rm{\hat j}}\left( {3 - 1} \right) + {\rm{\hat k}}\left( {2 - 1} \right)\)

\({\rm{\vec n}} = {\rm{\hat i}} - 2{\rm{\hat j}} + {\rm{\hat k}}\)

Projection of \(\left( {2{\rm{\hat i}} + 3{\rm{\hat j}} + {\rm{\hat k}}} \right){\rm{\;on\;to\;this}}\;{\rm{\vec n}}\)

\(\left| {\rm{a}} \right|\cos \theta = \frac{{\vec a\vec n}}{{\left| n \right|}}\)

\(\Rightarrow \left| {\rm{a}} \right|\cos {\rm{\theta }} = \left| {\frac{{\left( {2{\rm{\hat i}} + 3{\rm{\hat j}} + {\rm{\hat k}}} \right)\cdot\left( {{\rm{\hat i}} - 2{\rm{\hat j}} + {\rm{\hat k}}} \right)}}{{\sqrt {1 + 4 + 1} }}} \right|\)

\(\Rightarrow \left| {\rm{a}} \right|\cos {\rm{\theta }} = \left| {\frac{{2 - 6 + 1}}{{\sqrt 6 }}} \right|\)

\( \Rightarrow \left| {\rm{a}} \right|\cos {\rm{\theta }} = \frac{3}{{\sqrt 6 }}\)

\(\therefore \left| {\rm{a}} \right|\cos {\rm{\theta }} = \sqrt {\frac{3}{2}} \)

CONCEPT:

Perpendicular Distance of a Point from a Plane

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d and a point whose coordinate is, (x1, y1, z1)

Now, distance between the point and the plane = \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

CALCULATION:

Given: Equation of line \(\frac{{x - 3}}{1} = \frac{{y - 4}}{{\;2}} = \frac{{z - 5}}{1}\) and equation of plane is x - y + z - 5 = 0

First let's find out if they given line is parallel to the plane or not.

As we can see that, the direction ratios of the given line are: (1, 2, 1)

Similarly, the direction ratios of the normal to the given plane are: (1, -1, 1)

⇒ 1 × 1 - 2 × 1 + 1 × 1 = 0

So, the given line is parallel to the given plane.

Let's find out a point on the given line

As we can see that, point Q (3, 4, 5) lies on the line.

So, finding the distance between the point Q and the given plane is same as finding the distance between the given line and plane.

As we know that, distance between a point and a plane is given by: \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

Here, x1 = 3, y1 = 4 and z1 = 5

⇒ \(d = \left| {\frac{{3\times {1}\; + \;4 \times {(- 1)}\; + \;5 \times {1} - \ 5}}{{\sqrt {{1^2}\; + \;\;{(- 1)^2}\; + \;{1^2}} }}} \right| = \frac{1}{\sqrt {3}} \ units\)

Hence, option B is the correct answer.

Concept:

The minimum value can be obtained by the following steps

• Step1: Plot the constraints graphically
• Step2: Obtain the feasible region (the region which is common for all the constraints)
• Step3: Checking for the minimum value at the corner points, the point where the minimum value occurs is the optimum point. 

Calculation:

To minimize (3x + 5y)

Given

3x + 5y ≤ 15

4x + 9y ≤ 18

13x + 2y ≤ 2

x ≥ 0; y ≥ 0

• Now from the figure itself we use (A B C D) is the feasible reason clearly as we have to minimize (3x + 5y).
• The minimum will occur at [x = 0 and y = 0] point A

⇒ 3x + 5y = 0 Minimum and in all rest cases as x ≥ 0, y ≥ 0 [3x + 5y ≥ 0]

• Hence at (0, 0) 3x + 5y is minimum.

Mistake Points

•  Keep in mind that the minimum value is asked not the maximum.
• So don’t get confused otherwise at point (c) maximum will occur and the answer will be different which is incorrect.

Concept:

Area under a Curve by Integration

Find the area under this curve is by summing horizontally

In this case, we find the area is the sum of the rectangles, heights y = f(x) and width dx.

\(\therefore \;{\bf{A}}\; = \;\mathop \smallint \nolimits_{\bf{a}}^{\bf{b}} {\bf{ydx}}\; = \;\mathop \smallint \nolimits_{\bf{a}}^{\bf{b}} {\bf{f}}\left( {\bf{x}} \right){\bf{dx}}\)

Calculation:

\({\rm{Area\;under\;curve\;\;}} = {\rm{\;}}\mathop \smallint \nolimits_0^{\rm{\pi }} ({\rm{c}}\sin {\rm{x}}){\rm{dx}}\)

\(= c\mathop \smallint \nolimits_0^\pi \sin xdx\)

\(= c\left[ { - \cos x} \right]_0^{\rm{\pi }}\)

\(= - c\left[ {\cos \pi - \cos 0} \right]\)

\(= - c\left[ { - 1 - 1} \right]\)

\(= 2c\)

Concept:

Let x be the units of P and y are the unit of Q.

Given':

• Two models, P, and Q, of a product, earn profits of Rs.100 and Rs.80 per unit, respectively. 

⇒ Objective function Max. Z = 100 x + 80 y

• Production times for P and Q are 5 hours and 3 hours, respectively with the total production time available being 150 hours

⇒ 5x + 3y ≤ 150

• For a total batch size of 40,

⇒ x + y ≤ 40

Non-negativity constraints,

x > 0, y > 0

• So, the correct answer is option 2.

Additional Information Max. Z = 100 x + 80 y subjected to

5x + 3y ≤ 150,

x + y ≤ 40,

x > 0, y > 0 

• The shaded region is the feasible region. Finding the corner points coordinates of B.

As B is the intersection of

x + y = 40

5x + 3y = 150

• On solving these coordinates of B is (15, 25)
• Now, finding value of objective function at all corner points

Z(A) = Z(0, 40) = 3200

Z(B) = Z(15, 25) = 3500

Z(C) = Z(30, 0) = 3000

Z(O) = Z(0, 0) = 0

• So Maximum value is at B (15, 25). The no. of units of P is 15

Calculation:

Draw the graph of  y = x + 2 , x-axis and y-axis.

Let the required area be A.

Using the formula of the area under the curve, \(\rm A=\left | \int_{a}^{b}f(x)-g(x) dx \right |\)

\(\Rightarrow \rm A=\left | \int_{-2}^{0}(x+2) dx \right |=\left | \left [ \frac {x^{2}}{2}+2x \right ]_{-2}^{0} \right |\)

Substitute the limit to evaluate the area.

\(\Rightarrow \rm A=\left | {0+0-\frac{4}{2}} +4 \right |=2\)

Hence, the required answer is option 3.

Concept:

Natural number: Natural numbers are the positive integers or non

negative integers which start from 1 and end at infinity.

Consider the quadratic equation ax2 + bx + c = 0

Discriminant (D):

D = b² - 4ac

For real root

D ≥ 0

b² - 4ac ≥ 0

Calculation:

Given equation 

x2 + x + n = 0

(1)² - 4n ≥ 0

1  ≥  4n

n ≤ \(\rm \frac{1}{4}\)

This is possible for n = 0

∴ Probability of the required event is zero.

Formula used:

Probability of occurrence of the event:

P(E) =  \(\frac{n(E)}{n(S)}\)

Where,

n(E) = Number of favorable outcome

n(S) = Number of possible outcome

Calculation:

Total events

n(S) = 25 = 32

Number of outcomes having zero head

n(E̅) = 1 i.e.  {TTTTT}

Hence, the number of outcomes of obtaining at least one head is

n(E) = 32 - 1 = 31

Hence, the required probability

P(E) = \(\frac{31}{32}\)

Concept:

For a uniformly distributed random variable,

\(E[X]=\mathop \int \limits_{-\infty}^{\infty}x f_x(x)\ dx\)
Calculation:

\(E[\sin x]=\mathop \int \limits_{-\infty}^{\infty}sin x f_x(x)\ dx\)

\(=\dfrac{1}{2\pi}\mathop \int \limits_{-\pi}^{\pi}\sin x \ dx\)

\(=\dfrac{-1}{2\pi}[\cos x]^\pi_{-\pi}\)

\(\dfrac{-1}{2\pi}\left[\cos \pi-\cos (-\pi)\right]\)

E[sin x] = 0

Concept:

If the two lines \(\rm {x -x_1\over a}={y-y_1\over b}={z-z_1\over c}\) and \(\rm {x -x_2\over p}={y-y_2\over q}={z-z_2\over r}\) are coplanar, then 

\(\begin{vmatrix} x_1-x_2& y_1-y_2 & z_1-z_2\\ a& b &c \\ p& q &r \end{vmatrix} = 0\)

∴ Option 2 is correct