Mathematics Mock Test - 6

Answer: Option 3

Explanation:

Events A and B are mutually exclusive, This means events A and B cannot happen together. If A happens, it excludes B from happening, and vice-versa.

∴ If A and B are mutually exclusive,

Calculation:

Given: \(\ln \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) - {\rm{a}} = 0\)

\( \Rightarrow \ln \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{a}}\)

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {{\rm{e}}^{\rm{a}}}\)

\(\Rightarrow {\rm{\;}}\smallint \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \smallint {{\rm{e}}^{\rm{a}}}\)

On integrating both sides, we get

⇒ y = xe^a + c

Concept:

If two vectors are perpendicular to each other, their dot product is zero.

Calculations:

Consider a plane passing through a point A with position vector 'a' and perpendicular to 'b' to the given plane.

Let 'r' be the position vector of an arbitrary point P on the plane.

Then AP is perpendicular to \(\rm\vec b\).

As we know, If two vectors are perpendicular to each other, their dot product is zero.

⇒\(\rm \vec{AP} \cdot \vec{b} = 0\)

\(\therefore \rm \left( {\bar r - \bar a} \right).\bar b = 0\)

Concept:

• To multiply an ‘m × n’ matrix by an ‘n × p’ matrix, then 'n' must be the same, and the result is an m × p matrix.
• If A is a matrix of order m × n’ then the order of the transpose matrix is ‘n × m’

Calculation:

Given: [x y z] \(\left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]\,\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]\) 

⇒ Order will be (1 × 3) (3 × 3) (3 × 1) = (1 × 1)

⇒ Order will be → (1 × 3) (3 × 1) 

⇒ Order will be → (1 × 1) 

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are the same then the value of the determinant is zero.
• We can "split" a determinant along multiple rows or columns.

Calculation:

Given,

Determinant

\(\left| \begin{array}{l} {\begin{array}{*{20}{c}} {a + b}&{b + c}&{c + a}\\ {b + c}&{c + a}&{a + b}\\ {c + a}&{a + b}&{b + c} \end{array}} \end{array} \right|\)

This can be written as:

(b) \(\left| \begin{array}{l} {\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}} \end{array} \right|\) + \(\left| \begin{array}{l} {\begin{array}{*{20}{c}} b&c&a\\ c&a&b\\ a&b&c \end{array}} \end{array} \right|\) =

Using this property.

If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1

Here the determinant  \(\left| \begin{array}{l} {\begin{array}{*{20}{c}} b&c&a\\ c&a&b\\ a&b&c \end{array}} \end{array} \right|\) can be written as \(\left| \begin{array}{l} {\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}} \end{array} \right|\)  by interchanging Column 3 & 2

So, \(\left| \begin{array}{l} {\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}} \end{array} \right|\) + \(\left| \begin{array}{l} {\begin{array}{*{20}{c}} b&c&a\\ c&a&b\\ a&b&c \end{array}} \end{array} \right|\) = \(\left| \begin{array}{l} {\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}} \end{array} \right|\) + \(\left| \begin{array}{l} {\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}} \end{array} \right|\) = 2 \(\left| \begin{array}{l} {\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}} \end{array} \right|\) 

⇒  K = 2.

Explanation:

\(I=\mathop \smallint \limits_{ - 1}^1 {e^{\left| x \right|}}dx.\)

\(I= \mathop \smallint \limits_{ - 1}^0 {e^{ - x}}dx + \mathop \smallint \limits_0^1 {e^x}dx\)

\(I= \left[ { - {e^{ - x}}} \right]_{-1}^0 + \left[ {{e^x}} \right]_0^1\)

I = [-e^-0 + e¹] + [e¹ - e⁰]

I = -1 + e¹ + e - 1

Concept:

Here, we have to evaluate the function with a given condition.

Calculation:

Given:

\(f(x) = \frac{1}{{1 - x}}\)

\( \Rightarrow f\{ f(x)\} = \frac{{1 - x}}{{ - x}}\)

\( \Rightarrow f[f\{ f(x)\} ] = \frac{{ - x}}{{ - x - 1 + x}} = x\)

∴ Derivative of f[f{f(x)}] 1.

Fisher-Price Index is the Geometric Mean (G.M.) of Laspeyers and Paasche Index number.

 Key Points

• Fisher-Price index takes into account both current and base year.
• The Formulae for the Fisher-Price Index is :

         Fisher-Price index= \(\sqrt{(Laspeyres \ Price \ Index \ * \ Paasche \ Price \ Index)}\)

or,

Fisher-Price index= \(\sqrt{\frac{∑ (Current \ year \ Price \ * \ Base \ year \ Quantity)}{ ∑ (Base \ year \ Price \ * \ Base \ year \ Quantity)} * \frac{∑ (Current \ year \ Price \ * \ Current \ year \ Quantity)}{ ∑ (Base \ year \ Price \ * \ Current \ year \ Quantity)}}\) 

Where: 

Laspeyres Index Formula (LPI) =  \(\frac{∑ (Current \ year \ Price \ * \ Base \ year \ Quantity)}{ ∑ (Base \ year \ Price \ * \ Base \ year \ Quantity)}\)

Paasche Index Formula (PPI) = \( \frac{∑ (Current \ year \ Price \ * \ Current \ year \ Quantity)}{ ∑ (Base \ year \ Price \ * \ Current \ year \ Quantity)}\)

Additional Information

• Fisher-Price Index is called the ideal price index as it corrects the positive price bias of the Laspeyres Price Index (which only uses the base period basket) and the negative price bias of the Paasche Price Index (which only uses the current year basket) by taking the Geometric average of the two indexes.
• The normal result of comparison when calculating these 3 indices are:

          Laspeyres ≥ Fisher ≥ Paasche

CONCEPT:

Chain Rule of Derivatives:

• \(\rm \frac{d}{dx}f(g(x))=\frac{d}{d\ g(x)}f(g(x))\times \frac{d}{dx}g(x)\).
• \(\rm \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\).

CALCULATION:

Given:

Here y = t¹⁰ + 1 and x = t⁸ +1

∴ t⁸ = x - 1 

⇒ t² = (x - 1)^{1 / 4}

so, y = (x - 1)^{5 / 4} + 1

Differentiate both sides w. r. t. x,

\(\frac{{dy}}{{dx}} = \frac{5}{4}\)(x - 1)^{1 / 4}

Again, differentiate both sides w. r. t. x,

\(\frac{{{d^2}y}}{{d{x^2}}}\, = \frac{5}{{16}}\) (x - 1)^{-3 / 4} 

\(\frac{{{d^2}y}}{{d{x^2}}}\, = \frac{5}{{16\,{{\left( {x - 1} \right)}^{3/4}}}} = \frac{5}{{16{{({t^2})}^3}}}\, = \frac{5}{{16{t^6}}}\)

Concept:

The critical point of a function: Consider the function f(x) then the values at which f'(x) = 0 or not defined are known as critical points.

First derivative test: We say that function has extremum at some point x = a, if the first derivative at that point is 0.

Second derivative test: If x = a is a critical point of the function f(x) then we say that function has maxima if \(\rm f''(a) < 0\) similarly if \(\rm f''(0) > 0\)  then the function has minima at that point.

Calculation:

Given: Suppose that two numbers are x and y.

x + y = s

⇒ y = s - x

Then f (x) = xy = x (s - x) = xs - x²

∴ f' (x) = s - 2x

⇒ f'' (x) = - 2 < 0

f' (x) = 0 form maximum value of f (x)

\(\therefore x =\frac{S}{2}\ and\ y=\frac{S}{2}\)

Thus each number is half of the sum.

Concept:

The slope of the tangent to a curve y = f(x) is m = \(\rm dy\over dx\)

The slope of the normal = \(\rm -{1\over m}\) = \(\rm -{1\over {dy\over dx}}\)

Calculation:

Given:

The slope of the line will be the negative ratio of the coefficient of x and coefficient of y that is,

 \({m_{{\mathop{\rm Tan}\nolimits} gent}} = - \frac{b}{a}\)

also, y = be^-x/a

\(\therefore \frac{{dy}}{{dx}} = \frac{{ - b}}{a}{e^{ - x/a}}\)

\(\therefore \frac{{ - b}}{a} = \frac{{ - b}}{a}{e^{x/a}}\)

\( ⇒ {e^{x/a}} = 1\)

\( ⇒ - \frac{x}{a} = 0 \)

⇒ x = 0.

If x = 0

Then \(0 + \frac{y}{b} = 1\)

⇒ y = b,

⇒  The point is (0, b).

Concept:

• The face value or par value of a bond is the price at which a bond is sold to buyers at the time of issue.

Explanation:

• When the market price of the bond is less than its face value/ par value, the bond is selling at a discount.
• When the market price of the bond is more than its face value/ par value, the bond is selling at a premium.
• Hence the correct answer is option 1.

Concept:

Monotonic function: If a function is differentiable on the interval (a, b) and if the function is increasing/strictly increasing or decreasing/strictly decreasing, then the function is known as a monotonic function.

First derivative test:

• If \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} \ge 0\) for all x in (a, b) then, f(x) is an increasing function in (a, b).
• If \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} \le 0\) for all x in (a, b) then, f(x) is an decreasing function in (a, b).

For strictly increasing or decreasing:

• \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} > 0\) for strictly increasing.
• \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} < 0\) for strictly decreasing.

f(x) will be monotonically decreasing, if (x) < 0.

Calculation:

Given:

f(x) = cos x - 2px

⇒ f'(x) = - sin x - 2p < 0

⇒ sin x + 2p > 0

\(⇒ \frac{sin x}{2} + p > 0\)

⇒ -1 ≤ sin x ≤ 1

So, the above inequality will be only possible when,

⇒ p > 1/2

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(x) = x(x -1)2 

⇒ f(0) = 0

f(2) = 2 

⇒  f(0) ≠ f(2)

Thus mean value theorem is applicable

Then, \(f'(x) = \frac{{f(2) - f(0)}}{{2 - 0}}\)

\( \Rightarrow 3{x^2} - 4x + 1 = \frac{{2 - 0}}{{2 - 0}} = 1 \)

\(\Rightarrow 3{x^2} - 4x = 0\)

\( \Rightarrow x(3x - 4) = 0 \)

\(\Rightarrow x = 0,x = 4/3.\)

We will take only x = 4/3 ϵ (0, 2) 

Concept:

A function f(x) is said to be continuous at a point x = a, in its domain if  exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a ⇔ \(\rm \lim_{x\rightarrow a^{+}}f(x)=\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a}f(x)\)

Calculation:

\(\rm f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {Kx^2}&{if}&{x \le 2}\\ 3&{if}&{x > 2} \end{array}} \right.\) 

\(\rm \lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2}K x^{2}\)

⇒ \(\rm \lim_{x\rightarrow 2^{-}}f(x)=4 K\)                               

Similarly, 

\(\rm \lim_{x\rightarrow 2^{+}}f(x)=3 \)

Function is continuos at x = 2,

So, \(\rm \lim_{x\rightarrow 2^{+}}f(x)=\lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2}f(x)\)

⇒ 4K = 3 

⇒ K = 3/4

The correct option is 1.

Concept:

2 tan-1 x = sin-1 \(\frac{{2x}}{{1 + {x^2}}}\)

Calculation:

Given:

sin^-1 x = 2 tan^-1 x 

⇒  sin^-1 x = sin^-1 \(\frac{{2x}}{{1 + {x^2}}}\)

⇒  \(\frac{{2x}}{{1 + {x^2}}}\) = x 

⇒ x ³ - x = 0

\( ⇒ \) x(x + 1)(x - 1) = 0 

⇒  x = {-1, 1, 0}.

Concept:

We know that,

\(\left| {adj\;A} \right| = {\left| A \right|^{n - 1}}\)

The adjoint of a matrix (also called the adjugate of a matrix) is defined as the transpose of the cofactor matrix of that particular matrix. For matrix A, the adjoint is denoted as adj (A).

We know that, 

|A| = |A^T|

⇒ |adj A| = |(Adj A)T| 

(Adj A)T Is represents the determinant formed by its co-factor matrix A.

⇒ |adj A| = |(Adj A)T|  = |A|^n-1

​Calculation:

Given: 3 × 3 determinant is 3, 

|A| = 3

⇒ |(Adj A)T|  = (3)^(3-1) = 9

• The value of the determinant formed by its cofactor is 9.

CONCEPT:

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

CALCULATION:

\(Given \ \rm \vec P= 2 \hat a_x - \hat a _z \)

\(\vec Q = 2 \hat a_x - \hat a_y + 2 \hat a_z \)

\(component \ \rm of \ \rm \vec p \ \rm along \ \rm \vec Q=(\vec P \cdot\hat Q)\hat Q\)

\(\vec P \cdot \hat Q=(2 \hat a_x - \hat a _z )\cdot \frac{(2 \hat a_x - \hat a_y + 2 \hat a_z )}{3}\)

\(\vec P \cdot \hat Q=\frac{2}{3} \)

\((\vec P \cdot\hat Q)\hat Q=\frac{2}{3} \frac{(2 \hat a_x - \hat a_y + 2 \hat a_z )}{3}\)

\((\vec P \cdot\hat Q)\hat Q=\frac{2}{9}(2 \hat a_x - \hat a_y + 2 \hat a_z )\)

\((\vec P \cdot\hat Q)\hat Q=0.4444\hat a_x-0.2222\hat a_y+0.4444\hat a_z\)

Option 4 is correct.

Concept:

The equal matrix will have the corresponding elements as the same.

Calculation:

Given: kA = \(\left[ {\begin{array}{*{20}{c}} {\rm{0}}&{{\rm{3a}}}\\ {{\rm{2b}}}&{{\rm{24}}} \end{array}} \right]\)

⇒ k\(\left[ {\begin{array}{*{20}{c}} 0&2\\ 3&{ - 4} \end{array}} \right]\) = \(\left[ {\begin{array}{*{20}{c}} {\rm{0}}&{{\rm{3a}}}\\ {{\rm{2b}}}&{{\rm{24}}} \end{array}} \right]\)

⇒ 2k = 3a, 3k = 2b, -4k = 24

⇒ a = \(\frac{2k}{3}\), b = \(\frac{3k}{2}\), k = - 6

⇒ k = -6, a = -4, b = -9

Area of parallelogram with two vectors are given as

\({\rm{A}} = \left| {{\rm{\vec a}} \times {\rm{\vec b}}} \right|\)

\({\rm{\vec a}} \times {\rm{\vec b}} = \left| {\begin{array}{*{20}{c}} {\rm{i}}&{\rm{j}}&{\rm{k}}\\ 3&1&4\\ 1&{ - 1}&1 \end{array}} \right|\)

\({\rm{\vec a}} \times {\rm{\vec b}} = {\rm{i}}\left( {1 + 4} \right) - {\rm{j}}\left( {3 - 4} \right) + {\rm{k}}\left( { - 3 - 1} \right) = 5{\rm{i}} + {\rm{j}} - 4{\rm{k}}\)

\({\rm{A}} = \left| {{\rm{\vec a}} \times {\rm{\vec b}}} \right| = \sqrt {{5^2} + 1 + {4^2}} = \sqrt {42} \)

Concept:

Draw the constraints to find the feasible region:

• To draw the inequalities, first, draw the equation form of the inequalities.
• 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 (0,0) else the opposite side of (0,0).

Calculation:

Given: x + y ≥ 5

⇒ y ≥ – x + 5

• The given linear inequality covers the area (shaded region) above the straight line y = – x + 5 as shown below,

Similarly, x – y ≤ 3

⇒ – y ≤ – x + 3

⇒ y ≥ x – 3

• The given linear inequality covers the area (shaded region) above the straight line y = x – 3 as shown below,

• Hence, the solution of the two linear inequalities is the common area (shaded region) covered by the two inequalities as shown below,

∴ The solution to the given linear inequalities lies in the first and second quadrants.

Concept

Let the system of equations be,

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

\({\rm{\;}} ⇒ \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\\ {{{\rm{d}}_2}}\\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

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

⇒ If det (A) ≠ 0, system is consistent has 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

For no solution or infinitely many solutions

\(⇒ \left| {\begin{array}{*{20}{c}} α &1&1\\ 1&α &1\\ 1&1&α \end{array}} \right|\) = 0 

⇒ α = 1, α = -2

• But for α = 1, clearly, there are infinitely many solutions as column matrix B will become zero matrices.

⇒ ​(adj A). B = O

For no solution,

• When we put α = - 2 then matrix B will become a non zero column matrix with the same entries as -3 

⇒  (adj A). B ≠ O

• So, α = - 2 for no solution.

Concept:

Let y = f(a) then the area bounded by the curve, line x =a and x = b is given by 

\(\rm Area = \int_{a}^{b}f(x)dx\)

Calculations: 

Given, the curve \(y=a\sqrt{x}+bx\) passes through the point (1, 2)

Hence, the point (1, 2) will satisfy the curve \(y=a\sqrt{x}+bx\).

⇒2 = a(1) + b(1)

⇒ a + b = 2 ....(1)

Also, given the area bounded by the curve, line x = 4 and x - axis is 8 sq. unit

\(\rm Area = \int_{0}^{4}(a\sqrt x + bx)dx\)

⇒ 8 = \(\rm [a \dfrac {x^{\frac 3 2}}{\frac 3 2} + b \frac{x^2}{2}]_0^4\)

⇒ 8 = \(\rm \dfrac{2a}{3}. 8 + 8b \)

⇒ 2a + 3b = 3....(2)

Solve equation (1) and (2) simultaneously, we get 

a = 3 an d b = -1

Concept:

1. Integration by Substitution:

• If the given integration is of the form \(\smallint {\rm{g}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right){\rm{f'}}\left( {\rm{x}} \right){\rm{dx}}\) where \({\rm{g}}\left( {\rm{x}} \right)\) and \({\rm{f}}\left( {\rm{x}} \right)\) are both differentiable functions then we substitute \({\rm{f}}\left( {\rm{x}} \right) = {\rm{u}}\) which implies that \({\rm{f'}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{du}}\).
• Therefore, the integral becomes \(\smallint {\rm{g}}\left( {\rm{u}} \right){\rm{du}}\) which can be solved by general formulas.
• \(\rm \smallint \sec^2 x\cdot dx = \tan x + c\)

Solution:

Given: \(\rm \smallint \frac {\sec^2 ({ln{(x)}})}{x} dx\)

In the given problem substitute \(\ln \left( {\rm{x}} \right) = u\) therefore, \(\frac{{{\rm{dx}}}}{{\rm{x}}} = {\rm{du}}\).

The given integral becomes

\(\smallint \sec^2 {\rm{u}}\cdot du= \tan {\rm{u}} + {\rm{C}}\)

Resubstitute \({\rm{u}} = \ln \left( {\rm{x}} \right)\).

Calculation:

Given: A = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 4&5 \end{array}} \right]\) and f(t) = t² - 3t + 7

Now,

A² = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 4&5 \end{array}} \right]\)\(\left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 4&5 \end{array}} \right]\) = \(\left[ {\begin{array}{*{20}{c}} -7&{ - 12}\\ 24&17 \end{array}} \right]\)

Also, f(A) = A2 - 3A + 7

⇒ f(A) = \(\left[ {\begin{array}{*{20}{c}} -7&{ - 12}\\ 24&17 \end{array}} \right]\)- 3\(\left[ {\begin{array}{*{20}{c}} 1&{ - 2}\\ 4&5 \end{array}} \right]\)+ 7\(\left[ {\begin{array}{*{20}{c}} 1&0\\ { 0}&{1} \end{array}} \right]\) 

⇒ f(A) = \(\left[ {\begin{array}{*{20}{c}} -3&-6\\ {12}&{9} \end{array}} \right]\)

⇒ f(A) + \(\left[ {\begin{array}{*{20}{c}} 3&6\\ {-12}&{-9} \end{array}} \right]\) = \(\left[ {\begin{array}{*{20}{c}} -3&-6\\ {12}&{9} \end{array}} \right]\) + \(\left[ {\begin{array}{*{20}{c}} 3&6\\ {-12}&{-9} \end{array}} \right]\) 

⇒ f(A) + \(\left[ {\begin{array}{*{20}{c}} 3&6\\ {-12}&{-9} \end{array}} \right]\) = \(\left[ {\begin{array}{*{20}{c}} 0&0\\ { 0}&{0} \end{array}} \right]\)

Concept:

To form the differential equation of the given equation

• Differentiate the equation, the number of times as many as the constants are there.
• Find out the constants in terms of the variables.
• Substitute the variables in the original equation.

Calculation:

Given equation is y = ae-2x + be2x 

There are two constants a and b so differentiate two times

Differentiating w.r.t x, we get

y' = -2ae-2x + 2be2x

Differentiating again w.r.t x

y'' = 4ae-2x + 4be2x

y'' = 4(ae-2x + be2x)

y'' = 4y

y'' - 4y = 0

Calculation:

From the question, the line equation given is:

\(\frac{{x - 4}}{2} = \frac{{y - 5}}{2} = \frac{{z - 3}}{1}\)

Let’s assume the point on the line as ‘P’.

Now,

\(\frac{{x - 4}}{2} = \frac{{y - 5}}{2} = \frac{{z - 3}}{1} = r\)

⇒ x = 2r + 4; y = 2r + 5; and z = r + 3

The point on the line is given as:

P(x, y, z) = (2r + 4, 2r + 5, r + 3)

The point of intersection between the given line and plane is obtained.

On substituting point in the plane,

(2r + 4) + (2r + 5) + (r + 3) = 2

⇒ 5r = -10

∴ r = -2

Now, the point ‘P’ is:

∴ P(0, 1, 1)

Therefore, option (c) is the correct answer.

Concept:

(A + B)2 = (A + B) . (A + B)

⇒ (A + B)2 = A2 +  AB + B2 + BA

Calculations:

Here,

AB = \(\left[ {\begin{array}{*{20}{c}} {\rm{0}}&{\rm{1}}\\ {\rm{1}}&{\rm{0}} \end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}} {\rm{0}}&{{\rm{ - i}}}\\ {\rm{i}}&{\rm{0}} \end{array}} \right]\,{\rm{ = }}\,\left[ {\begin{array}{*{20}{c}} {\rm{i}}&{\rm{0}}\\ {\rm{0}}&{{\rm{ - i}}} \end{array}} \right]\)

And,

BA = \(\left[ {\begin{array}{*{20}{c}} {\rm{0}}&{\rm{-i}}\\ {\rm{i}}&{\rm{0}} \end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}} {\rm{0}}&{{\rm{ 1}}}\\ {\rm{1}}&{\rm{0}} \end{array}} \right]\,{\rm{ = }}\,\left[ {\begin{array}{*{20}{c}} {\rm{-i}}&{\rm{0}}\\ {\rm{0}}&{{\rm{ i}}} \end{array}} \right]\) = - AB

∴ AB + BA = O

⇒ (A + B)² = A² + B²

CONCEPT:

If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then    

   \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0\)

Concept: 

• Inequality: Two real expressions and numbers related by symbols ≥, ≤, < or > are inequalities.
• Examples: x - y ≥ 0 and x + y < 5.
• A linear inequation in two variables represents a line that divides the plane into two halves. A vertical line will divide the plane into left half-plane and right half-plane. A nonvertical line will divide the plane into the upper half-plane and lower half-plane. 

Calculation:

Given:

The pair of inequations are :

x + 2y ≥ 3 ....(1)

x - y ≥ -3 ....(2)

• To draw the line of inequation (1), we need to find two points for the corresponding equation. Putting x = 0 in 1 gives y = 3/2. Similarly putting y = 0, gives x = 3. Hence the two points on the line are (0, 3/2) and (3, 0). Join the points as well as extended them to either side to get a decent outlook of the line.
• Similarly, a line for inequation (2) will be made.  
• The lines of their respective equations are shown in the figure below:

• For x + 2y ≥ 3, the sum of x and y will always be greater than or equal to 3 which is only possible if we choose the region above the x + 2y = 3 line or the upper half-plane.
• Similarly for x - y ≥ 3, x should be greater than y, which means y should be smaller, which is only possible if we take the region below x - y = 3 line or the lower half-plane,
• The overlapped region or the feasible region will represent the region containing the solution to the inequations which is shown in green in the figure below:

So, the correct answer is option (2).

Additional Information

How to draw the inequalities:

• It is important to know the technique to solve linear inequations. Suppose there are two inequations represented by,

x - 2y ≥ 5

x + 2y ≤ 6

• The lines are shown below in the figure:

• For the inequation, x - 2y ≥ 5, the points on the line x - 2y = 5 will only satisfy the equation. The tilted line divides the X-Y plane into an upper and lower half-plane, hence y will be the important parameter since its impact on the inequation will decide which half-plane needs to be chosen to reach the solution. Since we need the value greater than 5 according to the inequation, we have to decrease the value of y and go down. Hence area under the line x - 2y = 5, the lower half-plane as well as points on the line will contain the solution of inequation along with the line as shown in the figure below with blue shaded region.

• Similarly, for inequation x + 2y ≤ 6, the points on equation line x + 2y = 6 will represent will satisfy the equation. Again, the tilted line divide the X-Y plane into the upper and lower half-plane, hence y will be the important parameter. Since we need the value less than 6 according to the inequation, we have to decrease the value of y. Hence area under the line x + 2y = 6, the lower half-plane as well as points on the line will contain the solution of inequation as shown in the figure below with the orange shaded region.a 

• The shaded regions in the above two figures show the solution for individual inequations. The overall solution for both the inequations will be the overlapped area of the shaded regions of the two figures along with a portion of lines in the overlapped regions. The overlapped region is also called the feasible region.

•  It is important to note that the region of the solution is unbounded, which means it will extend when we go further left, right, or downwards under the lines, but only a small portion of it has been shown in the figure due to space constraints.

Concept:

• Any company or business has to spend certain costs for running its business. These costs are classified as fixed costs and variable costs. 
• Fixed Cost: These are the expenses that are independent of the production output units. Even with zero production, this cost has to be paid regularly. 
• Variable Cost: These expenses are dependent on the production output and will increase with an increase in production. Some examples are the cost of raw material, packaging costs, energy consumption per unit cost, etc.

• Cost Function or Total cost (C(x)): It is represented in terms of the sum of variable cost V(x) for producing ‘x’ units and fixed cost k as below:

C(x) = V(x) + k ….(1)  

• Marginal Cost: It is the increase in total cost due to a unit increase in output. It is calculated by deriving the partial derivative of the cost function with respect to output. 

Calculation: 

Given:

• Number of units of milk produced = x units
• Per unit profit, p = 600 - 5x
• Revenue Function,

R(x)  = px

⇒ R(x) = (600 - 5x) × x

⇒ R(x) =600x - 5x²

• Marginal Cost, 

​ \(⇒ \frac{\mathrm{d} {R(x)}}{\mathrm{d} x} = \frac{\mathrm{d} (600x - 5x^{2})}{\mathrm{d} x} = 600 - 10x \)dC(x)dx" id="MathJax-Element-1-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">C(x)dxdC(x)dxdC(x)dx" id="MathJax-Element-10-Frame" role="presentation" style="position: relative;" tabindex="0">$\frac{\mathrm{d}C(x)}{\mathrm{d}x}$ d(x2+10x+12500)dx" id="MathJax-Element-2-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">d(x2+10x+12500)dxd(x2+10x+12500)dx" role="presentation" style="display: inline; position: relative;" tabindex="0">d(x2+10x+12500)dx" id="MathJax-Element-11-Frame" role="presentation" style="position: relative;" tabindex="0">d(x2+10x+12500)dx$\frac{\mathrm{d}(x^2+10x+12500)}{\mathrm{d}x}$dC(x)⇒dC(x)dx=d(5500+3x2)dx" id="MathJax-Element-4-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">dx=d(5500+3x2)dx⇒dC(x)dx=d(5500+3x2)dx" id="MathJax-Element-3-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">" id="MathJax-Element-12-Frame" role="presentation" style="position: relative;" tabindex="0">$\frac{\mathrm{}}{\mathrm{}}\frac{\mathrm{}\mathrm{}\mathrm{}{}^{\mathrm{}}}{\mathrm{}}$ [Math Processing Error]  

• Hence the correct answer is an option (2).

P(A∪B) = P(A) + P(B) – P(A∩B)

Given the two events are independent, so:

P(A∩B) = P(A).P(B)

Let the probability of event A happening be x. The probability of event B happening will also be x. (Given, the two events are equally likely)

P(A∪B) = x + x – x.x

∴ 2x – x² = 0.75

⇒ x² – 2x + 0.75  = 0

10x² – 20x + 7.5 = 0

10x² – 15x – 5x + 7.5 = 0

(5x – 2.5) (2x - 3) = 0

\(\Rightarrow x = \frac{1}{2}~or~\frac{3}{2}\)

As x can not be greater than 1, hence x = 0.5

Concept:

∫ f(x).dx = 1

Mean (μ) = ∫ x.f(x).dx

Calculation:

\({\rm{Mean\;}}\left( \mu \right) = \mathop \smallint \nolimits_0^6 x.f\left( x \right).dx\)

\(= \mathop \smallint \nolimits_0^2 k{x^2}.dx + \mathop \smallint \nolimits_2^4 2kx.dx + \mathop \smallint \nolimits_4^6 x\left( { - kx + 6k} \right).dx\)

\( = k\left[ {\frac{{{x^3}}}{3}} \right]_0^2 + 2k\left[ {\frac{{{x^2}}}{2}} \right]_2^4 - k\left[ {\frac{{{x^3}}}{3}} \right]_4^6 + 6k\left[ {\frac{{{x^2}}}{2}} \right]_4^6\)

\(\mu = \frac{{8k}}{3} + 12k - \frac{{152k}}{3} + 60k\)

\(\mu = 24k = 24 \times \frac{1}{8}\)

∴ μ = 3

Concept:

Binomial distribution The binomial distribution formula is,

:P (X = r) = \({{\bf{n}}_{{{\bf{C}}_{\bf{r}}}}}{{\bf{p}}^{\bf{r}}}{{\bf{q}}^{{\bf{n}} - {\bf{r}}}}\)

P(X) = probability of success on an individual trial

p = probability of “success”, r = number of success, q = probability of “failure” OR q = 1- p

n = number of trials

• \({{\rm{n}}_{{{\rm{C}}_{\rm{r}}}}} = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!{\rm{r}}!}}\)

Calculations:

Let p (fresh egg) \( = \frac{{90}}{{100}} = \frac{9}{{10}} = p\)

P(rotten egg) \( = \frac{{10}}{{100}} = \frac{1}{{10}} = q;\) n = 5, r = 5

So the probability that no egg is rotten,

\( = {}^5{C_5}{\left( {\frac{9}{{10}}} \right)^5}{\left( {\frac{1}{{10}}} \right)^0} = {\left( {\frac{9}{{10}}} \right)^5}.\) 

Concept:

\(\sum\limits_{k = 0}^4 {p(X = K) = 1 } \)

Calculation

\(⇒ \sum\limits_{k = 0}^4 {C{K^2} = 1}\)

⇒ C(1² + 2² + 3² + 4²) = 1

⇒ \(C = \frac{1}{{30}}.\) 

Concept:

Let * be a binary operation on a non-empty set S. If there exists an element e in S such that a * e = e * a = a ∀ a ∈ S. Then the element e is said to be an identity element of S with respect to *.

Calculation:

Given: A = {1, - 1, i, - i} and * is a operation on A such that a * b = ab ∀ a, b ∈ A.

Let e be the identity element of A with respect to *.

As we know that if e is an identity element of a non-empty set S with respect to a binary operation * then a * e = e * a = a ∀ a ∈ S.

Let a ∈ A and because e is the identity element of A with respect to given operation *.

.e a * e = a =  e * a ∀ a ∈ Q - {1}.

According to the definition of *, we have

⇒ a * e = ae = a

⇒ e = 1

Hence, 1 is the identity element of A with respect to given operation *.

Given

The index number for the price of diesel in 2015 was 125 and in 2016, it was 140, on a base year of 2011

Formula

% change = [(Final - initial)/initial] × 100

Calculation

% change = (140 - 125)/125] × 100

⇒ [15/125] × 100

∴ The percent increase in price of diesel from 2015 to 2016 is 12%

Given:

In a race of 200 m, A beats B by 45m or 9 seconds.

Concept used:

Time and Distance

Calculation:

A beats B by 45 m means in constant time t,

A runs 200 m while B runs 155m

Speed of B = (45/9) = 5 m/sec

B covers 155m in same time A covers 200 m

Time taken by B = (155/5) = 31 sec

A takes 31 sec.

Given∶

Joe invests $ 5000.

Jeff invests $ 6000.

Total profit = $ 880

Formula Used∶

Ratio of their investment = ratio of their profit

Calculation∶

Total profit = $ 880

Joe's share for managing the business, i.e.,

⇒ 25/2 % of 880 = 25/2 × 880/100 = $ 110

The remaining profit of Joe and Jeff as per their investment = 880 - 110 = $ 770

Ratio of amounts = 5000 ∶ 6000 = 5 ∶ 6

Sum of ratios = 5 + 6 = 11

Joe's share = 770 × 5/11 = $ 350

Joe's total share = 350 + 110 = $ 460

Jeff's share = 770 × 6/11 = $ 420.

∴ The correct option is (1).

Calculation:

Speed of A in still water = 15km/h

Let rate of stream be x km/h

Distance = d km

Then, rate of downstream = (15 + x) km/h

Rate of upstream = (15 – x) km/h

Time is taken to travel downstream = 2  × times taken to travel upstream

\(\frac{15 + x }{d} = 2 \times \frac{15 - x}{d}\)

⇒ 15 + x = 30 - 2x  

⇒ 3x = 15  

⇒ x = 5 Km/h

∴ Rate of stream = 5 km/hr

∴ The rate of downstream = (15 + 5) km/hr = 20km/hr

According to the question,

40% sugar is in the solution, then,

Quantity of the sugar in the mixture initially-

\(= 40 \times \frac{{300}}{{100}} = 120gm\)

Now, by the condition in the question:

\(\Rightarrow \frac{{120 + x}}{{300 + x}} = \frac{{50}}{{100}}\)

\(\Rightarrow \frac{{120 + x}}{{300 + x}} = \frac{1}{2}\)

⇒ 240 + 2x = 300 + x

⇒ x = 60 gm

Hence, the required answer is 60 gm.

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

• 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: 67 when divided by x, gives the value of remainder as 3.

• Hence x must be a factor of (67-3) = 64. Factors of 64 are {1, 2, 4, 8, 16, 32, 64}.
• Since x < 50, the largest value of x will be 32.
• So, the correct answer is option 3.

Concept:

• Identity Relation:

Let A be a non-empty set then the relation I_A = {(a, a): ∀ a ∈ A} on A is called the identity relation on A.

• Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive ⇔ (a, a) ∈ R, ∀ a ∈ A.

• Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation ⇔ (a, b) ∈ R ⇒ (b, a) ∀ a, b ∈ A.

• Anti–symmetric Relation:

Let R be a relation on a non-empty set A, then the relation R is said to be anti-symmetric

⇔ (a, b) ∈ R and (b, a) ∈ R ⇒ a = b ∀ a, b ∈ A.

• Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation ⇔ (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A.

• Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Note:

The identity relation on a non-empty set is always reflexive, symmetric, anti-symmetric and transitive.

Calculation:

Given: X = {a, b, c} and R is a relation on X such that R = {(a, a), (b, b), (c, c)}

As we can see that, (x, x) ∈ R ∀ x ∈ X ⇒ R is an identity relation on X.

We know that, the identity relation on a non-empty set is always reflexive, symmetric, anti-symmetric and transitive.

Concept:

A set which does not contain any element is called an empty set and it is denoted by ϕ.

Hence, option 4 is correct.