Mathematics Mock Test - 8

Explanation:

Option 1 ) Two vectors are collinear only if they are parallel to the same line whether their magnitudes and directions are the same or not.   FALSE

Option 2 ) Two collinear vectors don't necessarily have the same magnitudes.   FALSE

Option 3 ) Collinear vectors are parallel to the same line but may be opposite in directions hence may be unequal.  FALSE

Option 4 ) \(\overrightarrow a \)and -\(\overrightarrow a \) are opposite in directions but parallel to the same line, hence collinear.   TRUE

Concept:

1. Integration by parts: Integration by parts is a method to find integrals of products

The formula for integrating by parts is given by;

\(\Rightarrow \smallint {\rm{u\;vdx}} = {\rm{u\;}}\smallint {\rm{vdx}} - {\rm{\;}}\smallint \left( {\frac{{{\rm{du}}}}{{{\rm{dx}}}}{\rm{\;}}\smallint {\rm{vdx}}} \right){\rm{dx}}\)

Where u is the function u(x) and v is the function v(x)

2. ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

Calculation:

\(\rm \text {Let I} = \mathop \smallint \limits_1^e \sqrt x \ln \left( x \right)dx\\= \mathop \smallint \limits_1^e \ln \left( x \right) \sqrt x dx\)

Applying by parts rule, we get

\(\rm = \left[ {\ln \left( x \right) \times \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_1^e - \smallint \left[ {\frac{1}{x} \times \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]dx\)

\(\rm = \frac{2}{3} \left[ {\ln \left( x \right) \times x^{\frac{3}{2}} } \right]_1^e -\frac{2}{3} \smallint x^{\frac{1}{2}}dx\)

\(\rm = \left[ {\ln \left( x \right) \times {x^{\frac{3}{2}}} \times \frac{2}{3} - \frac{4}{9} \times {x^{\frac{3}{2}}}} \right]_1^e\)

\(= \frac{2}{9}\sqrt {{e^3}} + \frac{4}{9}\)

Concept:

Conditions for the subtraction of matrices:

Two matrices should be of the same order (number of rows = number of columns).

Add the corresponding element of other matrices.

Calculation:

Given:

\(f(x) = \left[ {\begin{array}{*{20}{c}} x&λ \\ {2λ }&x \end{array}} \right]\)

\(\Rightarrow f(λ x) = \left[ {\begin{array}{*{20}{c}} {λ x}&λ \\ {2λ }&{λ x} \end{array}} \right]\)

\(f(λ x) - f(x) = \left[ {\begin{array}{*{20}{c}} λ x &λ \\ {2λ }&{λ x} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} x&λ \\ {2λ }&x \end{array}} \right]\)

\(f(λ x) - f(x) = \left[ {\begin{array}{*{20}{c}} λ x -x &0\\ {0 }&{λ x - x} \end{array}} \right] \)

|f(λx) - f(x)| = (λx - x)²

|f(λx) - f(x)| = x²(λ - 1)²

Given:

\((\overrightarrow a + \overrightarrow b ).(\overrightarrow a - \overrightarrow b ) = 8\ \)

\(\left| {\overrightarrow a } \right| = 8\left| {\overrightarrow b } \right|\)

Calculation:

We have,

\((\overrightarrow a + \overrightarrow b ).(\overrightarrow a - \overrightarrow b ) = 8\ \)

⇒ \(\vec a.\vec a \ - \vec a . \vec b + \vec b . \vec a - \vec b. \vec b = 8 \)

⇒  \(\left|\vec a \right|^2 - \vec a .\vec b + \vec a . \vec b - \left|\vec b \right|^2 = 8\)        [∵ \(\vec b. \vec a = \vec a . \vec b\)]

⇒ \(\left| \vec a \right|^2 - \left| \vec b \right|^2 = 8\)

⇒ \((8\left| \vec b \right|)^2 - \left| \vec b \right|^2 = 8\)       [\(\left| {\overrightarrow a } \right| = 8\left| {\overrightarrow b } \right|\)]

⇒ \(64 \left| \vec b \right|^2 - \left| \vec b \right|^2 = 8\)

⇒ \(63 \left| \vec b \right|^2 = 8\)

⇒ \(\left| \vec b \right|^2 = \frac{8}{63}\)

⇒ \(\left| \vec b \right| = \sqrt {\frac{8}{63}} \)

⇒ \(\left| \vec b \right| = { \frac{2\sqrt2}{3\sqrt7}} \)

⇒ \(\left| {\overrightarrow a } \right| = 8\left| {\overrightarrow b } \right|\)

Put the value of \(\vec b \)

⇒ \(\left| \vec a \right| = 8 \times \frac{2\sqrt2}{3\sqrt7}\)

⇒ \(\left| \vec a \right| = \frac{16\sqrt2}{3\sqrt7}\) 

∴ \(\left| \vec a \right| = \frac{16\sqrt2}{3\sqrt7}\)  and \(\left| \vec b \right| = { \frac{2\sqrt2}{3\sqrt7}} \)

Calculation:

Given:

\(⇒ {\cos ^{ - 1}}\frac{3}{5} - {\sin ^{ - 1}}\frac{4}{5} = {\cos ^{ - 1}}x\)

\(⇒ {\cos ^{ - 1}}\frac{3}{5} - {\cos ^{ - 1}}\sqrt {1 - \frac{{16}}{{25}}} = \cos \)

\(⇒ {\cos ^{ - 1}}\frac{3}{5} - {\cos ^{ - 1}}\frac{3}{5} = {\cos ^{ - 1}}x\)

⇒ cos^-1 x = 0 ⇒ x = 1

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:

Minor of -4 = \(\left| {\begin{array}{*{20}{c}} { - 2}&3\\ -8&9 \end{array}} \right|\) = 6,

Minor of 9 = \(\left| {\begin{array}{*{20}{c}} { - 1}&-2\\ -4&-5 \end{array}} \right|\) = - 3

And cofactor of -4 = (-1)²⁺¹ (6) = -6

Cofactor of 9 = (-1)³⁺³ (-3) = -3.

Concept:

De-Morgan’s Laws: For any two sets A and B, we have

• (A ∪ B)’ = A’ ∩ B’
• (A ∩ B)’ = A’ ∪ B’

Power set: Let A be a non-empty set. Then, \(P\left( A \right) = \left\{ {B\;|\;B\; \subseteq A} \right\}\) is called as the power set of A.

 For any non-empty set A with n(A) = x. Then the total no. of subsets of set A is given by: 2^x

i.e n(P(A)) = 2^x.

Calculation:

Given:  n (U) = 1000, n (X) = 300, n (Y) = 400 and n (X ∩ Y) = 200

As we know that, n (X ∪ Y) = n (X) + n (Y) – n (X ∩ Y)

⇒ n (X ∪ Y) = 300 + 400 – 200 = 500

According to De-Morgans law : (X ∪ Y)’ = X ‘ ∩ Y’

⇒ n [(X ∪ Y)’] = n (X ‘ ∩ Y’)

We know that, n [(X ∪ Y)’] = n (U) – n (X ∪ Y)

⇒ n [(X ∪ Y)’] = 1000 – 500 = 500

∴ n (X ‘ ∩ Y’) = 500

As we know that, for any non-empty set A with n(A) = x. Then the total no. of subsets of set A is given by: 2^x i.e n(P(A)) = 2^x

⇒ n (P (X‘∩ Y’)) = 2⁵⁰⁰

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 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: 2kx - 2y + 3z = 0, x + ky + 2z = 0, 2x + kz = 0

For the homogeneous system of equations,

⇒ (adj A). B = 0 for all value of k and the column matrix B is zero matrix.

So, for non trivial solution,

|A|  = 0 

⇒ 2k(k²) + 2(k - 4) + 3(-2k) = 0

⇒ k³ - 2k - 4 =0 

⇒  (k - 2)(k² + 2k + 2) = 0 

∴ k = 2 (Real value).

Concept:

If a Matrix is nor - singular then its determinant is non - zero (\(\left| A \right| \neq 0\))

Calculation:

Given: Matrix is singular

\(\left[ {\begin{array}{*{20}{c}} {{\rm{ - x}}}&{\rm{x}}&{\rm{2}}\\ {\rm{2}}&{\rm{x}}&{{\rm{ - x}}}\\ {\rm{x}}&{{\rm{ - 2}}}&{{\rm{ - x}}} \end{array}} \right]\) ≠ 0

⇒ -x(- x² - 2x) - x(x² - 2x) + 2(-4 - x²) ≠ 0

⇒ 2x² - 8 ≥ 0 

⇒ x² ≥ 4

∴ x ≠ ± 2. 

Calculation:

y = f(f(f(x))) + (f(x))² 

\(\frac{{dy}}{{dx}}=f'(f(f(x)))f'(f(x))f'(x)+2f(x)f'(x)\)

= f'(1)f'(1)f'(1) + 2f(1)f'(1)

= 3 × 3 × 3 + 2 × 1 × 3

= 27 + 6 = 33

Concept:

\(\int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx = \int_{a}^{c} f(x) dx\)

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

Calculation:

Given, \(\displaystyle \int_{-3}^2 f(x)\:dx=\frac{7}{3}\:and\:\displaystyle \int_{-3}^9 f(x)\:dx=-\frac{5}{6}\)

Using property,

\(\displaystyle \int_{-3}^2 f(x)\:dx + \int_{2}^9 f(x)\:dx= \int_{-3}^9 f(x)\:dx\)

⇒ \({7 \over 3} + \int_{2}^9f(x) dx =- {5\over 6}\)

⇒ \(\displaystyle \int_2^9f(x)\:dx =- {5 \over 6} - {7 \over 3}\)

⇒ \(\displaystyle \int_2^9f(x)\:dx = - {19 \over 6}\)

∴ The correct answer is option (1).

Concept:

Equation of a line in cartesian form:- \(\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}\)

Equation of a line in vector form:- \(\vec r = \vec a + \lambda \vec b\) 

where, \(\vec a = x_1 \hat{i} + y_1\hat j + z_1 \hat k\) and \(\vec b = a \hat i + b \hat j + c \hat k\)

Calculation:

We have,

Cartesian equation:- \(\frac{{x + 5}}{3} = \frac{{y - 7}}{2} = \frac{{z + 3}}{2}\)

⇒ \(\frac{{x - (-5)}}{3} = \frac{{y - 7}}{2} = \frac{{z - (-3)}}{2}\)      ------- equation (1)

⇒ \(\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}\)           -------- equation (2)

Comparing (1) and (2)

⇒ x₁ = -5, y₁ = 7, z₁ = -3

⇒ a = 3, b = 2, c = 2

⇒ \(\vec a = -5\hat i + 7\hat j - 3 \hat k\)

⇒ \(\vec b = 3 \hat i + 2 \hat j + 2 \hat k\)

Equation of line in vector form:- \(\vec r = \vec a + \lambda \vec b\)

⇒ \(\vec r = (-5\hat i + 7\hat j - 3\hat k)+ \lambda (3\hat i + 2\hat j + 2\hat k)\)

∴ The vector equation for the line is \(\vec r = (-5\hat i + 7\hat j - 3\hat k)+ \lambda (3\hat i + 2\hat j + 2\hat k)\) 

Given:

Equation of the planes:- \(\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6\) and \(\vec{r}. (2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})=-5\)

Concept:

The vector equation of a plane passing through the intersection of planes \(\vec r . \vec{n_1} = d_1\) and \(\vec r . \vec{n_2} = d_2\) and also through the point (x₁, y₁, z₁) is \(\vec r( \vec{n_1} + λ \vec{n_2}) = d_1 + λ d_2\)

Calculation:

The plane passes through

⇒ \(\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6\)

Comparing it with \(\vec r . \vec{n_1} = d_1\),

⇒\(\)\(\vec{n_1} = \hat i + \hat j + \hat k \ and \ d_1 = 6\)

⇒ \(\vec{r}. (2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})=-5\)

⇒ \(- \vec r .(2\hat i + 3 \hat j + 4 \hat k) = 5\)

⇒ \(\vec r.(-2 \hat i - 3 \hat j - 4 \hat k) = 5\)

Comparing it with \(\vec r . \vec{n_2} = d_2\)

⇒ \(\vec{n_2} = - 2 \hat i - 3 \hat j - 4 \hat k \ and \ d_2 = 5\)

Now, putting the value of n₁, n₂, d₁ and d₂

Equation of plane is

⇒ \(\vec r[ (\hat i + \hat j + \hat k) + λ (- 2 \hat i - 3 \hat j - 4\hat k)] = 6 + λ 5\)

⇒ \(\vec r[ (\hat i + \hat j + \hat k)- λ (2 \hat i + 3 \hat j + 4\hat k)] = 6 +5 λ \)     ------ equation (1)

Now, Put \(\vec r = x\hat i + y \hat j + z \hat k\)

⇒ \((x \hat i + y \hat j + z \hat k).[ (\hat i + \hat j + \hat k)- λ (2 \hat i + 3 \hat j + 4\hat k)] = 5 λ + 6 \)

⇒ \((x \hat i + y \hat j + z \hat k).(\hat i + \hat j + \hat k)- λ (x\hat i + y \hat j + z \hat k).(2 \hat i + 3 \hat j + 4\hat k) = 5 λ + 6 \)

⇒ \((x × 1) + ( y × 1) + (z × 1) - λ[(x × 2 )+ ( y × 3) + (z × 4)] = 5 λ + 6 \)

⇒ x + y + z - λ[2x + 3y + 4z] = 5λ + 6

⇒ x + y + z - 2λx - 3λy -4λz = 5λ + 6

⇒ (1 - 2λ)x + (1 - 3λ)y + (1 - 4λ)z = 5λ + 6 ------- equation (2)   

Since the plane passes through (1, 1, 1)

Putting (1, 1, 1) in equation (2)

⇒ (1 - 2λ)x + (1 - 3λ)y + (1 - 4λ)z = 5λ + 6

⇒ (1 - 2λ) × 1 + (1 - 3λ) × 1 + (1 - 4λ) × 1 = 5λ + 6

⇒ 1 - 2λ + 1 - 3λ + 1 - 4λ = 5λ + 6

⇒ 3 - 9λ = 5λ + 6

⇒ -14λ = 3

⇒ \(λ = \frac{-3}{14}\)

Putting the value of λ in equation (1)

⇒ \(\vec r[ (\hat i + \hat j + \hat k)- (\frac{-3}{14}) (2 \hat i + 3 \hat j + 4\hat k)] = 6 +5 \times \frac{-3}{14}\)

⇒ \(\vec r[ (\hat i + \hat j + \hat k)+ (\frac{3}{14}) (2 \hat i + 3 \hat j + 4\hat k)] = 6 - \frac{15}{14}\)

⇒ \(\vec r[ \hat i + \hat j + \hat k+ \frac{6}{14} \hat i + \frac{9}{14} \hat j + \frac{12}{14}\hat k] = \frac{69}{14}\)

⇒ \(\vec r [(1 +\frac{6}{14}) \hat i + ( 1 +\frac{9}{14}) \hat j + (1 +\frac{12}{14})\hat k] = \frac{69}{14}\)

⇒ \(\vec r [\frac{20}{14} \hat i + \frac{23}{14} \hat j + \frac{26}{14}\hat k] = \frac{69}{14}\)

⇒ \(\vec r [\frac{1}{14} (20 \hat i + 23 \hat j + 26\hat k)] = \frac{69}{14}\)

⇒ \(\frac{1}{14} \vec r.(20 \hat i + 23 \hat j + 26\hat k) = \frac{69}{14}\)

⇒ \(\vec r.(20 \hat i + 23 \hat j + 26 \hat k) = 69\)

∴ The vector equation of the required plane is \(\vec r.(20 \hat i + 23 \hat j + 26 \hat k) = 69\)    

Given:

Equation of plane is \({\vec r} \cdot(2 \hat{\imath}-6 \hat {\jmath}-3 {\hat k})+1=0\)

Concept:

Vector equation of a plane at a distance 'd' from the origin and unit vector to normal form origin \(̂ n\) is \(\vec r. \hat n = d\)

Formula Used:

\(\vec n = \hat n = \frac{1}{|\vec n|}(\vec n)\)

Calculation:

We have,

⇒ \(⃗{r} \cdot(2 \hat{\imath}-6 \hat{\jmath}-3 \hat{k})+1=0\)

⇒ \(⃗{r} \cdot(2 \hat{\imath}-6 \hat{\jmath}-3 \hat{k}) = -1\)

Multiplying with -1 on both sides,

⇒ \(-\vec {r} \cdot(2 \hat{\imath}-6 \hat{\jmath}-3 \hat{k}) = -1 \times -1 \)

⇒ \({\vec r} \cdot(-2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k}) = 1\)     ----- equation (1)

So, \(\vec n = - 2 \hat i + 6 \hat j + 3 \hat k\)

Magnitude of \(\vec n\) = \(\sqrt{ (-2)^2 + 6^2 + 3^2}\)

⇒ \(|\vec n|\) = \(\sqrt{4 + 36 + 9}\) = \(\sqrt{49}\) = 7

Now, \(\hat n = \frac{1}{|\vec n|}(\vec n)\)

⇒ \(\hat n = \frac{1}{7}(- 2 \hat i + 6 \hat j + 3 \hat k)\)

⇒ \(\vec n = \frac{-2}{7} \hat i + \frac{6}{7} \hat j + \frac{3}{7} \hat k\)

∴ Direction cosines of the unit vector perpendicular to the given plane are \(\frac{-2}{7}, \frac{6}{7}, \frac{3}{7}\)

Concept:

If a plane is parallel to a line then the normal vector to the plane will be perpendicular to the direction of the line.

The direction of the line,

\(\vec b = (2,2,a)\)

The vector perpendicular to the given plane  2x - y + z = 0,

\(\vec \eta = (2,-1,1)\)

Calculation:

Given:

Since, the plane 2x - y + z = 0 is parallel to the line \(\frac{{x - 1/2}}{1} = \frac{{y - 2}}{{ - 2}} = \frac{{z + 1}}{a}\)

⇒ 2 (1) + (-1) (-2) + 1 (a) = 0 

⇒ a + 4 = 0 ⇒ a = - 4.

Concept:

• A(0, b, c) in yz - plane and B(a, 0, c) in zx - plane through O (origin) is px + qy + rz = 0.
• Since, It passes through A and  B. 

∴ 0p + qb + rc = 0 and pa + 0q + rc = 0 

Put the value of p and q in terms of r, a, b, and c to the equation of the plane,

⇒ q = -rc/b, p = -rc/a

⇒ (-rc/a)x + (-rc/b)y + rz = 0.

Required plane is,

⇒  bcx + cay - abz = 0.

Concept:

1. Let the given two position vectors p and q respectively and r be the vector dividing the segment PQ internally in the ratio m: n

1. Internal Section Formula: When the segment PQ is divided internally in the ratio m: n, we use this formula. ⇔ \({\rm{r}} = \left( {\frac{{{\rm{mq\;}} + {\rm{\;np}}}}{{{\rm{m}} + {\rm{\;n}}}}} \right)\)
2. External Section Formula: When the segment PQ is divided externally in the ratio m: n, we use this formula. ⇔ \({\rm{s}} = \left( {\frac{{{\rm{mq}} - {\rm{np}}}}{{{\rm{m}} - {\rm{\;n}}}}} \right)\)

Calculation:

Let the required ratio is k :1 then the point suing section formula of the line joining (-2, 4, 7) and (3, -5, 8),

\((\frac{3k -2}{k+1}, \frac{-5k +4}{k+1}, \frac{8k +7}{k+1})\)

The above point will lie on the plane so,

\(⇒1(\frac{3k -2}{k+1}) -2(\frac{-5k +4}{k+1}) +3 (\frac{8k +7}{k+1}) = 17\)

⇒ k = 3/10

Concept: 

• Two positive integers a and b are said to be congruence modulo m (m ≠ 0) if, a mod m = b mod m or (a-b) is divisible by m. The condition is denoted by the following:

a ≡ b(mod m) ….(1)

Calculation:

Given: 73 ≡ x mod 13 where 50 < x < 65

• On Comparing this with equation (1), we get,

a = 73, b = x and m = 13.

• As per the definition of congruence modulo and given information, 73 mod 13 = x mod 13 or (73 - x) is divisible by 13.
• Since x lies between 50 and 65, the only number that can satisfy the above criteria is 60, since (73 - 60) = 13 which is divisible by 13.
• Hence the value of x will be 60.
• So, the correct answer is option 1

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 + 8y needs to be maximized subject to the constraints,

x, y ≥ 0 ....(1), 2x + 3y ≥ 10 ....(2) and 5x + 2y ≥ 14 ....(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 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 5x + 2y = 14 with the y-axis where x = 0. At x = 0, y is equal to 7. Hence the corner point is (0, 7)
• One point is the intersection of line 2x + 3y = 10 with an x-axis where y = 0. At y = 0, x is equal to 5. Hence the corner point is (5, 0)
• Final point is the intersection point of lines 2x + 3y = 10 and 5x + 2y = 14. Solving the two equations. we get x = 2 and y = 2. The final corner point is (2, 2)
• 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 = 5x + 8y
  (0, 7)	56
  (5, 0)	25
  (2, 2)	26

• It is clear that the feasible region is unbounded.
• The maximum value of the objective function is obtained at (0,7).
• Since the feasible region is unbounded, it is necessary to confirm whether (0, 7) is the actual solution to the problem.
• Line 5x + 8y = 56 will be drawn along with the feasible region and the following figure will be obtained:

• We see that the line 5x + 8y = 56 passes through the corner point (0, 7).
• Since the line lies inside the feasible region, it can move upwards to achieve the maximum value of the objective function.
• Hence the solution of the LPP is unbounded. 
• So, the correct answer is option 2.  

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.

Given:

A die is thrown 3 times.

Events A and B stated:-

4 on the 3rd throw.

6 on the 1st and 5 on the 2nd throw.

Formula Used:

P(A|B) = P(A ∩ B)/P(B)

Calculation:

A die is thrown 3 times.

Total cases = 6 × 6 × 6 =216

A : 4 on the third throw.

B : 6 on the first and 5 on the second throw.

A = {(1, 1, 4), (1, 2, 4),....(1, 6, 4), (2, 1, 4), (2, 2, 4),....(2, 6, 4),

        (3, 1, 4), (3, 2, 4),....(3, 6, 4), (4, 1, 4), (4, 2, 4),....(4, 6, 4)

        (5, 1, 4), (5, 2, 4),.....(5, 6, 4), (6, 1, 4), (6, 2, 4),....(6, 6, 4)}

⇒ P(A) = \(\frac{36}{216}\)

B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}

⇒ P(B) = \(\frac{6}{216}\)

Thus A ∩ B = {(6, 5, 4)}

So, P(A ∩ B) = \(\frac{1}{216}\)

Now, P(A|B) = \(\frac{P(A ∩ B)}{P(B)}\)

⇒ P(A|B) = \(\frac{\frac{1}{216}}{\frac{6}{216}}\)

⇒ P(A|B) = \(\frac{1}{6}\)

∴ The probability of A knowing that B has already taken place is 1/6

Concept:

I. Greatest Integer Function: (Floor function)

The function f (x) = [x] is called the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x.

Domain of [x] is R and range is I.

II. \(\mathop \smallint \nolimits_a^b f\left( x \right)\;dx = \;\mathop \smallint \nolimits_a^c f\left( x \right)\;dx + \;\mathop \smallint \nolimits_c^b f\left( x \right)\;dx,\;where\;a < c < b\)

Calculation:

Given: \(g\left( x \right) = x \times \left[ {\frac{1}{x}} \right]\)

For the given integral \(\mathop \smallint \nolimits_{\frac{1}{3}}^1 g\left( x \right)\;dx\), when x lies between 1 / 3 and 1 / 2 i.e \(\frac{1}{3} < x < \frac{1}{2}\)

\(\Rightarrow 2 < \frac{1}{x} < 3\)

So, according to the above inequality: \(g\left( x \right) = x \times \left[ {\frac{1}{x}} \right] = 2x\)

Similarly for the given integral \(\mathop \smallint \nolimits_{\frac{1}{3}}^1 g\left( x \right)\;dx\), when x lies between 1/ 2 and 1 i.e \(\frac{1}{2} < x < 1\)

\(\Rightarrow 1 < \frac{1}{x} < 2\)

So, according to the above inequality: \(g\left( x \right) = x \times \left[ {\frac{1}{x}} \right] = x\)

As we know that, \(\mathop \smallint \nolimits_a^b f\left( x \right)\;dx = \;\mathop \smallint \nolimits_a^c f\left( x \right)\;dx + \;\mathop \smallint \nolimits_c^b f\left( x \right)\;dx,\;where\;a < c < b\)

\(\Rightarrow \;\mathop \smallint \nolimits_{\frac{1}{3}}^1 g\left( x \right)\;dx = \;\mathop \smallint \nolimits_{\frac{1}{3}}^{\frac{1}{2}} 2x\;dx + \;\mathop \smallint \nolimits_{\frac{1}{2}}^1 x\;dx\)

\(\Rightarrow \;\mathop \smallint \nolimits_{\frac{1}{3}}^1 g\left( x \right)\;dx = \frac{2}{2} \times \left( {\frac{1}{4} - \frac{1}{9}} \right) + \frac{1}{2} \times \left( {1 - \frac{1}{2}} \right) = \frac{{37}}{{72}}\)

Concept:

• Revenue Function (R(x)): It is represented in terms of the product of per unit selling price p and the number of units sold x as below:

R(x) = px ….(1)  

• Marginal Revenue: It is the increase in revenue due to a unit increase in output. It is calculated by deriving the partial derivative of the revenue function with respect to output.  
• 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 ….(2)  

• 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.

Marginal Profit =  Marginal Revenue - Marginal Cost  ….(3)

• Total profit reaches its maximum value when marginal revenue becomes equal to marginal cost, ie Marginal Cost = Marginal Revenue

Explanation: 

Given:

• Marginal cost indicates an increase in the cost when production is increased by 1 unit.
• Marginal revenue indicates an increase in the revenue when selling is increased by 1 unit.
• Below is the example of a curve where marginal revenue and marginal cost are plotted:  

• From the graph, in the region where the marginal cost curve is below the marginal revenue line, marginal cost < marginal revenue, an increase in cost by producing 1 more unit will be lower than the increase in revenue by selling 1 more unit. In this case, more units need to be produced to increase the overall profit until the marginal cost becomes greater than the marginal revenue.
• Hence, the correct answer is an option (2).

Additional Information  

• 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.

Concept:

Simple Events:- If an event E has only one sample point of a sample space, i.e., a single outcome of an experiment it is called a Simple or Elementary event.

Formula Used:

Probability of an event occurring = (No. of favorable outcomes)/(Total no. of outcomes)

Concept:

Here, 3 coins are tossed simultaneously so the total number of possible outcomes,

n(s) = 2³ = 8

The all possible outcomes are,

{HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

Calculation:

Number of outcomes when first Head comes,

n(H I^st) = 4

n (H ^st ∩ 2 consecutive Head) = 2

For the probability of two head-on-three coins when the first head has already appeared,

⇒ P(consecutive 2H / 1st Head) = \(\left[\frac{2}{4}\right]\) = 1/2

Let the speed of P be 5 m/s and that of Q be 4 m/s. Since they are running in the same direction, we have

Relative speed = 5 – 4 = 1 m/s

Hence, time taken to meet for the first time = 800 / 1 = 800 s

Hence, P and Q meet every 800 seconds.

Total time taken by P in completing the race = 10000/5 = 2000 s

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:

\(f(x) = \int_{ - 10}^x {(t^4 - 4){e^{ - 4t}}dt} \)

\(⇒ f'(x) = ({x^4} - 4){e^{ - 4x}}\)

Now,

\( f'(x) = 0\)

\( ⇒ x = \pm \sqrt 2 \pm \sqrt 2 \)

\(f''(x) = - 4({x^4} - 4){e^{ - 4x}+}4{x^3}{e^{ - 4x}}\)

⇒ At \(x = \sqrt 2 \) and \(x = -\sqrt 2 \) the given function has extreme value.

Concept:

All possible combinations of two children in four families,

B B, BG, G B, G G

Calculation:

The first child is a boy → (BB, BG)

if the first child is a boy then the other is also a boy → (B B)

P (Ist B / 2^st B) = \(\left[\frac{1}{2}\right]\)

Concept:

The equation of the tangent is \(\frac{{y\; - {{\rm{y}}_1}}}{{x - {{\rm{x}}_1}}} = m\) where m is the slope of the tangent

The normal line is a line that’s perpendicular to the tangent line and passes through the point of tangency.

The equation of the normal is \(\frac{{y\; - {{\rm{y}}_1}}}{{x - {{\rm{x}}_1}}} = - \frac{1}{m}\)

Calculation:

Given:

Let P(h, k) be a point on the curve y = x2 + 7x + 2, nearest to the line y = 3x - 3 then the tangent to the curve at P will be parallel to  y = 3x - 3

A tangent at P(h, k) will be parallel to given the 

\({\left. {\frac{{dy}}{{dx}}} \right|_{\left( {h,k} \right)}} = 2h + 7 = 3 \Rightarrow h = - 2\)

Point P lies on curve k = (-2)² - 7 × 2 + 2 = -8

Normal at P(-2, -8), Normal slope = \(- \frac{1}{3}\)

∴ x + 3y + 26 = 0

CONCEPT:

• Among the four components of the time series, the secular trend analysis (also known as trend analysis) depicts the long-term direction of the series.
• One of the most widely used in practice mathematical techniques of finding the trend values is the method of least squares.
• It plays an important role in finding the trend forecasts for the future economic and business time series data

The methods used to analyze the simple trend of the time series data are:

1. Graphical method
2. Semi averages method
3. Moving average method
4. Method of least square

EXPLANATION:

Trend analysis by moving average method:

• This method is used to draw a smooth curve for time series data.
• It is mostly used for eliminating the seasonal variations for a given variable.
• The moving average method helps to establish a trend line by eliminating the cyclical, seasonal, and random variations present in the time series.
• The period of the moving average depends upon the length of the time series data.
• As shown in the figure below, the red smooth curve is the trend cycle, which is noticeably smoother than the original data and captures the main movement of the time series ignoring the minor fluctuations.
• The order of the moving average determines the smoothness of the trend-cycle estimate.

Computation of Straight-line trend by using Method of Least squares:

• The method of least squares is a technique for finding the equation which best fits a given set of observations.
• In this technique, the sum of squares of deviations of the actual and computed values is the least and eliminates personal bias.

​

Calculation:

Let sin^-1 a = A, sin^-1 a = A, sin^-1 b = B, sin^-1 c = C

∴ sin A = a, sin B = b, sin C = c and A + B + C = π, then

We know the trigonometric identity,

⇒ sin 2A + sin 2B + sin2C = 4 sin A sin B sin C    .....(i)

LHS:

⇒ 2 sin (A + B). cos (A-B) + 2sin C cos C  

⇒ 2 sin (π - C). cos (A-B) + 2sin C cos C  

⇒ 2 sin C. cos (A-B) + 2sin C cos C  

⇒ 2 sin C (cos (A-B) + cos C)

⇒ 2 sin C (cos (A-B) + cos (π -(A+B))

⇒ 2 sin C (cos (A-B) - cos((A+B))

⇒ 2sin C (2sin A sin B)

⇒ 4 sin A sin B sin C = RHS

⇒ sin 2A + sin 2B + sin2C = 4 sin A sin B sin C

\(⇒ \sin A\cos A + \sin B\cos B + \sin C\cos C\)

= 2sin A sin B sin C

\(⇒ \sin A\sqrt {\left( {1 - {{\sin }^2}A} \right)} + \sin B\sqrt {(1 - {{\sin }^2}B)} + \sin C\sqrt {1 - {{\sin }^2}C} \)

\(= 2\sin A\sin B\sin C\)

\(⇒ a\sqrt {(1 - {a^2})} + b\sqrt {(1 - {b^2})} + c\sqrt {(1 - {c^2})} = 2abc,\)

Given:

Pipe P is 5 times fater than pipe Q

P can fill the cistern in 30 minutes

Calculation:

P = 5Q (Efficiency)

Pipe P's 1 minute work = 1/30

Pipe Q’s 1 minute work = 1 /150 

If both pipes open

Pipe (P + Q)’s 1 minute work = 1/30 + 1/150 = 6/150

= 1/25

∴ It will take 25 minute to fill the tank.

Calculation:

The vector perpendicular to a and b is

\(\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} \hat i&\hat j &\hat k\\ 1&1&0\\ 0&1&1 \end{array}} \right| = \hat i - \hat j + \hat k\)

Since the length of this vector is\(\sqrt 3 \) , the unit vector,

Perpendicular to a and b is \( \pm \frac{{a \times b}}{{\left| {a \times b} \right|}} = \pm \frac{1}{{\sqrt 3 }}(i - j - k)\)

Hence there are two such vectors.

Concept:

• If matrix A = [aij] then the adj A matrix is the transpose of the matrix [Aij] is the cofactor of the element aij.
• We can do this problem by property also as,

A. Adj A = |A|I

Calculation:

Given:

\([A = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&2\\ 0&2&{ - 3}\\ 3&{ - 2}&4 \end{array}} \right];adj\,A = \left[ {\begin{array}{*{20}{c}} {{C_{11}}}&{{C_{12}}}&{{C_{13}}}\\ {{C_{21}}}&{{C_{22}}}&{{C_{23}}}\\ {{C_{31}}}&{{C_{32}}}&{{C_{33}}} \end{array}} \right]^T\)

C₁₁ = 8 - 6 = 2, C12 = -(0 + 9) = - 9, C13 = 0 - 6 = - 6

C₂₁ = - (-8 + 4) = 4, C₂₂ = 4 - 6 = - 2, C23 = (-2 + 6) = - 4, 

C31 = + 6 - 4 = 2,C₃2 = (- 3,- 0) = 3, C₃3 = 2 - 0 = 2

\(\therefore adjA = {\left[ {\begin{array}{*{20}{c}} 2&{ - 9}&6\\ 4&{ - 2}&{ - 4}\\ 2&3&2 \end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}} 2&4&2\\ { - 9}&{ - 2}&3\\ { - 6}&{ - 4}&2 \end{array}} \right]\)

Now, multiply matrix A with Adj A to get final result which will be,

\(\therefore \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&2\\ 0&2&{ - 3}\\ 3&{ - 2}&4 \end{array}} \right] .Adj A= \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&2\\ 0&2&{ - 3}\\ 3&{ - 2}&4 \end{array}} \right] \left[ {\begin{array}{*{20}{c}} 2&4&2\\ { - 9}&{ - 2}&3\\ { - 6}&{ - 4}&2 \end{array}} \right] \)

\(A. Adj (A)= \left[ {\begin{array}{*{20}{c}} 8&0&0\\ 0&8&0\\ 0&0&8 \end{array}} \right]\)

Alternate MethodA (adj A) = \(\left| A \right|\) I = 81 =\(\left[ {\begin{array}{*{20}{c}} 8&0&0\\ 0&8&0\\ 0&0&8 \end{array}} \right]\).

Concept:

|x| = x, for x > 0 and |x| = -x fo x < 0

Area of square = (side)²

Calculation:

For x < 0, y = -x - 1 and y = x + 1

y = -x - 1, when x = 0, y = -1 ⇒ (0, -1) and when y = 0, x = -1 ⇒ (-1, 0)

And y = x + 1, when x = 0, y = 1 ⇒ (0, 1) and y = 0, x = -1 ⇒ (-1, 0)

For x > 0, y = x - 1 and y = -x + 1

y = x - 1, when x = 0, y = -1, (0, -1) and when y = 0, x = 1 ⇒ (1, 0)

And, y = -x + 1, when x = 0, y = 1, (0, 1) and when y = 0, x = 1 ⇒ (1, 0)

We get square, now length of side = distance between any two points

Consider points (1, 0) and (0, 1)

Distance = \(\sqrt{{(1-0)^2+(0-1)^2}}=√2\)

Area of square = (√ 2)² 

= 2 sq units

Hence, option (3) is correct.

Concept:

Solving logarithmic inequalities, it is important to understand that the direction of inequality changes if the base of the logarithmic is less than 1.

If, log_a x < log_a y

Then,

• x < y, when 'a' is more than 1.

• x > y, when 'a' is less than 1. 

Formula used:

log x + log y = log xy

• log x^a = a.log x
• log (x/y) = log x - log y
• log (e) = 1
• log (1) = 0
• log 10 = 1 

• log (1/x) = - log x

Calculation:

Domain: x² - 25 > 0 or x - 5 > 0

⇒ x > 5      ---(1)

Now, \(\log _{0.5}(x^{2}-25)-\log _{0.5}(x-5)<\log _{0.5}3\)

\(⇒ \log _{0.5}(x^{2}-25)<\log _{0.5}3+\log _{0.5}(x-5)\)

\(⇒ \log _{0.5}(x^{2}-25)<\log _{0.5}3(x-5)\)      (∵ log x + log y = log xy)

⇒ x² - 25 > 3 (x - 5)      (Direction changed as base is less than 1)

⇒ x2 - 25 > 3x - 15

⇒ x2 - 3x - 10 > 0

⇒ (x - 5) (x + 2) > 0

⇒ x > 5 or x > -2      ---(2)

From equation (1) & (2), we get,

x > 5

Hence, x > 5.

By the rule of alligation,

Calculation:

Given:

p (X ≥ 1) = 5/9

1 - P (x = 0) = 5/9

\(\rm 1-{^2}C_0p^0q^2=5/9\)

\(\rm q^2=1-\frac{5}{9}=\frac{4}{9}\)

\(\rm q=\frac{2}{3}\)

\(\rm p=1-\frac{2}{3}=1/3\)

Y ∼ (4, 1/3)

P (Y = 4) = \(\rm {^4}C_4p^4q^0=\frac{1}{81}\)

Also, 

V(2x - Y) = 2² V(x) + (-1)² V(Y)

= 4 n_x p_x q_x +  n_y p_y q_y

= \(4\times2\times \frac{1}{3}\times\frac{2}{3}+4\times\frac{1}{3}\times\frac{2}{3}\)

\( = \frac{16}{9}+\frac{8}{9}=\frac{24}{9}\)

• Face Value/ Par Value (F): It is the price at which the bond is sold to buyers at the time of issue.
• Premium: When the market price of the bond is higher than its face value, the bond is selling at a premium.
• Discount: When the market price of the bond is lower than its face value, the bond is selling at a discount.
• Nominal Rate of Interest/ Coupon Rate, (r): It is the rate at which a bond yields interest. It is the interest rate paid by a bond relative to its face value.
• Periodic dividend payment or coupon payment (C): Periodic payments are done under the bond, It is equal to the product of face value and the coupon yield.

C = Fr ....(1)

• Current Yield (i): It is the interest rate paid by a bond relative to its bond price.
• When a bond sells at discount, the Current Yield is greater than the Coupon Yield.
• When a bond sells at a premium, the Current Yield is lower than the Coupon Yield.
• When the bond sells at par, the Current Yield is equal to the Coupon Yield.
• If C is the periodic payment done after each period (Coupon payment), then the total purchase price 'V' of a bond (Present value of the bond),

V = \(\frac{C(1-(1+i)^{-n})}{i} + F(1+i)^{-n}\) ....(2)

Calculation:

Given: F = 3,500. 

• The nominal interest rate or Coupon Rate, r = 7%
• From equation (1),
• Periodic dividend payment or coupon payment,

C = Fr = 3,500 \(\times\) 0.07 = 245

• n = 8 years
• (1.08)-8    = 0.54 and \(\frac{1-(1.08)^{-8}}{0.08} \) = 5.75.
• Total purchase price of the bond will be given by the formula from equation (2),

V = \(\frac{C(1-(1+i)^{-n})}{i} + F(1+i)^{-n}\) 

\(\Rightarrow\) V = \(\frac{245(1-(1+0.08)^{-8})}{0.08} \) + 3,500(1 + 0.08)-8

\(\Rightarrow\) V = \(\frac{245(1-(1.08)^{-8})}{0.08} \) + 3,500(1.08)-8

\(\Rightarrow\) V = 245 × 5.75 + 3,500 × 0.54

\(\Rightarrow\) V = 1,408.75 + 1,890

\(\Rightarrow\) V = 3,298.75 Rs

• Hence the correct answer is option 4.

Additional Information

• The yield to maturity is the estimated annual rate of return for a bond assuming that the investor holds the asset until its maturity date and reinvests the payments at the same rate.1
• The coupon rate is the annual income an investor can expect to receive while holding a particular bond. 
• At the time it is purchased, a bond's yield to maturity and its coupon rate is the same.
• As economic conditions change, investors may demand the bond more or less. As the price of the bond changes, the yield to maturity of the bond will inversely change.
• Though bonds may be issued with variable rates tied to LIBOR, most bonds are issued with a fixed rate, causing the coupon rate and yield to often be different.

Calculation:

\(y = 1 + \frac{1}{x} + \frac{1}{{{x^2}}} + \frac{1}{{{x^3}}}... ....+ ... ∞\) , |x| > 1

The above is the infinite G.P. where  |x| > 1

⇒ 1/|x| < 1

So, the sum of the infinite G.P.

\(⇒ y = \frac{1}{{\left( {1-\frac{1}{x}} \right) }} = \frac{x}{{x -1 }} \)

\(⇒ \frac{{{y^2}}}{{{x^2}}} = \frac{1}{{{{( x-1)}^2}}} \)      ...(1)

Solving for [dy/dx] for

 \(⇒ y = \frac{1}{{\left( {1-\frac{1}{x}} \right) }} = \frac{x}{{x -1 }} \)

\(\therefore \frac{{dy}}{{dx}} = \frac{{(x-1)\times1 - x( 1)}}{{{{(x-1)}^2}}} = \frac{{x-1 - x}}{{{{(x-1)}^2}}} = \frac{-1}{{{{(x-1)}^2}}}\)

from (i), \(\frac{{dy}}{{dx}} = -\frac{{{y^2}}}{{{x^2}}}.\)

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, Shyam aims to meet his nutrient requirements while minimizing his expenses. x and y represent the cost per unit of product A and B. These represent the decision variables for the problem.
• The objective is to minimize the costs in this example.
• The cost of product A is Rs 50 per unit while the cost of product B is Rs 60 per unit. 
• Hence, the objective function is given by,

Z = 50x + 60y

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

Concept:

If a differential equation has the form f(x,y)dy = g(x,y)dx then it is said to be a homogeneous differential equation if the degree of f(x,y) and g(x, y) is the same.

Some useful formulas are;

\(\rm \int {1\over x} dx =ln\ x + C\)

\(\rm \int {1\over 1+x^2} dx =tan^{-1} x + C\)

Calculation:

\(\rm {{dy}\over {dx}} = {{x+y}\over {x-y}}\)

Taking y = vx where \(\rm {{dy}\over{dx}} = v+ x{{dv}\over{dx}}\) then we get

\(\rm v+ x{{dv}\over{dx}} = {{x+vx}\over {x-vx}}\)

\(\rm v+ x{{dv}\over{dx}} = {{x(1+v)}\over {x(1-v)}}\)

\(\rm x{{dv}\over{dx}} = {{1+v}\over {1-v}} -v\)

\(\rm x{{dv}\over{dx}} = {{1+v-v+v^2}\over {1-v}} \)

\(\rm {{dx}\over{x}} = {{1-v}\over {1+v^2}} dv\)

Integrating the above equation we get,

\(\rm \int{{dx}\over{x}} = \int{{1}\over {1+v^2}}dv- \int{{v}\over {1+v^2}}dv\)

\(\rm \log x = \tan^{-1} v-{1\over 2}log(1+v^2)+c\)

Now finally putting the value of (v = y/x) in the equation we get,

\(\rm \log x = tan^{-1} {y\over x}-{1\over 2}log(1+({y\over x})^2)+c\)

Concept:

The critical point of a function:

Consider the function \(\rm f(x)\) then the values at which \(\rm f'(x) = 0\) or not defined are known as critical points.

First derivative test:

We say that function has extremum at some point \(\rm x=a\) if the first derivative at that point is 0.

Second derivative test:

If \(\rm x=a\) 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: y = \({e^{\left( {2{x^2} - 2x + 1} \right)sin^2x}}\)

For minima or maxima, \(\frac{{dy}}{{dx}} = 0\)

∴ \({e^{\left( {2{x^2} - 2x + 1} \right)}}{\sin ^2}x\left[ {\left( {4x - 2} \right){{\sin }^2}x + 2\left( {2{x^2} - 2x + 1} \right)\sin x\cos x} \right] = 0\)

\(\Rightarrow \left[ {\left( {4x - 2} \right){{\sin }^2}x + 2\left( {2{x^2} - 2x + 1} \right)\sin x\cos x} \right] = 0\)

\(\Rightarrow 2\sin x\left[ {\left( {2x - 1} \right)\sin x + \left( {2{x^2} - 2x + 1} \right)\cos x} \right] = 0\)

\(\Rightarrow \sin x = 0\,also,f\left( 0 \right) > 0\)

∴ y is minimum for sin x = 0

Thus the minimum value of y = \({e^{\left( {2{x^2} - 2x + 10} \right)\left( 0 \right)\, = \,e\left( 0 \right)\, = \,1.}}\)

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: The function f be a twice differentiable function on (1, 6). 

f(2) = 8, f'(2) = 5, f'(x) ≥ 1, f"(x) ≥ 4, ∀ x ∈ (1, 6)

\(f''(x) = \frac{{f'\left( 5 \right) - f'\left( 2 \right)}}{{5 - 2}} ≥ 4\)       ----(i)

\( \Rightarrow f'(5) ≥ 17\)

\(f'\left( x \right) = \frac{{f\left( 5 \right) - f\left( 2 \right)}}{{5 - 2}} ≥ 1 \)       ----(ii)

\(\Rightarrow f\left( 5 \right) ≥ 11\)

∴ f'(5) + f(5) ≥ 28

CONCEPT:

• The angle of intersection between the two curves will be the same as the angle of the intersection or between the tangents on both the curves on a common point.
• If the slope of a tangent at a common point on the first curve is m1 and the slope of a tangent at a common point on the second curve is m2 then the angle of intersection,

\(⇒ tan θ = \frac{m_1 - m_2}{1 + m_1 m_2} \;\;\;\;\; \ldots \left( 1 \right)\)

CALCULATION:

Given:

First of all, find the point of intersection of these two curves.

⇒ sin x - cos x ⇒ x = π / 4

Now, 

\( y = \sin x ⇒ {\left( {\frac{{dy}}{{dx}}} \right)_{x = \pi /4}} = \frac{1}{{\sqrt 2 }} = m_1\)

\(y = \cos x ⇒ {\left( {\frac{{dy}}{{dx}}} \right)_{x = \pi /4}} = \frac{{ - 1}}{{\sqrt 2 }} = m_2\)

\(\Rightarrow \tan \theta = \frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}} = 2\sqrt 2\).

\( ⇒ \theta = {\tan ^{ - 1}}(2\sqrt 2 )\)

Concept:

​Parametric Form:

If f(x) and g(x) are the functions in x, then 

\(\rm df(x)\over dg(x)\) = \(\rm \frac{df(x)\over dx}{dg(x)\over dx}\) 

Calculation:

Put, t = tan θ ⇒ θ = tan^-1(t)

\(x = {\cos ^{ - 1}}\frac{1}{{\sqrt {1+ {{\tan }^2}θ } }}\)

⇒ x = cos^-1 (cos θ) = θ 

x = tan^-1 t 

⇒ \(\frac{{dx}}{{dt}} = \frac{1}{{1 + {t^2}}}\) -- (1)

Taking→  \(\sin y = \frac{t}{{\sqrt {1 + {t^2}} }}\)

as t = tan θ  

Thus, y = sin^-1  \(\frac{{\tan θ }}{{\sec θ }}\) = sin^-1 (sinθ) 

⇒ y = θ = tan^-1 t

\(\frac{{dy}}{{dt}} = \frac{1}{{1 + {t^2}}}\) --- (2)

So dy/dx = (dy/dt)/(dx/dt)

From (1) & (2)

\(\therefore \,\frac{{dy}}{{dx}} = 1\)

Concept:

Rolle's theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b) then,

f′(c) = 0 for some c ∈ [a, b].

Calculation:

From Rolle's theorem in (1, 26), f (1) = f (26) = 5,

In a given interval, the function satisfies all the conditions of Rolle's theorem, therefore in [1, 26], at least, there is a point for which f' (x) = 0 

Concept:

The product of two matrices is only possible when the number of columns of the first matrix is equal to the number of rows of the second matrix.

Calculations:

Given:

X = \(\left[ {\begin{array}{*{20}{c}} {\rm{3}}&{{\rm{ - 4}}}\\ 1&{{\rm{ - 1}}} \end{array}} \right] \) 

It is asked for Xⁿ

Let's Assume that n = 2 and solve it.

⇒ So X² = \(\left[ {\begin{array}{*{20}{c}} {\rm{5}}&{{\rm{ - 8}}}\\ 2&{{\rm{ - 3}}} \end{array}} \right] \).

Now tallying with the given options:

As n = 2

a) \(\left[ {\begin{array}{*{20}{c}} {\rm{3n}}&{{\rm{ - 4n}}}\\ n&{{\rm{ - n}}} \end{array}} \right]\)= \(\left[ {\begin{array}{*{20}{c}} {\rm{6}}&{{\rm{ - 8}}}\\ 2&{{\rm{ - 2}}} \end{array}} \right]\) ≠ \(\left[ {\begin{array}{*{20}{c}} {\rm{5}}&{{\rm{ - 8}}}\\ 2&{{\rm{ - 3}}} \end{array}} \right] \)

So Option (1) is not correct

b) \(\left[ {\begin{array}{*{20}{c}} {{\rm{2 + n}}}&{{\rm{5 - n}}}\\ {\rm{n}}&{{\rm{ - n}}} \end{array}} \right]\)= \(\left[ {\begin{array}{*{20}{c}} {{\rm{4}}}&{{\rm{3}}}\\ {\rm{2}}&{{\rm{ - 2}}} \end{array}} \right]\) ≠ \(\left[ {\begin{array}{*{20}{c}} {\rm{5}}&{{\rm{ - 8}}}\\ 2&{{\rm{ - 3}}} \end{array}} \right] \)

So Option (2) is not correct

c) \(\left[ {\begin{array}{*{20}{c}} {{{\rm{3}}^{\rm{n}}}}&{{{\left( {{\rm{ - 4}}} \right)}^{\rm{n}}}}\\ {{{\rm{1}}^{\rm{n}}}}&{{{\left( {{\rm{ - 1}}} \right)}^{\rm{n}}}} \end{array}} \right]\) = \(\left[ {\begin{array}{*{20}{c}} {{{\rm{9}}}}&{{{\left( {{\rm{ 16}}} \right)}}}\\ {{{\rm{1}}}}&{{{\left( {{\rm{ 1}}} \right)}}} \end{array}} \right]\)≠ \(\left[ {\begin{array}{*{20}{c}} {\rm{5}}&{{\rm{ - 8}}}\\ 2&{{\rm{ - 3}}} \end{array}} \right] \)

So So Option (3) is not correct

The matrices in (a), (b), and (c) do not tally with\(\left[ {\begin{array}{*{20}{c}} {\rm{5}}&{{\rm{ - 8}}}\\ 2&{{\rm{ - 3}}} \end{array}} \right]\).

∴ None of these is the correct Answer.

Concept:

• \(\frac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}} = \frac{1}{{1 + {x^2}}}\;\)
• Chain Rule:
• Let y = f(v) be a differentiable function of v and v = g(x) be a differentiable function of x then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} \cdot \frac{{dv}}{{dx}}\)

• If x1 < x2 then f(x1) <  f(x2) ∀ x1, x2 ∈ (a, b)
• Here, \(\frac{{dy}}{{dx}} > 0\;or\;f'\left( x \right) > 0\)

Similarly, f(x) is said to be a decreasing function:

• If If x1 < x2 then f(x1) > f(x2) ∀ x1, x2 ∈ (a, b).
• Here, \(\frac{{dy}}{{dx}} < 0\;or\;f'\left( x \right) < 0\)

Calculation:

Given:

\(y =\frac{1}{1 +x^2} \)

\(\frac{dy}{dx} =\frac{-2x}{(1 +x^2)^2} \)

To be decreasing,

\(\frac{dy}{dx} =\frac{-2x}{(1 +x^2)^2} <0\)

⇒ x > 0

Concept:

Here we will use the concept of the addition of matrix by adding corresponding elements.

Calculations:

\({D_1} = \left| {\begin{array}{*{20}{c}} 1&{15}&8\\ 1&{35}&9\\ 1&{25}&{10} \end{array}} \right|,\)

\({D_2} = \left| {\begin{array}{*{20}{c}} 2&{15}&8\\ 4&{35}&9\\ 8&{25}&{10} \end{array}} \right|\)

\({D_3} = \left| {\begin{array}{*{20}{c}} 3&{15}&8\\ 9&{35}&9\\ {27}&{25}&{10} \end{array}} \right|,\)

\({D_4} = \left| {\begin{array}{*{20}{c}} 4&{15}&8\\ {16}&{35}&9\\ {64}&{25}&{10} \end{array}} \right|\)

\({D_5} = \left| {\begin{array}{*{20}{c}} 5&{15}&8\\ {25}&{35}&9\\ {125}&{25}&{10} \end{array}} \right|\)

Now, all the corresponding elements will be added to get the final result.

The element fist row and first column = 1 + 2 +3 + 4 + 5 = 15

Similarly,

⇒  D₁ + D₂ + D₃ + D_{4 +} D₅  = \(\rm \left| {\begin{array}{*{20}{c}} {15}&{15}&8\\ {55}&{35}&9\\ {225}&{25}&{10} \end{array}} \right|\)

= 15(125) + 15(1475) - 8(6500)

= 1875 + 22125 - 52000 = - 28000.

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 - a)2 = 2bx               ....(1)

There are two constants a and b so differentiate two times

2(y - a)\(\rm dy\over dx\) = 2b

b = (y - a)\(\rm dy\over dx\)                                            ....(2)

Differentiating one more time w.r.t x

0 = \(\rm \left(dy\over dx\right)^2\) + (y - a)\(\rm d^2y\over dx^2\)

Multiply (y - a) on both sides, we get

(y - a)\(\rm \left(dy\over dx\right)^2\) + (y - a)²\(\rm d^2y\over dx^2\) = 0

From equation (1) and (2), (y - a)2 = 2bx  and (y - a)\(\rm dy\over dx\) = b

(y - a)\(\rm {dy\over dx} \times {dy\over dx}\) + (2bx) \(\rm d^2y\over dx^2\) = 0              

b\(\rm {dy\over dx}\) + (2bx) \(\rm d^2y\over dx^2\) = 0

2x\(\rm d^2y\over dx^2\) + \(\rm {dy\over dx}\)= 0