Mathematics Test-6

Given equation is:

\(x^{2} \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+y=\log x\)

After reducing, it becomes,

\(D(D-1) y-D y+y=z \Rightarrow\left(D^{2}-2 D-1\right) y=z\)

Now the auxiliary equation is \((D-1)^{2}\)

\(D=1,1\) [roots are real and equal]

\(\therefore\) Complimentary function is \(\left(C_{1}+C_{2} z\right) e^{\mathrm{z}}\)

Particular integral will be:

\(P . I .=\frac{1}{(D-1)^{2}} z=(1-D)^{-2} z=\left(1+2 D+3 D^{2}+\ldots\right) z\)

P.I. \(=z+2\)

Now complete solution will be:

\(y=C . F+P . I=\left(C_{1}+C_{2} z\right) e^{z}+z+2\)

Substituting, \(z=\log x\)

\(y=\left(C_{1}+C_{2} \log x\right) x+\log x+2\)

Let y = f(x) = x be the given equation.

\(\therefore \int \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int \mathrm{x} \mathrm{dx}=\frac{\mathrm{x}^{2}}{2}+\mathrm{C}\)

The curve (line) y = x is above the x-axis for x > 0, and below the x-axis for x < 0. Using the above concept for the area of a curve, we can say that the required area is:

\(I=\left|\int_{-1}^{0} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right|+\left|\int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right|\)

\(=\left|\left[\frac{\mathrm{x}^{2}}{2}\right]_{-1}^{0}\right|+\left|\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2}\right|\)

\(=\left|0-\frac{1}{2}\right|+\left|\frac{4}{2}-0\right|\)

\(=\frac{1}{2}+\frac{4}{2}\)

\(=\frac{5}{2}\)

Given: There are two positive numbers x and y such that x + y = 60

Here we have to determine the numbers x and y such that x + y = 60 and xy³ is maximum.

∵ x + y = 60

⇒ x = 60 - y

Now by substituting x = 60 - y in the expression xy³ we get

Let f(y) = (60 - y) × y³ = 60y³ - y⁴

Now let's find f'(x)

⇒ f'(y) = 180y² - 4y³

Now let's find the roots of the equation f'(y) = 0

⇒ 180y² - 4y³ = 0

⇒ y = 0 or 45

By neglecting y = 0 ∵ we need to maximize f(y)

So, y = 45

Now let's find out f''(y) i.e\(\frac{d^{2}(f(y))}{d y^{2}}\)

⇒ f''(y) = 360y - 12y²

Now evaluate the value of f''(y) at y = 45

⇒ f''(45) = (360 × 45) - (12 × 45²) = - 8100 < 0

As we know that according to second derivative test if f''(c) < 0 then x = c is a point of maxima

So, y = 45 is a point of maxima for f(y)

So, when y = 45 then x = 60 - y = 15

Therefore, the required numbers are 15 and 45.

As we know that, if each term of any row or of any column is sum of two quantities, then the determinant can be expressed as the sum of the two determinants of same order.

\(\Rightarrow\left|\begin{array}{lll}b+c & a+b & a \\ c+a & b+c & b \\ a+b & c+a & c\end{array}\right|=\left|\begin{array}{lll}b & a+b & a \\ c & b+c & b \\ a & c+a & c\end{array}\right|+\left|\begin{array}{ccc}c & a+b & a \\ a & b+c & b \\ b & c+a & c\end{array}\right|\) .....(i)

Now, first let's solve\(\left|\begin{array}{lll}b & a+b & a \\ c & b+c & b \\ a & c+a & c\end{array}\right|\)

\(\Rightarrow\left|\begin{array}{lll}b & a+b & a \\ c & b+c & b \\ a & c+a & c\end{array}\right|=\left|\begin{array}{lll}b & a & a \\ c & b & b \\ a & c & c\end{array}\right|+\left|\begin{array}{ccc}b & b & a \\ c & c & b \\ a & a & c\end{array}\right|\)

As we know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

\(\Rightarrow\left|\begin{array}{lll}b & a+b & a \\ c & b+c & b \\ a & c+a & c\end{array}\right|=0\) .....(ii)

Now let's solve\(\left|\begin{array}{ccc}c & a+b & a \\ a & b+c & b \\ b & c+a & c\end{array}\right|\)

\(\Rightarrow\left|\begin{array}{lll}c & a+b & a \\ a & b+c & b \\ b & c+a & c\end{array}\right|=\left|\begin{array}{ccc}c & a & a \\ a & b & b \\ b & c & c\end{array}\right|+\left|\begin{array}{ccc}c & b & a \\ a & c & b \\ b & a & c\end{array}\right|\)

As we know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

\(\Rightarrow\left|\begin{array}{lll}c & a+b & a \\ a & b+c & b \\ b & c+a & c\end{array}\right|=\left|\begin{array}{ccc}c & b & a \\ a & c & b \\ b & a & c\end{array}\right|=a^{3}+b^{3}+c^{3}-3 a b c\) .....(iii)

Now, by substituting (ii) and (iii) in equation (i), we get

\(\Rightarrow\left|\begin{array}{lll}b+c & a+b & a \\ c+a & b+c & b \\ a+b & c+a & c\end{array}\right|=a^{3}+b{ }^{3}+c^{3}-3 a b c\)

Median:

Case 1: If the number of observations (n) is even

Median \(=\frac{\text { Value of }\left(\frac{\mathrm{n}}{2}\right)^{\text {th }} \text { observation }+\text { Value of }\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }} \text { observation }}{2}\)

Case 2: If the number of observations (n) is odd

Median = Value of \(\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}\) observation

Mode: The mode is the value that appears most frequently in a data set.

Given data are

2, 3, x, 2x - 4, 5, 8

Number of observations = 6 = even

So,

Median \(=\frac{1}{2}(x+2 x-4)\) .....(i)

But according to question, Median is 7.

\(\Rightarrow \frac{1}{2}(x+2 x-4)=7\)

⇒ 3x - 4 = 14

⇒ x = 6

So, observations are

2, 3, 6, 8, 5, 8

Since, 8 comes two times.

So, mode of observations is 8.

Concept:

• Probability of both A and B = P(A ∩ B)
• If A and B are independent, then, P(A ∩ B) = P(A).P(B)
• Probability of A or B = P(A ∪ B)
• If A and B are disjoint, then, P(A ∪ B) = P(A) + P(B)

Calculation:

E₁ represents the event that a card with the number 1 is drawn.

And the card with the number 1 can be drawn from either of the four boxes

A₁, A₂, A₃, or A₄

Given, The probability of selection of box \(A_{i} = \frac{i}{10}\)

Since, Total number of cards in the box A_i = i

\(\Rightarrow\) The probability of drawing a card with the number 1 from \(A_{i}\)

\(\Rightarrow \frac{1}{i}\)

Let The probability of selecting box A_i and choosing card with number 1 = P(A_i ∩ E₁)

\(\Rightarrow P(A_{i} ∩ E_{1}) = \frac{i}{10} \times \frac{1}{i} = \frac{1}{10} = 0.1\)

So, P(A₁ ∩ E₁) = P(A₂ ∩ E₁) = P(A₃ ∩ E₁) = P(A₄ ∩ E₁) = 0.1

Now, P(E₁) = P(A₁ or A₂ or A₃ or A₄)

⇒ P(E₁) = P(A₁ ∩ E₁) + P(A₂ ∩ E₁) + P(A₃ ∩ E₁) + P(A₄ ∩ E₁)

\(\Rightarrow P(E_{1}) = 0.1 + 0.1 + 0.1 + 0.1 = 0.4 = \frac{4}{10} = \frac{2}{5}\)

Given:

Series is 0.6 + 0.66 + 0.666 + … to n terms

Formula:

Sum of first \(n\) terms of G.P \(=\frac{a\left(r^{n}-1\right)}{(r-1)}, r \neq 1\)

Calculation:

0.6 + 0.66 + 0.666 + ….....+ n terms

⇒ 0.6 (1+ 1.1 + 1.11 + ...........+ n terms)

\(\Rightarrow 0.6\left(1+\frac{11}{10}+\frac{111}{100}+\ldots n\right.\) terms \()\)

\(\Rightarrow 0.6\left(\frac{9}{9}+\frac{99}{90}+\frac{999}{900}+\ldots n\right.\) terms \()\)

\(\Rightarrow \frac{0.6}{9}\left((10-1)+\frac{100-1}{10}+\frac{1000-1}{100}+\ldots n\right.\) terms \()\)

\(\Rightarrow \frac{6}{90}\left((10-1)+\left(10-\frac{1}{10}\right)+\left(10-\frac{1}{100}\right)+\ldots n\right.\) terms \()\)

\(\Rightarrow \frac{1}{15}\left((10+10+10+\ldots)-\left(1+\frac{1}{10}+\frac{1}{100}+\ldots\right)\right)\)

\(\Rightarrow \frac{1}{15}\left(10 n-\left(\frac{1\left(1-\left(\frac{1}{10}\right)^{n}\right)}{1-\frac{1}{10}}\right)\right)\)

\(\Rightarrow \frac{1}{15}\left(10 n-\frac{10}{9}\left(1-\left(\frac{1}{10}\right)^{n}\right)\right)\)

\(\therefore \frac{1}{15}\left(10 n-\frac{10}{9}+\frac{1}{9 \times 10^{n-1}}\right)\)

The hyperbola taking \(Y\)-axis as reference is of the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\) has:

• Centre is given by: (0, 0)
• Vertices are given by: (0, ±a)
• Foci are given by: (0, ± c)

Given:

Foci \(=(0, \pm \sqrt{10})\)

Point on the hyperbola is \((2,3)\).

Calculation:

Foci \(=(0, \pm \sqrt{10})\)

so, ae \(=\sqrt{10}\)

We know that, \(b^{2}=a^{2} e^{2}-a^{2}\)

\(\Rightarrow b^{2}=(\sqrt{10})^{2}-a^{2}\)

\(\Rightarrow b^{2}=10-a^{2}\)

Substituting above in the equation of hyperbola,

\(\Rightarrow \frac{y^{2}}{a^{2}}-\frac{x^{2}}{10-a^{2}}=1\)

⇒ (2, 3) points satisfy the above equation

\(\Rightarrow \frac{3^{2}}{a^{2}}-\frac{2^{2}}{10-a^{2}}=1\)

\(\Rightarrow a^{4}-23 a^{2}+90=0\)

\(\Rightarrow\left(a^{2}-5\right)\left(b^{2}-18\right)=0\)

\(\Rightarrow a^{2}=5,18\)

From equation (i)

b² = 10 - 5 = 5

Again b² = 10 - 18 = - 8 (Not possible)

The required equation of hyperbola is:

\(\Rightarrow \frac{y^{2}}{5}-\frac{x^{2}}{5}=1\)

\(\Rightarrow {y}^{2}-{x}^{2}=5\)

Let A = [a_ij] be a square matrix.

Steps for obtaining adjoint of the matrix

1. We need to calculate cofactors of all elements

⇒ C_ij or A_ij = (−1)^i+j M_ij

Here, M_ij = Minor of matrix A

2. Transpose of cofactor is called adjoint of the matrix

⇒ Adj (A) = C^T

Calculation:

The adjoint of matrix A is the transpose of the cofactor matrix of A.

Cofactor of \(1=A_{11}=\left|\begin{array}{ll}2 & 0 \\ 6 & 1\end{array}\right|=2\)

Cofactor of \(0=A_{12}=-\left|\begin{array}{cc}5 & 0 \\ -1 & 1\end{array}\right|=-5\)

Cofactor of \(0=A_{13}=\left|\begin{array}{cc}5 & 2 \\ -1 & 6\end{array}\right|=32\)

Cofactor of \(5=A_{21}=-\left|\begin{array}{ll}0 & 0 \\ 6 & 1\end{array}\right|=0\)

Cofactor of \(2=A_{22}=\left|\begin{array}{cc}1 & 0 \\ -1 & 1\end{array}\right|=1\)

Cofactor of \(0=A_{23}=\left|\begin{array}{cc}1 & 0 \\ -1 & 6\end{array}\right|=-6\)

Cofactor of \(-1=A_{31}=\left|\begin{array}{ll}0 & 0 \\ 2 & 0\end{array}\right|=0\)

Cofactor of \(6=A_{32}=-\left|\begin{array}{ll}1 & 0 \\ 5 & 0\end{array}\right|=0\)

Cofactor of \(1=A_{33}=\left|\begin{array}{ll}1 & 0 \\ 5 & 2\end{array}\right|=2\)

\(\therefore\) Cofactor matrix \(=\left[\begin{array}{ccc}2 & -5 & 32 \\ 0 & 1 & -6 \\ 0 & 0 & 2\end{array}\right]\)

\(\operatorname{Adj}(A)=\left[\begin{array}{ccc}2 & 0 & 0 \\ -5 & 1 & 0 \\ 32 & -6 & 2\end{array}\right]\)

Given:

Expansion is \((1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\ldots \ldots \ldots\)

Concept:

When three terms ‘a’, ‘b’ and ‘c’ are in AP, then means: 2b = a + c

\((1+x)^{n}={ }^{n} C_{0} x^{0}+{ }^{n} C_{1} x^{1}+{ }^{n} C_{2} x^{2}+\ldots{ }^{n} C_{2} x^{n}\)

Calculation:

We have \((1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\ldots \ldots \ldots\)

Here, Coefficient of 2nd term (a) = nx,

Coefficient of 3rd term (b) \(=\frac{n(n-1)}{2 !}\)

Coefficient of 4 th term \((c)=\frac{n(n-1)(n-2)}{3 !}\)

Now, According to the concept used

\(\Rightarrow(n-1)=1+\frac{(n-1)(n-2)}{6}\)

\(\Rightarrow 6(n-1)-6=(n-1)(n-2)\)

\(\Rightarrow 6 n-6-6=n^{2}-3 n+2\)

\(\Rightarrow n^{2}-9 n+14=0 \)

\(\Rightarrow(n-2)(n-7)=0\)

\(n=2\) is invalid as \((1+x)^{2}\) doesn't have \(4^{\text {th }}\) term in it.

\(\therefore\) The value of \({n}\) is 7.

Formula:

\(C(n, r)=\frac{n !}{r ! \times(n-r) !}\), where \(r \leq n\)

Given: C (n, r) : C (n, r + 1) = 1 : 2 and C (n, r + 1) : C (n, r + 2) = 2 : 3

\(\Rightarrow \frac{C(n, r)}{C(n, r+1)}=\frac{\frac{n !}{r ! \times(n-r) !}}{\frac{n !}{(r+1) ! \times(n-r-1) !}}=\frac{1}{2}\)

⇒ 3r + 2 = n .....(i)

Similarly,

\(\Rightarrow \frac{C(n, r+1)}{C(n, r+2)}=\frac{\frac{n !}{(r+1) ! \times(n-r-1) !}}{\frac{n !}{(r+2) ! \times(n-r-2) !}}=\frac{2}{3}\)

⇒ 5r + 8 = 2n .....(ii)

By, solving equation (i) and (ii), we get

⇒ 5r + 8 = 2 × (3r + 2)

⇒ r = 4

It is given that \(\frac{\mathrm{x}}{2 \mathrm{x}+1} \geq \frac{1}{4}\). We have the following two cases:

Case 1: \(2 x+1>0 \Rightarrow x>-\frac{1}{2}\).

Multiplying both sides of \(\frac{x}{2 x+1} \geq \frac{1}{4}\) by \(4(2 x+1)\), gives us:

= 4x ≥ 2x + 1

Subtracting 2x from both sides, we get:

= 2x ≥ 1

Dividing both sides by 2, we get:

\(\Rightarrow \mathrm{X} \geq \frac{1}{2}\)

Combining this with our assumption 2x + 1 > 0, we get:

\(\mathrm{X}>-\frac{1}{2}\) and \(\mathrm{X} \geq \frac{1}{2}\)

\(\Rightarrow \mathrm{X} \geq \frac{1}{2}\)

Case 2: \(2 x+1<0 \Rightarrow x<-\frac{1}{2}\).

Multiplying both sides of \(\frac{x}{2 x+1} \geq \frac{1}{4}\) by \(4(2 x+1)\) (which is -ve in this case), will give us:

= 4x ≤ 2x + 1

Subtracting 2x from both sides, we get:

= 2x ≤ 1

\(\Rightarrow \mathrm{x} \leq \frac{1}{2}\)

Combining this with our assumption, we get:

\(\mathrm{x}<-\frac{1}{2}\) and \(\mathrm{x} \leq \frac{1}{2}\)

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

Considering both the cases together, we can say that \(\mathrm{x} \geq \frac{1}{2}\) or \(\mathrm{x}<-\frac{1}{2}\).

\(\Rightarrow \mathrm{x} \in\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)\)

Given, the line \(\frac{\mathrm{x}-5}{2}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}+4}{1}\) and the plane

The angle between the line \(\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{a}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{c}}\) to the normal to the plane \(\mathrm{Ax}+\mathrm{By}+\) \(\mathrm{Cz}=\mathrm{D}\) is given by,

\(\sin \theta=\left|\frac{\mathrm{Aa}+\mathrm{Bb}+\mathrm{Cc}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}} \cdot \sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{C}^{2}}}\right|\)

So, angle between the line \(\frac{\mathrm{x}-5}{2}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}+4}{1}\) and the plane \(3 \mathrm{x}-4 \mathrm{y}-\mathrm{z}+5=0\) is

⇒ Here a = 2, b = -1, c = 1 and A = 3, B = -4 and C = -1

\(\sin \theta=\left|\frac{(2)(3)+(-1)(-4)+(1)(-1)}{\sqrt{2^{2}+(-1)^{2}+1^{2}} \cdot \sqrt{3^{2}+(-4)^{2}+(-1)^{2}}}\right|\)

\(\sin \theta=\frac{9}{\sqrt{156}}\)

\(\theta=\sin ^{-1}\left(\frac{9}{2 \sqrt{39}}\right)\)

So, Acute angle between the line \(\frac{\mathrm{x}-5}{2}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}+4}{1}\) and the plane \(3 x-4 y-z+5=0\) is \(\theta=\sin ^{-1}\left(\frac{9}{2 \sqrt{39}}\right)\)

The intercept form of the line is given by:\(\frac{x}{a}+\frac{y}{b}=1\)

Where a is the x-intercept and b is the y-intercept

Calculation:

Maximize Z = 5x + 10y

Subject to, x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0

Converting the given in equations into equations

x + 2y = 120, x + y = 60, x - 2y = 0

\(x+2 y=120 \Leftrightarrow \frac{x}{120}+\frac{y}{60}=1\)

\(x+y=60 \Leftrightarrow \frac{x}{60}+\frac{y}{60}=1\)

\(x-2 y=0 \Leftrightarrow 2 y=x\) (Passing through origin)

The corner points of the feasible region are A(60, 30), B(40, 20),

C(60,0) and D(120, 0)The values of Z at these corner points are asfollows:

Corner point z = 5x + 10y

The maximum value of Z is 600 at all the points on the line segmentjoining (120, 0) and (60, 30).

Given: \(f(x) = 2^{|x|}\) and g(x) = [x] where [.] denotes greatest integer function

Here, we have to find out the value of f o g \(\left(-\frac{ 3}{2}\right)\)

\(\Rightarrow f o g \left(- \frac{17}{2}\right) = f( g\left(- \frac{17}{2}\right))\)

\(\because g(x) = [x]\), So \(g \left(- \frac{17}{2}\right) = \left[- \frac{17}{2}\right] = -9\)

\(\Rightarrow f o g \left(- \frac{17}{2}\right) = f(- 9)\)

\(\because f(x) = 2^{|x|}\) So, \(f(- 9) = 2^{|- 9|} = 29 = 512\)

So, f o g \(\left(- \frac{17}{2}\right) = 512\)