Mathematics Test - 15

Let the total number of shots be \(x\). Then,

Shots fired by \(A=\frac{5}{8} x\)

Shots fired by \(B=\frac{3}{8} x\)

Killing shots by \(A=\frac{1}{3}\) of \(\frac{5}{8} x=\frac{5 x}{24} ;\)

Shots missed by \(B=\frac{1}{2}\) of \(\frac{3}{8} x=\frac{3 x}{16}\).

\(\therefore \frac{3 x}{16}=27\) or \(x=\left(\frac{27 \times 16}{3}\right)=144\)

Birds killed by \(\mathrm{A}=\frac{5 x}{24}=\left(\frac{5}{24} \times 144\right)=30\)

Given:

\(3+\frac{9}{2}+6+\frac{15}{2}+\ldots .\)

\(d=\frac3 2\)

\(n=25\)

\(a=3\)

The sum of n terms of an A.P. with first term a and common difference d is given by:

\(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2} \times[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]\)

\(S_{25}=\frac{25}{2}[2 a+(25-1) \times d]\)

\(=\frac{25}{2}\left[2 \times 3+(24) \times \frac{3}{2}\right]\)

\(=\frac{25}{2} \times 42\)

\(=25 \times 21\)

\(=525\)

Given:

\(Z=6 x+2 y\)

\(x+2 y \geq 3\)

\( x+4 y \geq 4\)

\(3 x+y \geq 3\)

\( x \geq 0, y \geq 0\)

To draw the feasible region, construct table as follows:

Shaded portion XABCDY is the feasible region, whose vertices are \(\mathrm{A}(4,0), \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}(0,3)\).

\(B\) is the point of intersection of the lines \(x+4 y=4\) and \(x+2 y=3\)

Solving the above equations, we get:

\(\mathrm{x}=2, \mathrm{y}=\frac{1}{2} \)

\( \therefore \mathrm{B} =\left(2, \frac{1}{2}\right)\)

\(C\) is the point of intersection of the lines \(x+2 y=3\) and \(3 x+y=3\)

Solving the above equations, we get:

\(\mathrm{x}=\frac{3}{5}, y=\frac{6}{5} \)

\( \therefore C =\left(\frac{3}{5}, \frac{6}{5}\right)\)

Here, the objective function is \(Z=6 x+2 y\).

\(Z\) at \(A(4,0)=6(4)+2(0)=24\)

\(\mathrm{Z}\) at \(\mathrm{B}\left(2, \frac{1}{2}\right)=6(2)+2\left(\frac{1}{2}\right)=12+1=13\)

\(\mathrm{Z}\) at \(\left.C\left(\frac{3}{5}, \frac{6}{5}\right)=6\left(\frac{3}{5}\right)+2\left(\frac{6}{5}\right)=\frac{18}{5}\right)+\frac{12}{5}=6\)

\(\therefore Z\) at \(D(0,3)=6(0)+2(3)=6\)

\(\therefore Z\) has minimum value 6 at \(C\left(\frac{3}{5}, \frac{6}{5}\right)\) and \(D(0,3)\).

\(\therefore Z\) is minimum when, \(x=\left(\frac{3}{5}\right), y=\left(\frac{6}{5}\right), Z=6\) and \(x=0, y=3, Z=6\).

Concept:

The conjugate of a complex number is the other complex number having the same real part and opposite sign of the imaginary part.

Calculation:

Let \(z=\frac{1+i}{1-i}\)

Multiply \(1+i\) on both numerator and denominator

\(z=\frac{1+\mathrm{i}}{1-\mathrm{i}} \times \frac{1+\mathrm{i}}{1+\mathrm{i}}\)

\(z=\frac{(1+\mathrm{i})^{2}}{1^{2}-\mathrm{i}^{2}}\)

\(z=\frac{1+2 \mathrm{i}+\mathrm{i}^{2}}{1+1}\)

\(z=\frac{2 \mathrm{i}}{2}=\mathrm{i}\)

Conjugate of z.

\(\overline{\mathrm{z}}=-\mathrm{i}\)

Concept:

The equation of ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)\)

length of Latus rectum of ellipse \(=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}\)

Length of minor axis \(=2 \mathrm{~b}\).

Eccentricity \(=e=\sqrt{1-\left(\frac{b}{a}\right)^{2}}\)

Calculations:

Given, the latus rectum of an ellipse is equal to half of the minor axis.

Suppose, the equation of ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

length of Latus rectum of ellipse \(=\frac{2 b^{2}}{a}\)

Length of minor axis \(=2 b\)

Given, the latus rectum of an ellipse is equal to half of the minor axis

\(\Rightarrow \frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\mathrm{b}\)

\(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{1}{2}\)

If the equation of ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) then eccentricity \(=e=\sqrt{1-\left(\frac{b}{a}\right)^{2}}\)

\(\Rightarrow e=\sqrt{1-\left(\frac{1}{2}\right)^{2}}\)

\(\Rightarrow e=\frac{\sqrt{3}}{2}\)

If the latus rectum of an ellipse is equal to half of the minor axis, then what is its eccentricity e \(=\frac{\sqrt{3}}{2}\)

Given that,

f(x) = |x| - 1 ----(1)

Step: 1

Step: 2

Step: 3

Clearly, we can see that,

f(x) is continuous at x = 1 and

f(x) is not differentiable at x = 0 because there is a corner at x = 0.

Given that:

Lines passes through the origin and \((1,-3,4)\).

We know that:

Direction ratios: The direction ratios of line \(A B\) joining \(A\left(x_{1}, y_{1}, z_{1}\right)\) and B \(\left(x_{2},\right.\), \(\left.y _{2}, z _{2}\right)\) are:

\( x _{2}- x _{1}= a , y _{2}- y _{1}= b\) and \(z _{2}- z _{1}= c\)

Therefore,

Direction ratios of the line through \((0,0,0)\) and \((1,-3,4)\) are \((1-0),(-3-0)\) and \((4-0)\) i.e., \((1,-3,4)\)

Also, we know that:

Equation of a line: The equation of a line with direction ratio \((a, b, c)\) that passes through the point \(\left(x_{1}, y_{1}, z_{1}\right)\) is given by the formula:

\(\frac{x- x _{1}}{ a }=\frac{y- y _{1}}{ b }=\frac{z- z _{1}}{ c }\)

Now, equation of the line through the point \((0,0,0)\) and direction ratios \((1,-3,4)\) is given by:

\(\frac{ x -0}{1}=\frac{ y -0}{-3}=\frac{ z -0}{4}\)

Therefore, the equation of the line in the Cartesian form is:

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

Given:

Vectors î - 2xĵ - 3yk̂ and î + 3xĵ + 2yk̂ are orthogonal to each other,

⇒ (î - 2xĵ - 3yk̂).(î + 3xĵ + 2yk̂) = 0

⇒ 1 - 6x² - 6y² = 0

⇒ 6x² + 6y² = 1

⇒ x² + y² =$\frac{1}{6}$

So, locus of the point (x, y) is circle.

The number of ways to select \(r\) things out of \(n\) given things wherein \(r \leq n\) is given by: \({ }^{n} C_{r}=\frac{n !}{r ! \times(n-r) !}\)

Given: There are 5 subjects and for a student to pass in an examination the students has to pass in each of the 5 subjects.

Here, we have to find in how many ways a student can fail in the examination.

In order to fail in the examination the student can fail in 1 or 2 or 3 or 4 or 5 subjects out of the 5 subjects in each case.

Case 1: Student fails in any 1 subject out of the 5 subjects.

No. of ways in which student can fail in any 1 subject out of the 5 subjects \(={ }^{5} C_{1}\)

Case 2: Student fails in any 2 subjects out of the 5 subjects.

No. of ways in which student can fail in any 2 subjects out of the 5 subjects \(={ }^{5} C_{2}\)

Case 3: Student fails in any 3 subjects out of the 5 subjects.

No. of ways in which student can fail in any 3 subjects out of the 5 subjects \(={ }^{5} C_{3}\)

Case 4: Student fails in any 4 subjects out of the 5 subjects.

No. of ways in which student can fail in any 4 subjects out of the 5 subjects \(={ }^{5} C_{4}\)

Case 5: Student fails in all 5 subjects.

No. of ways in which student can fail all 5 subjects. \(={ }^{5} C_{5}\)

\(\therefore\) Total number of ways in which a student can fail in an examination \(={ }^{5} C_{1}+{ }^{5} C_{2}+{ }^{5} C_{3}+{ }^{5} C_{4}+{ }^{5} C_{5}=31\)