Mathematics Test - 17

Maximize

\(\mathrm{Z}=(\mathrm{x}+\mathrm{y})\)

\(\Rightarrow \mathrm{x}-\mathrm{y} \leq-1, \mathrm{y}-\mathrm{x} \leq 0, \mathrm{x}, \mathrm{y} \leq 0\)

put \(\mathrm{x}=0\) in \(\mathrm{x}-\mathrm{y}=-1 ..............\) (i)

\(\mathrm{y}=1\)

put \(\mathrm{x}=1\) in equation (i)

\(\mathrm{y}=2\)

put \(\mathrm{x}=-1\)

\(-1-\mathrm{y}=-1\)

\(\mathrm{y}=0\)

\(\mathrm{y}-\mathrm{x}=0\).......(ii)

put \(\mathrm{x}=0\) in equation (ii)

\(\mathrm{y}=0\)

put \(\mathrm{x}=1\)

\(\mathrm{y}=1\)

put \(\mathrm{x}=2\)

\(\mathrm{y}=2\)

put \(\mathrm{x}=-1\)

\(\mathrm{y}=-1\)

No feasible region, hence, no maximum value.

The fundamental principal of multiplication:

Let us suppose there are two tasks A and B such that task A can be done in m different ways following which the second task B can be done in n different ways.

Then the number of ways to complete the task A and B in succession respectively is given by: m \(\times\) n ways

Here, we have to form a 5 letter code using the first 6 letters of the English alphabet such that no letter is letter repeated.

Number of ways to choose \(1^{\text {st }}\) letter for the code \(=6\)

Number of ways to choose \(2^{\text {nd }}\) letter for the code \(=5\)

Number of ways to choose \(3^{\text {rd }}\) letter for the code \(=4\)

Number of ways to choose \(4^{\text {th }}\) letter for the code \(=3\)

Number of ways to choose \(5^{\text {th }}\) letter for the code \(=2\)

The number of ways to form a 5 letter code using first 6 letters of english alphabet \(=6 \times 5 \times 4 \times 3 \times 2=720\).

Given,

Equation of curve: \(\sqrt{ x }+\sqrt{ y }=1\)

\(\therefore \sqrt{ y }=1-\sqrt{ x }\)

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

\(=1-2 \sqrt{ x }+ x\)

Now let's find the limits,

\(\text { If } x=0\)

\(\sqrt{x}+\sqrt{y}=1\)

\(\Rightarrow \sqrt{y}=1\)

\(\Rightarrow y=1\)

And if, \(y=0, x=1\)

So the curve cut \(Y\)-axis at \((0,1)\) and \(X\)-axis at \((1,0)\)

\(\therefore x=0, x=1\)

Now,

\(\text { Area }= A =\int_{0}^{1} ydx\)

\(=\int_{0}^{1}(1-2 \sqrt{ x }+ x ) dx\)

\(=\left[ x -\frac{2( x )^{\frac{3}{2}}}{\frac{3}{2}}+\frac{ x ^{2}}{2}\right]_{0}^{1}\)

\(=\left[\left(1-\frac{4}{3}+\frac{1}{2}\right)-0\right]\)

\(=\left[-\frac{1}{3}+\frac{1}{2}\right]\)

\(=\frac{1}{6} \text { square unit }\)

Given:

\(f(x)=\frac{4 x^{3}-3 x^{2}}{6}-2 \sin x+(2 x-1) \cos x\)

\(f^{\prime}(x)=\left(2 x^{2}-x\right)-2 \cos x+2 \cos x-\sin x(2 x-1)\)

\(=(2 x-1)(x-\sin x)\)

\(\text { for } x>0, x-\sin x>0\)

\(x<0, x-\sin x<0\)

for \(x \in(-\infty, 0] \cup\left[\frac{1}{2}, \infty\right), f ( x ) \geq 0\)

for \(x \in\left[0, \frac{1}{2}\right], f^{\prime}(x) \leq 0\)

\(\Rightarrow f( x )\) increases in \(\left[\frac{1}{2}, \infty\right)\)

Let \(P(h, k)\) be any point on the locus. Then

Given: \(PA = PB\)

\(\Rightarrow {PA}^{2}={PB}^{2}\)

\(\Rightarrow(h-2)^{2}+(k-3)^{2}=(h+1)^{2}+(k-2)^{2}\)

\(\Rightarrow h^{2}-4 h+4+k^{2}-6 k+9=h^{2}+2 h+1+k^{2}-4 k+4\)

\(\Rightarrow-4 h-6 k+9=2 h-4 k+1\)

\(\Rightarrow 6 h+2 k=8\)

The equation of the locus of a point equidistant from the point \(A(2,3)\) and \(B(-1,2)\) is \(6 x+2 y=8\)

Given:

Arithmetic mean \(=(\mu)=16\)

variance \(=\sigma^{2}=4\)

We know that:

Standard deviation \(=\sigma=\sqrt{\text { variance }}\)

\(C V=\) Coefficient of variation \(=\frac{\sigma}{\mu}\)

\(\sigma=\) Standard deviation

\(\mu=\) Mean

Standard deviation \(=\sigma=\sqrt{4}=2\)

\(CV =\left(\frac{\sigma}{\mu}\right) \times 100\)

\(\Rightarrow C V=\left(\frac{2}{16}\right) \times 100\)

\(\therefore\) Coefficient of variation is \(12.5\)

Given: The standard deviation and the mean of 16 values are 15.6 and 20.5 respectively

Here, we have to find the coefficient of variation for the same observations.

As we know, Coefficient of variation \(=\frac{\text { standard deviaiton }}{\text { Mean }} \times 100\)

Coefficient of variation \(=\frac{15.6}{20.5} \times 100=76 \%\) (approximately)

Given: \(\cot A+\cot B=2 \cot C\)

\(\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}=\frac{2 \cos C}{\sin C}\)

We know that, \(\operatorname{Cos} A=\frac{b^{2}+c^{2}-a^{2}}{2 b e}, \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}\) and\(\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)

\(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=m\)

\(\Rightarrow \sin A=\frac{a}{m}, \sin B=\frac{b}{m}, \sin C=\frac{c}{m}\)

By substitution \(2 c^{2}=2\left(a^{2}+b^{2}-c^{2}\right)\)

\(\Rightarrow 2 c^{2}=a^{2}+b^{2}\)

Let the daughter's age be$x$ years.

Then, father's age$=\left(3x\right)$ years.

Mother's age$=(3x-9)$ years;

Son's age$=(x+7)$ years.

So,

$\Rightarrow x=23$

$\Rightarrow 2x+14=3x-9$

$\Rightarrow x=23$

Therefore Mother's age$=(3x-9)=(69-9)$ years$=60$ years

Trace of a matrix:

Trace of a matrix is the sum of elements on the main diagonal.

The trace is only defined for a square matrix \((n \times n)\).

Let \(A\) be \(n \times n\) matrix.

\(\operatorname{tr}(A)=\sum_{n=1}^n A_n\)

Given: \(A=\left[\begin{array}{ccc}1 & -5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9\end{array}\right]\)

Trace of matrix \(=\) Sum of elements on the main diagonal

\(= 1+7+9\)

\(= 17\)