Mathematics Test - 60

Given:

\(\frac{x}{4}>\frac{x}{2}+1\)

\(\Rightarrow \frac{x}{4}-\frac{x}{2}>1\)

\(\Rightarrow \frac{x-2 x}{4}>1\)

\(\Rightarrow \frac{-x}{4}>1\)

\(\Rightarrow \frac{-x}{4}>1\)

\(\Rightarrow-x>4\)

\(\Rightarrow x<-4\)

\(\begin{aligned} & \vec{a}=\hat{i}-\hat{j}+2 \hat{k} \\ & \vec{a} \times \vec{b}=2 \hat{i}-\hat{k} \\ & \vec{a} \cdot \vec{b}=3 \\ & |\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2=|\vec{a}|^2 \cdot|\vec{b}|^2 \\ & \Rightarrow 5+9=6|\vec{b}|^2 \\ & \Rightarrow|b|^2=\frac{7}{3} \\ & |\vec{a}-\vec{b}|=\sqrt{|\vec{a}|^2+\cdot|\vec{b}|^2-2 \vec{a} \cdot \vec{b}}=\sqrt{\frac{7}{3}} \\ & \text { projection of } \vec{b} \text { on } \vec{a}-\vec{b}=\frac{\vec{b} \cdot(\vec{a}-\vec{b})}{|\vec{a}-\vec{b}|} \\ & =\frac{\vec{b} \cdot \vec{a}-|\vec{b}|^2}{\vec{a}-\vec{b} \mid}=\frac{3-\frac{7}{3}}{\sqrt{\frac{7}{3}}} \\ & =\frac{2}{\sqrt{21}}\end{aligned}\)

The mean deviation is the sum of the deviations from the average divided by the number of items in the series.

\(MD =\left(\frac{1}{n}\right)\sum \mid x - M \mid\)

\(\therefore M\) is Mean or the median or mode.

If \(f^{\prime}(x)>0\) at each point in an interval then the function is said to be increasing.

If \(f ^{\prime}( x )<0\) at each point in an interval I, then the function is said to be decreasing.

Here, derivative of \(e^{x}\) is \(e^{x}\).

and it is greater than zero at any interval

\(\therefore e^{ x }\) is an increasing function.

Given:

\(\lim _{x \rightarrow 0} \frac{3^{x}+3^{-x}-2}{x}\)

We know that:

\(\lim _{x \rightarrow a}[f(x)+g(x)]=\lim _{x \rightarrow a} f(x)+\lim _{x \rightarrow a} g(x)\)

\(\lim _{x \rightarrow 0} \frac{\left(a^{x}-1\right)}{x}=\log a\)

\(\log m^{n}=n \log m\)

Therefore,

\(\lim _{x \rightarrow 0} \frac{3^{x}+3^{-x}-2}{x}\)

\(=\lim _{x \rightarrow 0} \frac{3^{x}-1+3^{-x}-1}{x}\)

\(=\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}+\lim _{x \rightarrow 0} \frac{3^{-x}-1}{x}\)

\(=\lim _{x \rightarrow 0} \frac{3^{x}-1}{x}+\lim _{x \rightarrow 0} \frac{\left(3^{-1}\right)^{x}-1}{x}\)

\(=\log 3+\log \left(3^{-1}\right)\)

\(=\log 3-\log 3\)

\(=0\)

Let number of first kind of cake is \(\mathrm{X}\) and another kind of cake is \(\mathrm{Y}\).

So, total flour required \(=200 \mathrm{X}+100 \mathrm{Y} \mathrm{g}\) 

And total fat required \(=25 \mathrm{X}+50 \mathrm{Y} \mathrm{g}\)

Since, maximum flour available is \(5 \mathrm{~kg}=5000 \mathrm{~g}\)

\(\therefore 200 \mathrm{X}+100 \mathrm{Y} \leq 5000\)

\(\Rightarrow 2 \mathrm{X}+\mathrm{Y} \leq 50\).....(1)

Also, maximum fat available is \(1 \mathrm{~kg}=1000 \mathrm{~g}\)

\(\therefore 25 \mathrm{X}+50 \mathrm{Y} \leq 1000\)

\(\Rightarrow \mathrm{X}+2 \mathrm{Y} \leq 40\)..........(2)

Since, quantity of cakes can't be negative.

\(\therefore \mathrm{X} \geq 0, \mathrm{Y} \geq 0\).....(3)

We have to maximize number of cakes that can be made.

So, Objective function is \(\mathrm{Z}=\mathrm{X}+\mathrm{Y}\)

After plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner point \(\mathrm{A}(\mathrm{0}, 20)\)

Value of \(Z=0+20=20\)

Corner point \(\mathrm{B}(20,10)\),

Value of \(Z=20+10=30\)

Corner point \(\mathrm{C}(25,0)\),

Value of \(Z=25+0=25\)

So, maximum cake that can be made is \(30\), where first kind of cake will be \(20\) and second kind of cake will be \(10\).

Given here,

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

We know that:

If \(f(x)=x^{n}\), then: 

\(f^{\prime}(x)=n x^{n-1}\)

Differentiating (1) with respect to x, we get:

\(\Rightarrow f^{\prime}(x)=1-\left(\frac{-1}{x^{2}}\right)\)

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

Putting x = -1 in above, we get:

\(f'(-1)=1+\frac{1}{(-1)^{2}}\)

\(=1+1\)

\(=2\)

Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem, we find that the value of the objective function for a maximization problem will likely be less than that for the simplex solution.

As we know to find the maximum or minimum value of the objective function we used to solve it for an initial basic feasible solution by the simplex method.For this, so we will convert the problem into the standard form which involves objective function and constraints equations in terms of x and y whose values we will determine from the initial basic feasible solution and using it further when we will get the optimization problem we will see that the value of the objective function for a maximization problem will likely be less than that for the simplex solution.

Given:

The lines \(\frac{x-3}{3}=\frac{y+4}{2}=\frac{z-1}{\lambda}\) and \(\frac{x+1}{3}=\frac{y-2}{2}=\frac{z}{1}\) are coplanar.

Here, we have to find the value of \(\lambda\).

As we know that, if two lines \(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\) and \(\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}\) arecoplanar then

\(\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|=0\)

Here, \(x_{1}=3, y_{1}=-4, z_{1}=1, a_{1}=3, b_{1}=2, c_{1}=\lambda\)

Similarly, \(x_{2}=-1, y_{2}=2, z_{2}=0, a_{2}=3, b_{2}=2\) and \(c_{2}=1\)

\(\left|\begin{array}{ccc}-4 & 6 & -1 \\ 3 & 2 & \lambda \\ 3 & 2 & 1\end{array}\right|=0\)

\(-4 \times(2-2 \lambda)-6 \times(3-3 \lambda)-1 \times(6-6)=0\)

\(-8+8 \lambda-18+18 \lambda=0\)

\(\lambda=1\)