Mathematics Test - 8

Let set A and B have mm and n elements, respectively.

2^m−2ⁿ = 56

2ⁿ(2^m−n−1) = 56 = 8×7 = 2³×7

Comparing both sides, we get

2ⁿ = 2³ and 2^m−n = 7

⇒ n = 3 and 2^m−n = 8

⇒ 2^m−n = 2³⇒ m-n = 3

⇒ m−3 = 3 ⇒ m = 6

Number of the elements in A is 6.

Given,

\(3 x+y+3 z=8\) and \(9 x+3 y+9 z=15\)

Now,

Divide \(9 x+3 y+9 z=15\) by \(3\)

We get,

\(3 x+y+3 z=5\)

Now,

We know that,

Distance between two parallel plane \(a x+b y+c z+d_{1}=0\) and \(a x+b y+c z+d_{2}=0\) is \(\left|\frac{{d}_{1}-{d}_{2}}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}\right|\)

So,

Distance between \(3 x+y+3 z=8\) and \(3 x+y+3 z=5\)

\(=\left|\frac{8-5}{\sqrt{3^{2}+1^{2}+3^{2}}}\right| \)

\(=\frac{3}{\sqrt{19}}\)

Given:

\(f(x)=x^{3}+3 x^{2}+3 x-7\)

We know that:

\(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\)

Derivative of a constant, i.e.,

\(\frac{\mathrm{d}(\text { constant })}{\mathrm{dx}}=0\)

Therefore, 

\(\frac{d f(x)}{d x}=3 x^{2}+6 x+3\)

Putting \(x=2\) in above, we get:

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=3(2)^{2}+6(2)+3\)

\(=3(4)+12+3\)

\(=12+12+3\)

\(\frac{\mathrm{d} \mathrm{f}(\mathrm{x})}{\mathrm{dx}}=27\)

The value of \(\frac{\mathrm{d} \mathrm{f}(x)}{\mathrm{dx}}\) at \(\mathrm{x}=2\) is 27.

\(\mathrm{Z}=3 \mathrm{x}+2 \mathrm{y}\)

Subject to the constraints

\(x+2 y \leq 10 \)

\(3 x+y \leq 15\)

Convert these inequalities into equations

\(\mathrm{x}+2 \mathrm{y}=10 \).....(i)

\(3 \mathrm{x}+\mathrm{y}=15 \text {......... (ii) }\)

From (i), we get

\(\mathrm{x}=0 \) when \( \mathrm{y}=5\) and \(\mathrm{y}=0\) when \(\mathrm{x}=10\)

So, the points \((0,5)\) and \((10,0)\) lie on the line given in (i).

From (ii), we get the points \((0,15)\) and \((5,0)\)

Let's plot these point and we get the graph in which, shaded part shows the feasible region.

Lines (i) and (ii) intersect at \((4,3)\) and other corner points of the region are \((0,5),(5,0)\) and \((0,0)\).

To find the maximum value of \(Z\), we need to find the value of \(Z\) at the corner points.

\(\begin{array}{lc}\text { Corner points } & \mathrm{Z}=3 \mathrm{x}+2 \mathrm{y} \\ (0,0) & 0 \\ (5,0) & 15 \\ (0,5) & 10 \\ (4,3) & 18\end{array}\)

Thus, \(Z\) is maximum at \((4,3)\) and its maximum value is 18.

Given that:

\(\left|\begin{array}{lcc}-a^{2} & a b & a c \\ a b & -b^{2} & b c \\ a c & b c & -c^{2}\end{array}\right|\)

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

\(=0+2 a^{2} b^{2} c^{2}+2 a^{2} b^{2} c^{2}\)

\(=4 a^{2} b^{2} c^{2}\)

\(f(x)=2 x-\tan ^{-1}x -\log \left\{x+\sqrt{x^{2}+1}\right\}\)

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

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

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

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

Function is increasing monotonically.

\(\Rightarrow \frac{1+2 x^{2}-\sqrt{{x}^{2}+1}}{1+{x}^{2}}>0\)

\(\Rightarrow 1+2 x^{2}-\sqrt{x^{2}+1}>0\)

\(\Rightarrow 1+2 x^{2}>\sqrt{x^{2}+1}\)

Squaring on both sides,

\(\Rightarrow\left(1+2 x^{2}\right)^{2}>x^{2}+1\)

\(\Rightarrow 4 x^{4}+3 x^{2}>0\)

For all \(x \in R\)

The unit vector u in the direction of a vector \(\overrightarrow{{A}}\) is given by: \(\hat{{u}}=\frac{\overrightarrow{{A}}}{|\vec{A}|}\), where \(|\overrightarrow{{A}}|\) is the magnitude of the vector \(\overrightarrow{{A}}\). The magnitude of a vector \(\overrightarrow{{A}}\) is given as: \(|\overrightarrow{{A}}|=\sqrt{\overrightarrow{{A}} \cdot \overrightarrow{{A}}}\). \(|{a} \hat{{i}}+{b} \hat{{j}}+{ck}|=\sqrt{({a} \hat{{i}}+{b} \hat{{j}}+{c} \hat{{k}}) .({a} \hat{{i}}+{b} \hat{{j}}+{ck})}=\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}\)

We have \(\overrightarrow{{A}}=2 \hat{{i}}-\hat{{j}}+2 \hat{{k}}\) and \(\overrightarrow{{B}}=-\hat{{i}}-2 \hat{{j}}+2 \hat{{k}}\).

\(\therefore \overrightarrow{{A}}-\overrightarrow{{B}}=(2 \hat{{i}}-\hat{{j}}+2 \hat{{k}})-(-\hat{{i}}-2 \hat{{j}}+2 \hat{{k}})\)

\(\Rightarrow \overrightarrow{{A}}-\overrightarrow{{B}}=3 \hat{{i}}+\hat{{j}}\)

The magnitude of \(\overrightarrow{{A}}-\overrightarrow{{B}}\) will be:

\(|3 \hat{{i}}+1 \hat{{j}}+0 \hat{{k}}|=\sqrt{3^{2}+1^{2}+0^{2}}=\sqrt{10}\)

Now, the unit vector along the direction of \(\overrightarrow{{A}}-\overrightarrow{{B}}\) will be:

\(\hat{{u}}=\frac{\overrightarrow{{A}}-\overrightarrow{{B}}}{|\overrightarrow{{A}}-\overrightarrow{{B}}|}\)

\(\Rightarrow \hat{{u}}=\frac{3 \hat{\mathbf{i}}+\hat{{j}}}{\sqrt{10}}\)

Here, for solving this LPP problem, converting all inequality constraints to equality constraints.

\(x=4.....(1)\)

\(y=6......(2)\)

\(3 x+2 y=18.....(3)\)

Now, plotting these on the graph,

From the figure we can see,

Point 'O' is origin i.e., \((0,0)\).

Co-ordinate of \(A=(4,0)\)

Point 'B' is the intersection point of\(x=4\) and\(3 x+2 y=18\), then

\(x=4\) putting in the equation \((3)\), we get

\(3 \times 4+2 y=18\)

\(y=3\)

Co-ordinate of \(B=(4,3)\)

Point 'C' is the intersection point of \(y=6\) and \(3 x+2 y=18\), then

\(y=6\) putting in the equation \((3)\), we get

\(3 x+2 \times 6=18\)

\(x=2\)

Co-ordinate of \(C=(2,6)\)

Co-ordinate of \(D=(0,6)\)

The shaded region will be optimum region.

Now, finding value of objective function at all points

\(\mathrm{Z}(O)=0+0=0\)

\(\mathrm{Z(A)=Z}(4,0)=6 \times 4+0=24\)

\(\mathrm{Z(B)=Z}(4,3)=6 \times 4+10 \times 3=54\)

\(\mathrm{Z(C)=Z}(2,6)=2 \times 6+6 \times 10=72\)

\(\mathrm{Z(D)=Z}(0,6)=0+6 \times 10=60\)

So, the maximum value of objective function is \(72\).

Let \(\triangle=\left|\begin{array}{ccc}\mathrm{x}+2 & \mathrm{x}+3 & \mathrm{x}-1 \\ \mathrm{x}+6 & \mathrm{x}+8 & \mathrm{x}+4 \\ \mathrm{x}+9 & \mathrm{x}+11 & \mathrm{x}+7\end{array}\right|\)

By applying \(\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}\)

\(=\left|\begin{array}{lll}\mathrm{x}+2 & 1 & -3 \\ \mathrm{x}+6 & 2 & -2 \\ \mathrm{x}+9 & 2 & -2\end{array}\right|\)

By applying \(\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}+\mathrm{C}_{2}\)

\(=\left|\begin{array}{ccc}x+2 & 1 & -2 \\ x+6 & 2 & 0 \\ x+9 & 2 & 0\end{array}\right|\)

Expanding along \(\mathrm{C}_{3}\), we get

\(=-2[2(x+6)-2(x+9)]\)

\(=-4[x+6-x-9]\)

\(=12\)

Let \(\overrightarrow{ a }= a _{1} \overrightarrow{ i }+ b _{1} \overrightarrow{ j }+ c _{1} \overrightarrow{ k }, \overrightarrow{ b }= a _{2} \overrightarrow{ i }+ b _{2} \overrightarrow{ j }+ c _{2} \overrightarrow{ k }\) and \(\overrightarrow{ c }= a _{3} \overrightarrow{ i }+ b _{3} \overrightarrow{ j }+ c _{3} \overrightarrow{ k }\) be the three vectors.

Condition for coplanarity:

\(\overrightarrow{ a } \cdot(\vec{b} \times \vec{c})=\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|=0\)

Here, \(2 \hat{ i }-\hat{ j }+\hat{ k }, \hat{ i }+2 \hat{ j }-3 \hat{ k }\) and \(3 \hat{ i }+ m \hat{ j }+5 \hat{ k }\) are coplanar.

\(\left|\begin{array}{rrr}2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & m & 5\end{array}\right|=0\)

\(\Rightarrow 2(10+3 m)+1(5+9)+1(m-6)=0\)

\(\Rightarrow 20+6 m+14+m-6=0\)

\(\Rightarrow 7 m+28=0\)

\(\Rightarrow m=-4\)

\(\therefore\) The value of \(m\) is \(-4\).