Mathematics Test - 17

In this question, we need to evaluate the given equation. For this, we will use the partial fraction method and apply the defined integral formulae.

Let the given integral be I so, we can write:

\(I=\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\)

Now, by following the partial fraction method, we can write the function \(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\) as:

\(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)}\)

\(=\frac{A(x+1)(x+2)+B(x+2)+C(x+1)^{2}}{(x+1)^{2}(x+2)}\)

\(=\frac{A\left(x^{2}+3 x+2\right)+B x+2 B+C\left(x^{2}+2 x+1\right)}{(x+1)^{2}(x+2)}\)

\(=\frac{x^{2}(A+C)+x(3 A+B+2 C)+(2 A+2 B+C)}{(x+1)^{2}(x+2)}\)

Now, comparing the coefficients of both sides of the above equation, we get:

\(\Rightarrow A+C=1\quad\quad\)......(i)

\(\Rightarrow 3 A+B+2 C=1\quad\quad\).....(ii)

\(\Rightarrow 2 A+2 B+C=1\quad\quad\).....(iii)

Solving the above equations to determine the value of the constants A, B and C.

From the equation (i), we can write \(A=1-C\quad\quad\)....(iv)

Substituting the value from the equation (iv) in the equation (ii) and (iii), we can write

\(\Rightarrow 3 A+B+2 C=1\)

\(\Rightarrow 3(1-C)+B+2 C=1\)

\(\Rightarrow 3-3 C+B+2 C=1\)

\(\Rightarrow B-C=-2\quad\quad\)..........(v)

Similarly,

\(\Rightarrow 2 A+2 B+C=1\)

\(\Rightarrow 2(1-C)+2 B+C=1\)

\(\Rightarrow 2-2 C+2 B+C=1\)

\(\Rightarrow 2 B-C=-1\quad\quad\).........(vi)

Solving equations (v) and (vi).

From the equation (v), we get:

\(\Rightarrow B-C=-2\)

\(\Rightarrow B=-2+C\quad\quad\)........(vii)

Substituting the value of the equation (vii) in the equation (vi) we get,

\(\Rightarrow 2 B-C=-1\)

\(\Rightarrow 2(-2+C)-C=-1\)

\(\Rightarrow-4+2 C-C=-1\)

\(\Rightarrow C=3\)

Substituting the value of C as 3 in the equation (vii), we get,

\(\Rightarrow B=-2+C\)

\(=-2+3\)

\(=1\)

Again, substituting the value of C as 3 in the equation (iv), we get,

\(\Rightarrow A=1-C\)

\(=1-3\)

\(=-2\)

Therefore, the values of the constant A, B and C are -2, 1 and 3 respectively.

So, the given integral function can be re-written as:

\(\Rightarrow \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x=\int\left(\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)}\right) d x\)

\(=\int\left(\frac{-2}{(x+1)}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)}\right) d x\)

Now, applying the property of the integration function.

\(\int(A+B+C) d x=\int A d x+\int B d x+\int C d x\) in the above function.

\(\Rightarrow \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x=\int\left(\frac{-2}{(x+1)}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)}\right) d x\)

\(=\int \frac{-2}{(x+1)} d x+\int \frac{1}{(x+1)^{2}} d x+\int \frac{3}{(x+2)} d x\)

Again, applying the property of the integration function \(\int \frac{d x}{x}=\ln x\) in the above equation.

\(\Rightarrow \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x=\int \frac{-2}{(x+1)} d x+\int \frac{1}{(x+1)^{2}} d x+\int \frac{3}{(x+2)}\)

\(=-2 \ln |x+1|+\int(x+1)^{-2} d x+3 \ln |x+2|+c\)

\(=-2 \ln |x+1|+3 \ln |x+2|-\frac{1}{(x+1)}+c\)

Therefore, we can see that the value of the integral \(\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\) is \(-2 \ln |x+1|+3 \ln |x+2|-\frac{1}{(x+1)}+c\) where, \(\tau\) is the integral constant.

The given equation is-

$-x^2+5x-6>0$

$\Rightarrow x^2-5x+6<0$

$\Rightarrow x^2-3x-2x+6<0$

$\Rightarrow x(x-3)-2(x-3)<0$

$\Rightarrow (x-3)(x-2)<0$

So, (2,3) is the solution to the given inequality.

Given,

For line I

\(x=a y+b \)

\(\Rightarrow y=\frac{x-b}{a} \ldots(1) \)

\(z=c y+d \)

\(\Rightarrow y=\frac{z-d}{c} \ldots(2)\)

From (1) and (2),

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

which can also be written as,

\(\frac{x-a}{a}=\frac{y-0}{1}=\frac{z-d}{c} \ldots(3)\)

Similarly,

For line II

\(\Rightarrow y=\frac{x-b^{\prime}}{a^{\prime}} \ldots(4) \)

\(z=c^{\prime} y+d^{\prime} \)

\(\Rightarrow y=\frac{z-d^{\prime}}{c^{\prime}} \ldots(5)\)

we can write,

\(\frac{x-b^{\prime}}{a^{\prime}}=\frac{y-0}{1}=\frac{z-d^{\prime}}{c^{z}} \text {... (6) }\)

As we know if two lines are perpendicular then \(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)

From equation (3) and (6), we get \(a a^{\prime}+c c^{\prime}+1=0\)

Given:

\(\overrightarrow{\mathrm{a}}=-4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}\) and\(\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}\)

\(\overrightarrow{\mathrm{c}}=\mathrm{ma}+\mathrm{n} \overrightarrow{\mathrm{b}}\)

\(2 \hat{\imath}+3 \hat{\jmath}=m(-4 \hat{\imath}+2 \hat{j})+n(2 \hat{\imath}+\hat{\jmath})\)

\(2 \hat{\imath}+3 \hat{\jmath}=(-4 m+2 n) \hat{\imath}+(2 m+n) \hat{\jmath}\)

Equating the scalar coefficients, we get:

\(-4 m+2 n=2\) ..(1)

\(2 m+n=3\) ...(2)

Multiplying equation (2) by 2 and adding to equation (1), we get:

\(4 n=8\)

\(n=2\)

Using either of the equations above, we also get:

\(\mathrm{m}=\frac{1}{2}\)

\(\therefore m+n=2+\frac{1}{2}=\frac{5}{2}\)

Mean = 10

Mean \(=\frac{(1+4+9+X+12+14+15+16)}{8}\)

\(\Rightarrow 10=\frac{(71+x)}{8}\)

\(\Rightarrow 71+x=80\)

\(\Rightarrow x=80-71\)

\(\Rightarrow x=9\)

The data are 1, 4, 9, 9, 12, 14, 15 and 16.

∵ 9 occurs 2 times in th given data.

∴ Mode = 9 (maximum frequency)

Given:

\(A=\{1,2,3,4,5,7,8,9\}\) and \(B=\{2,4,6,7,9\}\)

As we know that, \(A \cap B=\{x: x \in A\) and \(x \in B\}\)

\(\Rightarrow A \cap B=\{2,4,7,9\}\)

As we can see that, the number of elements present in \(A \cap B=4\) i.e \(n(A \cap B)=4\)

As we know that; If \(A\) is a non-empty set such that \(n(A)=m\) then the numbers of proper subsets of \(A\) are given by \(2^{m}-1\).

So, the number of proper subsets of \(A \cap B=2^{4}-1=15\)

Given:

The polygon given is a heptagon.

Therefore,the no. of sides of the given polygon \({n}=7\).

As we know that, no. of diagonals that can be drawn by joining the angular points of apolygon having \(n\) sides is given by: \({ }^{n} C_{2}-n=\frac{n \times(n-3)}{2}\)

\(\Rightarrow \text { No. of diagonals }=\frac{7(7-3)}{2}\)

Therefore,

No. of diagonal of heptagon \(=14\)