Mathematics Test - 28

According to the question,

We have to find the value of \(\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]\).

As we know, the matrix \(\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\) has 1 row and 3 columns, therefore the order of the matrix is \(1 \times\)3.

Also, since the matrix \(\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{ b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]\) has 3 rows and 3 columns, therefore the order of the matrix is \(\mathrm{3} \times \mathrm{3}\).

We know that, if \(\mathrm{A}\) is a matrix with order \({m} \times {n}\) and \(\mathrm{B}\) is a matrix with order \({n} \times {p}\), then the matrix \(\mathrm{AB}\) has the order \({m} \times {p}\).

Now we have to compute the order of \(\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]\), so the product of the orders of the matrices to get the result that is:

\((1 \times 3)(3 \times 3)\)

\(=(1 \times 3)\)

Therefore,

\(\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]\)

\(=[\mathrm{ax}+\mathrm{h} \mathrm{y}+\mathrm{g} \mathrm{z} \quad \mathrm{hx}+\mathrm{by}+\mathrm{fz} \quad \mathrm{gx}+\mathrm{fy}+\mathrm{cz}]\)

It is known that the distance between a point \(p\left(x_{1}, y_{1}, z_{1}\right)\) and a plane \(A x+B y+C z=D\), is given by, \(d=\left|\frac{A x_{1}+B y_{1}+C z_{1}-D}{\sqrt{A^{2}+B^{2}+C^{2}}}\right|\quad\quad\)...(1)
(a) The given point is (0,0,0) and the plane is \(3 x-4 y+12 z=3\).
\(d=\left|\frac{3 \times 0-4 \times 0+12 \times 0-3}{\sqrt{(3)^{2}+(-4)^{2}+(12)^{2}}}\right|=\left|\frac{3}{\sqrt{169}}\right|=\frac{3}{13}\)
(b) The given point is (3,-2,1) and the plane is \(2 x-y+2 z+3=0\).
\(d=\left|\frac{2 \times 3+(-2) \times(-1)+2 \times 1+3}{\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}}\right|=\left|\frac{13}{3}\right|=\frac{13}{3}\)
(c) The given point is (2,3,-5) and the plane is \(x+2 y-2 z=9\).
\(d=\left|\frac{2+2 \times 3-2(-5)-9}{\sqrt{(1)^{2}+(2)^{2}+(-2)^{2}}}\right|=\left|\frac{9}{3}\right|=3\)
(d) The given point is (-6,0,0) and the plane is \(2 x-3 y+6 z-2=0\).
\(d=\left|\frac{2(-6)-3 \times 0+6 \times 0-2}{\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}}}\right|=\left|\frac{-14}{\sqrt{49}}\right|=\frac{14}{7}=2\)

Given equation of the circle is:

\(x^{2}+y^{2}-4 x-6 y+11=0\)

\(\therefore \quad 2 g=-4\) and. \(2 f=-6\)

So, the centre of the circle is \(C(-g,-f) \equiv C(2,3)\) \(A(3,4)\) is one end of the diameter. Let the other end of the diameter be \(B\left(x_{1}, y_{1}\right)\)

Here, mid point of \(A B\) is \(C\).

\(\therefore 2=\frac{3+x_{1}}{2}\) and \(3=\frac{4+y_{1}}{2}\)

\(\Rightarrow x_{1}=1\) and \(y_{1}=2\)

So, the coordinates of other end of the diameter are (1,2).

Given: \(\lim _{x \rightarrow-\infty}\left[\frac{x^{4} \sin \left(\frac{1}{x}\right)+x^{2}}{\left(1+|x|^{3}\right)}\right]\).

We know, when \(x\) tends to \(-\infty, \frac{1}{x}\) tends to \(0\).

We also know, we can write \(x^{4}\) as \(\frac{x^{3}}{\frac{1}{x}} \ldots \) (i) and \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\ldots\)(ii).

Now, let the value of the limit be \(\mathrm{L}\).

So, \(L=\lim _{x \rightarrow-\infty}\left[\frac{x^{4} \sin \left(\frac{1}{x}\right)+x^{2}}{\left(1+|x|^{3}\right)}\right]\)

Now, from (i), we can write \(L=\lim _{x \rightarrow-\infty}\left[\frac{\frac{x^{3}}{\frac{1}{x}} \sin \left(\frac{1}{x}\right)+x^{2}}{\left(1+|x|^{3}\right)}\right]\)

\(\Rightarrow L=\lim _{x \rightarrow-\infty}\left[\frac{x^{3} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}}+x^{2}}{\left(1+|x|^{3}\right)}\right]\)

Now, from (ii), we know \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\)

So, we can write \(L=\lim _{x \rightarrow-\infty}\left[\frac{x^{3}+x^{2}}{\left(1+|x|^{3}\right)}\right]\)

Now, as \(x \rightarrow-\infty\), the value of \(x\) is negative.

So, by the definition of modulus function, we can say \(|x|=-x\).

\(\Rightarrow|x|^{3}=-x^{3}\)

\(\Rightarrow 1+|x|^{3}=1-x^{3}\)

So, we get the limit as \(L=\lim _{x \rightarrow-\infty}\left[\frac{x^{3}+x^{2}}{\left(1-x^{3}\right)}\right]\)

Now, the function consists of polynomials.

So, we will take the variable with the highest power, i.e., \(x^3\), common from the numerator as well as the denominator.

So, we get:

\(L=\lim _{x \rightarrow-\infty}\left[\frac{x^{3}\left(1+\frac{1}{x}\right)}{x^{3}\left(\frac{1}{x^{3}}-1\right)}\right]\)

\(\Rightarrow L=\lim _{x \rightarrow-\infty}\left[\frac{\left(1+\frac{1}{x}\right)}{\left(\frac{1}{x^{3}}-1\right)}\right]\)

Substituting \(x=-\infty\) in the limit, we get:

\(L=\frac{1+\frac{1}{-\infty}}{\frac{1}{-\infty^{3}}-1}\)

\(\Rightarrow L=\frac{1+0}{0-1}\)

\(\Rightarrow L=-1\)

I implies II means ≡ I → II

\(X^{-1}=\frac{\operatorname{Adj}(\mathrm{X})}{|X|}\)

If \(|X| \neq 0\) then \(X^{-1}\)

\(|X|=\frac{\operatorname{Adj}(X)}{X^{-1}}\)

If \(X^{-1}\) then \(|X| \neq 0\) also \(\mid\) Adj \(\left.X|=| X\right|^{n-1}\) then \(\mid\) Adj \(X \mid \neq 0\)

If \(X^{-1}\) then \(|X| \neq 0\)

I implies II and II implies I

∴ Both I and II are equivalent.

Note:

\(X^{-1}\) means \(X\) is invertible

|X| determinant of \(X\)

Given differential equation\(\frac{d y}{d x}=e^{x-y}+x^{2} e^{-y}\)

On solving, we get-

\(e^{y} d y=\left(e^{x}+x^{2}\right) d x\)

Integrationon both sides

\(\int e^{y} d y=\int\left(e^{x}+x^{2}\right) d x\)

\(e^{y}=e^{x}+\frac{x^{3}}{3}+c\)

\(3\left(e^{y}-e^{x}\right)=x^{3}+c^{\prime}\)

\(f: \mathrm{R} \rightarrow \mathrm{R}\) is defincd as \(f(x)=x^{4}\).

Let \(x, y \in \mathrm{R}\) such that \(f(x)=f(y)\).

\(\Rightarrow x^{4}=y^{4}\)

\(\Rightarrow x=\pm y\)

\(\therefore f(x)=f(y)\) does not imply that \(x=y\).

For example \(f(1)=f(-1)=1\)

\(\therefore f\) is not one-one.

Consider an element 2 in co-domain \(\mathrm{R}\). It is clear that there does not exist any \(x\) in domain \(\mathrm{R}\) such that \(f(x)=2\).

\(\therefore f\) is not onto.

Thus, function \(f\) is neither one \(-\) one nor onto.

Clearly, there are 2 ways of answering each of the 5 questions i.e true or false.

\(\therefore\) Total number of different sequence of answers \(=2 \times 2 \times 2 \times 2 \times 2=32\)

There is only one correct sequence.

So, the maximum number of sequences except for the all correct answer sequence \(=32-1=31\)

\(\because\) Different students have given different sequence of answers, so the maximum possible number of students \(=31\)

If \(f: \mathrm{R} \rightarrow \mathrm{R}\) be given by \(f(x)=\left(3-x^{3}\right)^{\frac{1}{3}},\)

\(f(x)=\left(3-x^{3}\right)^{\frac{1}{3}}\)

\(\therefore f o f(x)=f(f(x))=f\left(\left(3-x^{3}\right)^{\frac{1}{3}}\right)=\left[3-\left(\left(3-x^{3}\right)^{\frac{1}{3}}\right)^{3}\right]^{\frac{1}{3}}\)

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

\(f o f(x)=x\)

Solving \(y^{2}=6(x-1)\) and \(y^{2}=3 x\) we get

\(6(x-1)=3 x\)

\(\Rightarrow x=2\)

So, \(y=\pm \sqrt{6}\)

\(y^{2}=6(x-1) \Rightarrow x=1+\frac{y^{2}}{6}\) and

\(y^{2}=3 x \Rightarrow x=\frac{y^{2}}{3}\)

Area \(=\int_{-\sqrt{6}}^{\sqrt{6}}\left(1+\frac{y^{2}}{6}-\frac{y^{2}}{3}\right) d y\)

\(=2 \int_{0}^{\sqrt{6}}\left(1-\frac{y^{2}}{6}\right) d y\)

\(=2\left[y-\frac{y^{3}}{18}\right]_{0}^{\sqrt{6}}\)

\(=2 \times \frac{2 \sqrt{6}}{3}=\frac{4 \sqrt{6}}{3}\)

Area \(=\frac{4 \sqrt{6}}{3}\)