Mathematics Test - 11

If f(x) is continuous at x = 3

Then$\underset{x\to 3^{-}}{\lim }f\left(x\right)=f\left(3\right)=\underset{x\to 3^{+}}{\lim }f\left(x\right)$

Here$f\left(x\right)=\frac{x^2-9}{x^2-2x-3}=\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}=\frac{x+3}{x-1}$

Now,

$\underset{x\to 3^{-}}{\lim }f\left(x\right)\underset{h\to 0}{\lim }f\left(3-h\right)\underset{h\to 0}{\lim }\frac{3-h+3}{3-h+1}=\underset{h\to 0}{\lim }\frac{6-h}{4-h}$

= 6/4 = 3/2 = 1.5

$\Rightarrow f\left(3\right)=\underset{x\to 3^{-}}{\lim }f\left(x\right)=1.5$

The indefinite integral of function \(f(x)\) with respect to \(g(x)\) is defined as \(\int f(x) d g(x)\).

So, indefinite integral of \(\sin (x)\) with respect to \(\cos (x)\) is:

\(=\int \sin (x) d(\cos (x))\)

To solve this above integral. 

We have to convert the above equation into \(\int f(x) d x\) this form. 

For that firstly we have to find the differentiation of \(\cos ({x})\).

So, we can re-write the above equation as:

\(=\int \sin (x) \frac{d(\cos (x))}{d x} d x\quad\quad\) [multiplying numerator and denominator by \({dx}]\)

We know that the \(\frac{d(\cos (x))}{d x}=-\sin (x)\)

Putting the value of \(\frac{d(\cos (x))}{d x}=-\sin (x)\)

We get:

\(=\int \sin (x) \times(-\sin (x)) \times d x\)

\(=\int-\sin ^{2}(x) d x\)

We know that: \(\left(\cos (2 x)=1-2 \sin ^{2}(x)\right)\) and \(\sin ^{2}(x)=\frac{1-\cos (2 x)}{2}\)

Now, putting the value of \(\sin ^{2}(x)\) we get,

\(=-\int\left(\frac{1-\cos (2 x)}{2}\right) d x\)

\(=-\frac{1}{2} \int(1-\cos (2 x)) d x\)

\(=-\frac{1}{2}\left(\int 1 d x-\int \cos (2 x) d x\right)\)

We know that \(\int \cos (x) d x=\sin (x)\) so,

\(=-\frac{1}{2}\left(x-\frac{\sin (2 x)}{2}\right)+c\)

\(=\frac{\sin (2 x)}{4}-\frac{x}{2}+c\)

Let "a" and "b" are any two odd positive integers. In general form it can be written as \(a=2 m+1\) and \(b=2 n+1\). Where \(m\) and \(n\) are any positive integers.
Consider, \(\frac{a+b}{2}=\frac{(2 m+1)+(2 n+1)}{2}\) \(=\frac{2 m+2 n+2}{2}\) \(=(m+n+1)\). Here the results are positive integers.
Now, \(\frac{a-b}{2}\) \(=\frac{(2 m+1)-(2 n+1)}{2}\) \(=\frac{2 m-2 n}{2}=(m-n)\)
But \(a>b\) \(\Rightarrow (2 m+1)>(2 n+1)\) \(\Rightarrow m>n\) or \(m-n>0\) \(\Rightarrow \frac{a-b}{2}>0\) Hence, \(\frac{a-b}{2}\) is also a positive integer.
Now, we have to prove that of the numbers \(\frac{a+b}{2}\) and \(\frac{a-b}{2}\) is odd and another is even numbers. Consider, \(\frac{a+b}{2}-\frac{a-b}{2}\)
\(\frac{a+b-a+b}{2}=\frac{2 b}{2}=b\) Which are odd positive integers as we let it above (equation 1), It is already proved above that \(\frac{a+b}{2}\) and \(\frac{a-b}{2}\) are positive integers (equation 2)
And we know that the difference between an odd number and even number is always an odd number.
From (equation 1 ) and (equation 2 ) we can conclude that one of the integers \(\frac{a+b}{2}\) and \(\frac{a-b}{2}\) is even and other is odd.

Let \(m\) be the slope of required line.

\(\therefore\left|\frac{m-(-1)}{1+m(-1)}\right|=1\)

\(\Rightarrow \frac{m+1}{1-m}=\pm 1\)

\(\Rightarrow m+1=1-m\)

and \(m+1=-1+m\)

\(\Rightarrow m=0\) and \(m=\infty\)

\(\therefore\) Equation of line through \((1,1)\) is \(y-1=0, x-1=0\)

Thus, option \(x-1=0, y-1=0\) is correct.

Given Standard deviation of series \(x_{1}, x_{2}, x_{3} \ldots \ldots ., x_{n}\) is \(\sigma\)

\(\frac{1}{n} \sqrt{\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2}}=\sigma\)

Where \(\bar{x}=\frac{x_{1}+x_{2}+x_{3} \ldots \ldots+x_{n}}{n}\)

Variance of this series \(=\frac{1}{n} \sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2}=\sigma^{2}\)

Now for the series \(\mathrm{ax}_{1}, \mathrm{ax}_{2}, \mathrm{ax}_{3} \ldots \ldots . ., \mathrm{ax}_{\mathrm{n}}\)

Mean \(=\mathrm{a}\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3} \ldots \ldots+\mathrm{x}_{\mathrm{n}}}{\mathrm{n}}\right)=\mathrm{a} \overline{\mathrm{x}}\)

Variance \(=\frac{1}{n} \sum_{1}^{n}\left(a x_{i}-a \bar{x}\right)^{2}\)

\(\mathrm{V}=\frac{\mathrm{a}^{2}}{\mathrm{n}} \sum_{1}^{\mathrm{n}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}\)

\(V=a^{2} \times\left[\frac{1}{n} \sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2}\right]\)

\(V=a^{2} \sigma^{2}\)

We know that,

\(\int_{a}^{c} f(x)=\int_{a}^{b} f(x)+\int_{b}^{c} f(x)\)

\(f(x)=|\sin x|\)

\(f(x)=-\sin x,-\frac{\pi} 2 \leq q x<0\) and

\(f(x)=\sin x, 0 \leq q x<\frac{\pi}2\)

Given:

\(=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\sin x| d x\)

\(=\int_{-\frac{\pi}{2}}^{0}-\sin x d x+\int_{0}^{\frac{\pi}{2}} \sin x d x\)

\(=[\cos x]_{\frac{-\pi}{2}}^{0}+[-\cos x]_{0}^{\frac{\pi}{2}}\)

\(=1-0+(0-(-1))\)

\(=1+1\)

\(=2\)

The given equation is,

$a{x}^2+2bx+c=0$

Here, addition of roots$=\alpha +\beta =\frac{-2b}{a}$

Multiplication of roots$=\alpha \beta =\frac{c}{a}$

$\Rightarrow \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$

$=\frac{\frac{4b^2}{a^2}-2\frac{c}{a}}{{\displaystyle \frac{c}{a}}}$

$=\frac{4b^2-2ac}{ac}$

Given equation \(x e^{x}\).

Let, \(f(x)=x e^{x}\)

Now, differentiate the above equation w.r.t \(x\) and using the property that \(\frac{d}{d x} m n=m \frac{d}{d x} n+n \frac{d}{d x} m\) so we have,

\(\Rightarrow \frac{d}{d x} f(x)=x \frac{d}{d x} e^{x}+e^{x} \frac{d}{d x} x\)

Now, as we know that \(\frac{d}{d x} x^{n}=n x^{n-1}, \frac{d}{d x} e^{x}=e^{x}\) so we have,

\(\Rightarrow f^{\prime}(x)=x e^{x}+e^{x}(1)\)

\(\Rightarrow f^{\prime}(x)=e^{x}+x e^{x}\)

Now, again differentiate it w.r.t \(x\) we have,

\(\Rightarrow \frac{d}{d x} f^{\prime}(x)=\frac{d}{d x}\left(e^{x}+x e^{x}\right)\)

\(\Rightarrow \frac{d}{d x} f^{\prime}(x)=\frac{d}{d x} e^{x}+\frac{d}{d x} x e^{x}\)

\(\Rightarrow f^{\prime \prime}(x)=e^{x}+e^{x}+x e^{x}\)

\(\Rightarrow f^{\prime \prime}(x)=2 e^{x}+x e^{x}\)

Now, again differentiate w.r.t \(x\) we have,

\(\Rightarrow \frac{d}{d x} f^{\prime \prime}(x)=\frac{d}{d x}\left(2 e^{x}+x e^{x}\right)\)

\(\Rightarrow f^{\prime \prime \prime}(x)=2 e^{x}+e^{x}+x e^{x}\)

\(\Rightarrow f^{\prime \prime \prime}(x)=3 e^{x}+x e^{x}\)

Similarly,

\(\Rightarrow f^{n}(x)=n e^{x}+x e^{x} \quad \quad \ldots \ldots \ldots .(1)\)

Now, according to the question we have to find out the condition when \(n^{\text {th }}\) derivative vanishes.

\(\Rightarrow f^{n}(x)=0\)

So, from equation (1) we have,

\(\Rightarrow f^{n}(x)=n e^{x}+x e^{x}=0\)

\(\Rightarrow n e^{x}+x e^{x}=0\)

\(\Rightarrow x e^{x}=-n e^{x}\)

\(\Rightarrow x=-n\)

So, this is the required condition.