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# Tutor profile: Ratan C.

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Ratan C.
Assistant Professor (Passionate about teaching Mathematics)
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## Questions

### Subject:Pre-Calculus

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Question:

Find the range and the domain of the following function : $$\frac {x}{x^2-2x+1}$$

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Ratan C.

Definition : The domain of the function is the set of input values for which the function is real and defined. Let $$y=\frac {x}{x^2-2x+1}$$ Letting the denominator of $$y$$ equals to zero. That is $$x^2-2x+1=0$$ $$\implies (x-1)^2=0 \implies x-1=0 \implies x=1$$ Thus the only undefined point (singularity) is $$x=1$$ Therefore the domain of the function is set of all real numbers except $$x=1$$. i.e. Domain : $$(-\infty,1)\cup (1,\infty)$$ OR $$\mathbb{R}-\{1\}$$ Definition : The range of a function is the set of all possible outputs of the function. $$y=\frac {x}{x^2-2x+1}\quad \implies y\cdot (x^2-2x+1)=x$$ $$\implies yx^2-2xy+y-x=0 \quad \implies yx^2-(2y+1)x+y=0$$ which is a quadratic equation in $$x$$ of the form $$ax^2+bx+c=0$$, where $$a=y,b=-2y-1,c=y$$ Now the discriminant of the above quadratic equation is $$b^2-4ac=(-2y-1)^2-4y\cdot y=4y^2+4y+1-4y^2=4y+1$$ The range is the set of $$y$$ for which the discriminant is greater than or equal to zero. $$\implies 4y+1\geq 0 \quad \implies y\geq -\frac 1 4$$ i.e. Range : $$\left[-\frac 1 4 , \infty \right]$$

### Subject:Calculus

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Question:

Evaluate the derivative of $$\sin^2(x)$$ using first principle.

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Ratan C.

The derivative of $$f(x)$$ with respect to $$x$$ is defined as : $( f’(x)=\lim_{h\to 0} \frac {f(x+h)-f(x)} {h}$) Let $$f(x)=\sin^2(x)$$ Then $$f(x+h)=\sin^2(x+h)$$ $(\therefore \lim_{h\to 0} \frac {f(x+h)-f(x)} {h} =\lim_{h\to 0}\frac {\sin^2(x+h)-\sin^2(x)}{h}………\left(\frac 0 0 form \right)$) Using L’Hopital’s rule, $( \lim_{h\to 0} \frac {f(x+h)-f(x)} {h} =\lim_{h\to 0}\frac {\frac{d}{dh}\left(\sin^2(x+h)-\sin^2(x)\right)}{\frac{d}{dh}\left(h \right)}$) $(\therefore \lim_{h\to 0} \frac {f(x+h)-f(x)} {h} =\lim_{h\to 0}\frac {2\sin(x+h)\cdot \cos(x+h)-0}{1}$) $(\therefore \lim_{h\to 0} \frac {f(x+h)-f(x)} {h} =2\sin(x+0)\cdot \cos(x+0)$) $(\therefore \lim_{h\to 0} \frac {f(x+h)-f(x)} {h} =2\sin(x) \cos(x)$) $(\implies f’(x)=2\sin(x) \cos(x)$)

### Subject:Differential Equations

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Question:

Solve the following initial value problem : $( \frac{dy}{dx}-\frac 1 x y=1,\quad y(1)=2$)

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Ratan C.

Consider the initial value problem, $( \frac{dy}{dx}-\frac 1 x y=1,\quad y(1)=2$) The given differential equation is a first-order linear differential of the form $( \frac{dy}{dx}+P(x) y=Q(x)$) Here $$P(x)=-\frac 1 x$$ and $$Q(x)=1$$ Now, $$\int P(x)dx=\int -\frac 1 x dx=-\ln(x)=\ln(x^{-1}) =\ln\left(\frac 1 x \right)\cdots (\because a\ln(b)=\ln(b^a)$$ $$\therefore$$ the integrating factor $$\mu(x)$$ is given by $(\mu(x)=e^{\int P(x)dx}= e^{\ln\left(\frac 1 x \right)}=\frac 1 x \cdots (\because e^{\ln(a)}=a$) Now, the general solution is given by $( y\cdot\mu(x)=\int \mu(x)\cdot Q(x)dx+c$) $(y\cdot \frac 1 x =\int \frac 1 x \cdot 1 dx +c$) $(y\cdot \frac 1 x =\ln(x) +c$) $(y=x\ln(x) +cx$) Using the initial condition, $$y(1)=2$$ we have $( y(1)=1\cdot \ln (1)+c\cdot 1 \implies 2=0+c\implies c=2$) Thus, the required solution of the given initial value problem is $(y=x\ln(x) +2x$)

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