a) Find the equations of the tangent and normal to the circle x^2 + y^2 2x 4y + 3 = 0 at (2, 3). b) Evaluate: x 2 x 6 xx 2 OR A function f(x) is defined as follows: f(x) = cases 2x+3 & for x < 1 \\…

a) Equations of the tangent and normal to the circle at (2, 3)

Step 1: Rewrite the circle equation in standard form The given circle equation is: $$x^2+y^2-2x-4y+3=0$$

Complete the square for $x$ and $y$: $$x^2-2x+y^2-4y=-3$$ $$(x^2-2x+1)+(y^2-4y+4)=-3+1+4$$ $$(x-1{)}^2+(y-2{)}^2=2$$ This represents a circle with:

• Center $C(1,2)$
• Radius $r=\sqrt{2}$

Step 2: Find the slope of the radius to the point (2, 3) The slope $m_{CR}$ of the line joining the center $C(1,2)$ and the point $P(2,3)$ is: $$m_{CR}=\frac{3-2}{2-1}=1$$

Step 3: Find the slope of the tangent The tangent at $P$ is perpendicular to the radius $CP$. Thus, the slope $m_t$ of the tangent satisfies: $$m_t\cdot m_{CR}=-1⟹m_t=-1$$

Step 4: Equation of the tangent Using the point-slope form at $P(2,3)$: $$y-3=-1(x-2)⟹y=-x+5$$ Or in standard form: $$x+y-5=0$$

Step 5: Equation of the normal The normal is perpendicular to the tangent, so its slope $m_n$ satisfies: $$m_n\cdot m_t=-1⟹m_n=1$$ Using the point-slope form at $P(2,3)$: $$y-3=1(x-2)⟹y=x+1$$ Or in standard form: $$x-y+1=0$$

Final Answer:

• Equation of the tangent: $x+y-5=0$
• Equation of the normal: $x-y+1=0$

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b) Evaluate: ${\lim }_{x\to 2}\frac{\sqrt{x}-\sqrt{6-x}}{x-2}$

Step 1: Direct substitution Substituting $x=2$ gives: $$\frac{\sqrt{2}-\sqrt{4}}{0}=\frac{\sqrt{2}-2}{0}$$ This is an indeterminate form $\frac{0}{0}$, so we rationalize the numerator.

Step 2: Rationalize the numerator Multiply numerator and denominator by the conjugate $\sqrt{x}+\sqrt{6-x}$: $$\underset{x\to 2}{\lim }\frac{(\sqrt{x}-\sqrt{6-x})(\sqrt{x}+\sqrt{6-x})}{(x-2)(\sqrt{x}+\sqrt{6-x})}$$ Simplify the numerator: $$(\sqrt{x}{)}^2-(\sqrt{6-x}{)}^2=x-(6-x)=2x-6$$ So the limit becomes: $$\underset{x\to 2}{\lim }\frac{2x-6}{(x-2)(\sqrt{x}+\sqrt{6-x})}$$

Step 3: Factor the numerator $$2x-6=2(x-3)$$ Thus: $$\underset{x\to 2}{\lim }\frac{2(x-3)}{(x-2)(\sqrt{x}+\sqrt{6-x})}$$

Step 4: Apply L'Hôpital's Rule (or simplify) Since direct substitution still gives $\frac{-2}{0}$, we can simplify the expression differently. Rewrite the original limit as: $$\underset{x\to 2}{\lim }\frac{\sqrt{x}-\sqrt{6-x}}{x-2}=\underset{x\to 2}{\lim }\frac{\sqrt{x}-\sqrt{6-x}}{(x-2)}\cdot \frac{\sqrt{x}+\sqrt{6-x}}{\sqrt{x}+\sqrt{6-x}}$$ $$=\underset{x\to 2}{\lim }\frac{x-(6-x)}{(x-2)(\sqrt{x}+\sqrt{6-x})}=\underset{x\to 2}{\lim }\frac{2x-6}{(x-2)(\sqrt{x}+\sqrt{6-x})}$$ Factor numerator: $$=\underset{x\to 2}{\lim }\frac{2(x-3)}{(x-2)(\sqrt{x}+\sqrt{6-x})}$$ Now, observe that $x\to 2$ does not cancel the denominator. Instead, we can use the fact that: $$\underset{x\to 2}{\lim }\frac{\sqrt{x}-\sqrt{6-x}}{x-2}=\frac{d}{dx}\left(\sqrt{x}+\sqrt{6-x}\right){|}_{x=2}$$ Differentiate the expression inside the limit: $$\frac{d}{dx}\left(\sqrt{x}+\sqrt{6-x}\right)=\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{6-x}}$$ Evaluate at $x=2$: $$\frac{1}{2\sqrt{2}}-\frac{1}{2\sqrt{4}}=\frac{1}{2\sqrt{2}}-\frac{1}{4}$$

Step 5: Simplify the result $$\frac{1}{2\sqrt{2}}-\frac{1}{4}=\frac{\sqrt{2}}{4}-\frac{1}{4}=\frac{\sqrt{2}-1}{4}$$

Final Answer: $$\boxed{\frac{\sqrt{2}-1}{4}}$$

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OR

Is the function continuous at $x=1$? If not, how can you make it continuous?

Given function: $$f(x)=\{\begin{array}{ll}2x+3& \text{for }x<1\\ 4& \text{for }x=1\\ 6x-1& \text{for }x>1\end{array}$$

Step 1: Check continuity at $x=1$ For continuity at $x=1$, the following must hold:

1. $f(1)$ must exist.
2. ${\lim }_{x\to 1^{-}}f(x)={\lim }_{x\to 1^{+}}f(x)=f(1)$.

Step 2: Evaluate $f(1)$ $$f(1)=4$$

Step 3: Evaluate left-hand limit (LHL) $$\underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{-}}{\lim }(2x+3)=2(1)+3=5$$

Step 4: Evaluate right-hand limit (RHL) $$\underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }(6x-1)=6(1)-1=5$$

Step 5: Compare limits and function value

• LHL = 5
• RHL = 5
• $f(1)=4$

Since LHL = RHL ≠ $f(1)$, the function is not continuous at $x=1$.

Step 6: Make the function continuous To make $f(x)$ continuous at $x=1$, we must redefine $f(1)$ so that: $$f(1)=\underset{x\to 1}{\lim }f(x)=5$$

Final Answer: The function is not continuous at $x=1$. To make it continuous, redefine $f(1)=5$.