a) If a = i + 2 j k and b = i j + k , find the projection of b on a . [2] b) Evaluate, using L' Hopital rule: x 0 x x xx^3 . [2] c) Evaluate: x(x a)(x b) dx . [2]

a) Projection of $\vec{b}$ on $\vec{a}$

Given: $$\vec{a}=\hat{i}+2\hat{j}-\hat{k},\vec{b}=\hat{i}-\hat{j}+\hat{k}$$

Step 1: Recall the formula for the projection of $\vec{b}$ on $\vec{a}$: $$\text{Projection of }\vec{b}\text{ on }\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|}$$

Step 2: Compute the dot product $\vec{a}\cdot \vec{b}$: $$\vec{a}\cdot \vec{b}=(1)(1)+(2)(-1)+(-1)(1)=1-2-1=-2$$

Step 3: Compute the magnitude of $\vec{a}$: $$|\vec{a}|=\sqrt{1^2+2^2+(-1{)}^2}=\sqrt{1+4+1}=\sqrt{6}$$

Step 4: Substitute into the projection formula: $$\text{Projection}=\frac{-2}{\sqrt{6}}=-\frac{2\sqrt{6}}{6}=-\frac{\sqrt{6}}{3}$$

Final Answer: $$\boxed{-\frac{\sqrt{6}}{3}}$$

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b) Evaluate ${\lim }_{x\to 0}\frac{x-\sin x\cos x}{x^3}$ using L'Hôpital's Rule

Step 1: Check the limit type: $$\underset{x\to 0}{\lim }\frac{x-\sin x\cos x}{x^3}=\frac{0-0}{0}\text{(indeterminate form }\frac{0}{0}\text{)}$$ L'Hôpital's Rule applies.

Step 2: Differentiate numerator and denominator:

• Numerator: $f(x)=x-\sin x\cos x$ $$f^{\prime }(x)=1-\left[\cos x\cdot \cos x+\sin x\cdot (-\sin x)\right]=1-({\cos }^{2}x-{\sin }^{2}x)=1-\cos 2x$$
• Denominator: $g(x)=x^3$ $$g^{\prime }(x)=3x^2$$

Step 3: Apply L'Hôpital's Rule: $$\underset{x\to 0}{\lim }\frac{1-\cos 2x}{3x^2}$$ This is still $\frac{0}{0}$, so apply L'Hôpital's Rule again.

Step 4: Differentiate numerator and denominator again:

• Numerator: $f^{\prime }(x)=1-\cos 2x$ $$f^{\prime \prime }(x)=2\sin 2x$$
• Denominator: $g^{\prime }(x)=3x^2$ $$g^{\prime \prime }(x)=6x$$

Step 5: Apply L'Hôpital's Rule a second time: $$\underset{x\to 0}{\lim }\frac{2\sin 2x}{6x}=\underset{x\to 0}{\lim }\frac{\sin 2x}{3x}$$ This is still $\frac{0}{0}$, so apply L'Hôpital's Rule a third time.

Step 6: Differentiate numerator and denominator once more:

• Numerator: $f^{\prime \prime }(x)=2\sin 2x$ $$f^{\prime \prime \prime }(x)=4\cos 2x$$
• Denominator: $g^{\prime \prime }(x)=6x$ $$g^{\prime \prime \prime }(x)=6$$

Step 7: Apply L'Hôpital's Rule a third time: $$\underset{x\to 0}{\lim }\frac{4\cos 2x}{6}=\frac{4\cdot 1}{6}=\frac{2}{3}$$

Final Answer: $$\boxed{\frac{2}{3}}$$

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c) Evaluate $\int \frac{x}{(x-a)(x-b)}dx$

Step 1: Use partial fraction decomposition: $$\frac{x}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}$$ Multiply both sides by $(x-a)(x-b)$: $$x=A(x-b)+B(x-a)$$

Step 2: Solve for $A$ and $B$:

• Let $x=a$: $$a=A(a-b)⟹A=\frac{a}{a-b}$$
• Let $x=b$: $$b=B(b-a)⟹B=\frac{b}{b-a}$$

Step 3: Rewrite the integrand: $$\frac{x}{(x-a)(x-b)}=\frac{a}{(a-b)(x-a)}+\frac{b}{(b-a)(x-b)}=\frac{a}{a-b}\cdot \frac{1}{x-a}-\frac{b}{a-b}\cdot \frac{1}{x-b}$$

Step 4: Integrate term by term: $$\int \frac{x}{(x-a)(x-b)}dx=\frac{a}{a-b}\int \frac{1}{x-a}dx-\frac{b}{a-b}\int \frac{1}{x-b}dx$$ $$=\frac{a}{a-b}\ln |x-a|-\frac{b}{a-b}\ln |x-b|+C$$

Step 5: Simplify the expression: $$=\frac{a\ln |x-a|-b\ln |x-b|}{a-b}+C$$

Final Answer: $$\boxed{\frac{a\ln |x-a|-b\ln |x-b|}{a-b}+C}$$