# Tutor profile: Shuheer N.

## Questions

### Subject: Pre-Calculus

Question: Solve the following equation for x. $$3\sqrt x=x-4$$

Solution: It is given that $$3\sqrt x=x-4$$ Squaring both sides, we get $$9x=(x-4)^2$$ $$\implies 9x=x^2+16-8x$$ $$\implies x^2-17x+16=0$$ Splitting the middle term $$\implies x^2-16x-x+16=0$$ $$\implies x(x-16)-1(x-16)=0$$ $$\implies (x-16)(x-1)=0$$ This gives $$x-16=0 \implies x=16$$ And $$x-1=0 \implies x=1$$ As both values are positive , hence they are our solution.

### Subject: Basic Math

Qustion: What type of number 1 is?

Answer: 1 is neither prime nor composite.

### Subject: Calculus

Question: Showthat $$\frac{1}{17} $$≤$$\int_{1}^{2}$$$$\frac{1}{1+x^4}dx$$≤$$\frac{7}{24}$$

Solution: From$$1≤x≤2$$ We have, $$x^4≤2^4=16 \implies$$ $$1+x^4≤17\implies\frac{1}{1+x^4}≥\frac{1}{17}$$ Taking integral both sides from 1 to 2, we get $$\int_{1}^{2}\frac{1}{1+x^4}dx≥\int_{1}^{2}\frac{1}{17}dx=\frac{1}{17}[x]_{1}^{2}=\frac{1}{17}$$ Hence $$\frac{1}{17}≤\int_{1}^{2}\frac{1}{1+x^4}dx$$ Now for $$1≤x≤2$$ we can have $$1+x^4≥x^4\implies\frac{1}{1+x^4}≤\frac{1}{x^4}$$ Taking integral from 1 to2 both sides, we have $$\int_{1}^{2}\frac{1}{1+x^4}dx≤\int_{1}^{2}\frac{1}{x^4}dx=[\frac{-1}{3x^3}]_{1}^{2}=\frac{-1}{24}+\frac{1}{3}=\frac{7}{24}$$ Thus we have our result $$\frac{1}{17}≤\int_{1}^{2}\frac{1}{1+x^4}dx≤\frac{7}{24}$$

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