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Shubham M.

Engineer. Entreprenuer. Tutor.

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Pre-Calculus

TutorMe

Question:

FInd the domain of function: $$f(x)=\log_{10}(\log_{10}(\log_{10}(\log_{10}x)))$$

Shubham M.

Answer:

$$\log_{10}a$$ exists if $$a>0$$ Hence, $$f(x)=\log_{10}(\log_{10}(\log_{10}(\log_{10}x)))$$ exists if: $$(\log_{10}(\log_{10}(\log_{10}x)))>0$$ and $$x>0$$ $$(\log_{10}(\log_{10}x))>10^0=1$$ and $$x>0$$ $$(\log_{10}x)>10^1=10$$ and $$x>0$$ $$x>10^{10}$$ and $$x>0$$ Hence, domain of the given function is $$(10^{10},\infty)$$

Calculus

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Question:

Find: $$\lim_{x\to1}$$$$\frac{\sqrt{1-cos2(x-1)}}{(x-1)}$$

Shubham M.

Answer:

We know that $$2(sin (a))^2=1-cos(2a)$$ Hence, $$1-cos2(x-1)=2(sin(x-1))^2$$ Also, $$\sqrt{(sin(x-1))^2}=|sin(x-1)|$$ Hence, we have $$\lim_{x\to1}$$$$\frac{\sqrt{1-cos2(x-1)}}{(x-1)} = \lim_{x\to1}$$$$\frac{\sqrt{2}|sin(x-1)|}{(x-1)}$$ Due to nature of this problem, we may get different left hand limit and right hand limit. So we will calculate both. Left hand limit $$=\lim_{x\to1^-}$$$$\frac{\sqrt{2}|sin(x-1)|}{(x-1)}$$ $$=\lim_{h\to0}$$$$\frac{\sqrt{2}|sin (-h)|}{-h}$$ $$=\lim_{h\to0}$$$$\frac{\sqrt{2}sin (h)}{-h}$$ $$=-\sqrt{2}\lim_{h\to0}$$$$\frac{sin (h)}{h}$$ $$=-\sqrt{2}$$ Right hand limit $$=\lim_{x\to1^+}$$$$\frac{\sqrt{2}|sin(x-1)|}{(x-1)}$$ $$=\lim_{h\to0}$$$$\frac{\sqrt{2}|sin (h)|}{h}$$ $$=\lim_{h\to0}$$$$\frac{\sqrt{2}sin (h)}{h}$$ $$=\sqrt{2}\lim_{h\to0}$$$$\frac{sin (h)}{h}$$ $$=\sqrt{2}$$ As Left hand limit and Right hand limit are not equal, limit of the given expression does not exist.

Algebra

TutorMe

Question:

Find all the solutions of the inequality: $$x^4-6x^3+10x^2+1<0$$

Shubham M.

Answer:

The best approach to solve any problem of this kind is to look for ways to factorize the given inequality in such a way that we are left with an expression that can be easily solved. Given inequality: $$x^4-6x^3+10x^2+1 < 0$$ We can remove $$x^2$$ common from the first 3 terms of the expression to simplify it: $$x^2(x^2-6x+10)+1 < 0$$ One more trick that we can employ is to try and find ways to get expressions which are complete squares and take it apart from the expression. We can see that: $$x^2-6x+9 = (x-3)^2$$ Hence, $$x^2-6x+10 = (x-3)^2+1$$ Now original inequality can be written as: $$x^2((x-3)^2+1)+1 < 0$$ This can be further simplified to: $$x^2(x-3)^2+x^2+1 < 0$$ As square of any real expression is greater than or equal to zero, we can see that the left hand side of the inequality will always be greater than zero. Hence, no solutions exist for the given inequality.

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