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Tutor profile: Peeyush R.

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Peeyush R.
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Questions

Subject: Pre-Calculus

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Question:

Question $$(1)$$ Find the domain for the function $$f(x)=\frac{4x-20}{\sqrt{36-x^2}}$$

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Peeyush R.
Answer:

Solution $$(1)$$ The function $$f$$ is defined wherever $$36-x^2\ge 0$$ (so the square root is defined) and $$36-x^2 \neq 0$$ (so the fraction is defined). Solving the inequality $$36-x^2>0$$ or $$(6-x)(6+x)>0$$ gives $$-6<x<6$$, So, the domain of $$f$$ is $$(-6,6)$$.

Subject: Trigonometry

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Question:

Question $$(2)$$ If $$\left (\tan\alpha + \cot\alpha \right ) =4$$. Find the value of $$\log_{15}(\tan^{4}{\alpha} + \cot^{4}{\alpha} +31)$$

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Peeyush R.
Answer:

Solution $$(2)$$- Given $$\left (\tan\alpha + \cot\alpha \right )=4$$ On squaring both sides We get, $$\tan^{2}\alpha + \cot^{2}\alpha +2= 16$$ $$\tan^{2}\alpha + \cot^{2}\alpha = 14$$ Again squaring both sides $$\tan^{4}\alpha + \cot^{4}\alpha +2= 196$$ $$\tan^{4}\alpha + \cot^{4}\alpha = 194$$ Adding $$31$$ both sides we finally get: $$\tan^{4}\alpha + \cot^{4}\alpha + 31 = 225$$ Now, $$\log_{15}(\tan^{4}\alpha + \cot^{4}\alpha +31)$$ Substitute the corresponding value: $$\log_{15}(225)$$ $$\log_{15}(15^{2})$$ $$2$$ Therefore, the value of $$\log_{15}(\tan^{4}{\alpha} + \cot^{4}{\alpha} +31)$$ is $$2$$

Subject: Calculus

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Question:

Question $$(3)$$ If $$f'(x) $$ denotes the derivative of function $$f(x)$$ and $$f"(x)$$ denotes the derivative of $$f'(x)$$ where $$f(x)$$ is defined as $$\sin\left(\frac{\pi}{2}(x^{2}\right)$$. Find the anti dervative given below.

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Peeyush R.
Answer:

$$\int_{0}^{1}\frac{f"(x)dx}{\sqrt{f'(x)}}$$ Solution $$(3)$$-$$\int_{0}^{1}\frac{f"(x)dx}{\sqrt{f'(x)}}$$ $$f(x)= \sin\left(\frac{\pi}{2}(x^{2}\right).$$ On differentiating both sides with respect to $$x$$ We get, $$f'(x)dx=\pi x \cos\left(\frac{\pi}{2}(x^{2}\right)..$$ Note: The integration of the form $$\frac{f'(x)dx}{\sqrt{f(x)}}$$ is $$2\sqrt{f(x)}.$$ Therefore, The final value will be $$2\left [ \sqrt{f'(1)}-\sqrt{f'(0)} \right ]$$ $$2(0-0)=0$$ The value of the integral is $$0$$

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