Peeyush R.

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Pre-Calculus

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Question:

Find the domain of the function given below \frac{x+1}{\sqrt{(x-5)(x-3)^{2}(x+2)^{3}}}

Peeyush R.

Answer:

For the domain of this function, the denominator should not be zero and the square root function should always be positive. Therefore, on the number line mark points - 2,3,5. the value of this function becomes negative in the interval [-2,5]. so the domain is R -{[-2,5]}

Trigonometry

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Question:

If \left (tan\alpha + cot\alpha \right ) =4. Find \left log_{15}(tan^{4}\alpha + cot^{4}\alpha +31)

Peeyush R.

Answer:

\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) log_{15}(225)) log_{15}(15^{2}) 2 FINAL ANSWER IS 2

Calculus

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Question:

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(\frac{\prod}{2}(x^{2})). Find the anti dervative given below. \int_{0}^{1}\frac{f"(x)dx}{\sqrt{f'(x)}}

Peeyush R.

Answer:

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

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