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Tutor profile: Deepnika J.

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Deepnika J.
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Questions

Subject: Trigonometry

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Question:

If $$asinx + q\space cos(c + x) + q\space cos(c - x) = \alpha$$, $$\alpha > a$$ , then find out the minimum value of $$|cosc|$$.

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Deepnika J.
Answer:

We are given that; $$asinx + qcos(c + x) + qcos(c - x) = \alpha$$ Taking $$q$$ as a common factor from the second and third terms; $$asinx + q[cos(c + x) + cos(c - x)] = \alpha$$ Expand $$cos(c + x) \space \& \space cos(c - x)$$ as follows: $$asinx + q[(cosc \space cosx - sinc \space sinx) + (cosc \space cosx + sinc \space sinx)] = \alpha$$ We are left with the following expression: $$asinx + q[cosc \space cosx + cosc \space cosx] = \alpha$$ $$asinx + 2q \space cosc \space cosx = \alpha$$ Now, if we let $$2q \space cosc = b$$, then we have; $$asinx + bcosx = \alpha$$ Concept: We should know that the value of the expression $$(asinx + bcosx)$$ lies in the interval $$[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$$. Hence, $$-\sqrt{a^2 + b^2} \leq (asinx + bcosx) \leq \sqrt{a^2 + b^2}$$ $$\implies -\sqrt{a^2 + b^2} \leq \alpha \leq \sqrt{a^2 + b^2}$$ $$\implies \alpha \leq \sqrt{a^2 + b^2}$$ $$\implies \alpha \leq \sqrt{a^2 + 4q^2(cos c)^2}$$ $$\implies \alpha^2 \leq a^2 + 4q^2(cos c)^2$$ $$\implies \frac{\alpha^2 - a^2}{4q^2} \leq (cos c)^2$$ $$\implies \sqrt{\frac{\alpha^2 - a^2}{4q^2}} \leq |cos c|$$ Therefore, the minimum value of $$|cos c|$$ is equal to $$\sqrt{\frac{\alpha^2 - a^2}{4q^2}}$$.

Subject: Calculus

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Question:

Let $$f'(x) = sin(x^2)$$ and $$y = f(x^2 + 1)$$ then find out the value of $$\frac{dy}{dx}$$ at $$x = 1$$.

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Deepnika J.
Answer:

It is given that; $$y = f(x^2 + 1)$$ Using chain rule for differentiation; $$dy = f'(x^2 + 1) \space2x \space dx$$ Substituting the value of $$f'(x^2 + 1)$$ from the given relation $$f'(x) = sin(x^2)$$ ; $$\frac{dy}{dx} = sin[(x^2 + 1)^2] \space 2x$$ At $$x = 1$$; $$\frac{dy}{dx} = sin[(1^2 + 1)^2] \space 2(1)$$ $$ \implies \frac{dy}{dx} = 2 \space sin\space4$$

Subject: Algebra

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Question:

Find the value of $$ x^{\log_x a \space \log_a y \space \log_y z \space} $$.

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Deepnika J.
Answer:

Let $$A = x^{\log_x a \space \log_a y \space \log_y z \space} $$ Taking logarithm of both sides: $$\log A = \log (x^{\log_x a \space \log_a y \space \log_y z \space}) $$ Using properties of logarithm; $$\log A = (\log_x a \space \log_a y \space \log_y z \space) \log x$$ $$\implies \log A = \frac{\log a}{\log x} \space \frac{\log y}{\log a} \space \frac{\log z}{\log y} \space \log x$$ $$\implies \log A = \log z$$ $$\implies A = z$$ Hence, $$x^{\log_x a \space \log_a y \space \log_y z \space} = z$$

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