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# Tutor profile: Deepak D.

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Deepak D.
Research Assistant for 1 year; former Academic Secretary in Student Council; IIT Gandhinagar
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## Questions

### Subject:LaTeX

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Question:

It is a markup language which makes documents etc. No questions here.

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Deepak D.

### Subject:Calculus

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Question:

let $$\alpha(a)$$ and $$\beta(a)$$ be the roots of the equation: $$(\sqrt[3]{1+a} - 1) x^2 + (\sqrt{1+a} - 1) x + (\sqrt[6]{1+a} - 1) = 0$$ where $$a > -1$$ Then find $$\displaystyle \lim_{a \to 0^+} \alpha(a)$$ and $$\displaystyle \lim_{a \to 0^+} \beta(a)$$.

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Deepak D.

Step 1: $$a > -1 \implies a + 1 > 0$$ Step 2: The highest root we have in equation is the 6th root of $$(a+1)$$. Let's assume $$a+1 = k^6$$ (change of variable from $$a$$ to $$k$$) Step 3: Accordingly, the limit variable will change. When $$a \to 0^+$$, $$k \to 1$$. Now we have our new limits to evaluate (hopefully, easier!) Step 4: Substitute $$k^6$$ in the given equation to get: $$(k^2 -1) x^2 + (k^3 -1) x + (k -1) = 0$$ Now factorize it by taking the common multiple $$(k-1)$$ out - $$\implies (k-1) \left( (k+1)x^2 + (k^2 + 1 + k)x + 1 \right) = 0$$ and now $$k \to 1$$ Step 5: The roots $$\alpha$$ and $$\beta$$ are the roots of the quadratic equation above when the limit is exercised. Therefore, $$2x^2 + 3x + 1 = 0 \\ ( 2x + 1) (x + 1) = 0$$ ANSWER the roots at the given limits are $$- \dfrac{1}{2}$$ and $$-1$$.

### Subject:Physics

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Question:

A motorboat going downstream overcame a raft at a point A. t= 60 minutes later it turned back and after some time passes a raft at a distance of 6 km from the point A. Find the flow velocity assuming the duty of the engine to be constant.

Inactive
Deepak D.

There are two methods to solve this problem. First one is a conventional and approaches the problem from the frame of reference of an observer standing on the shore and looking at the whole incident. The second method is more thoughtful and simple, once understood. We shift the frame of reference from the observer at rest to an observer on the raft. Since any frame either at rest or moving at a constant velocity is termed inertial, our answers match. Follow with me to see the beauty: Although the raft is moving on the river with velocity equal to flow velocity the observer sitting on it treats it at rest. To the same observer, the motorboat crosses him at t = 0. To think from the eyes of this observer is the only tricky part here - so always keep this in mind. The person sees that the motorboat takes a U-turn at t = 60 min or 1 hr. It is given in the question that engine duty of motorboat is constant which means that it travels with same speed irrespective of downstream or upstream. Therefore, in the frame of the observer (who feels he is at rest) on raft, the motorboat will reach to him in another 1 hr after taking the U-turn. Let is recollect now. The motorboat takes 2 hrs to reach back to raft after taking a U-turn. In the meantime, the raft is also moving. It has actually moved a distance of 6 km in that time. All these are in frame of reference of the observer on shore (because the raft guy doesn't feel himself moving at all). Since time is independent of any inertial frame we choose, we are left with just this small arithmetic: Speed of flow = speed of raft = $$\dfrac{\text{distance covered by raft}}{\text{time taken by raft}} = \dfrac{6}{2} = 3 \ km/h$$ And, yes! That's our answer folks.

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