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# Tutor profile: Baiju T.

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Baiju T.
Associate Professor in Mathematics
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## Questions

### Subject:Trigonometry

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Question:

Prove that no value of $sec\theta$ can satisfy the equation$6 sec^{2}\theta -5 sec\theta +1=0$

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Baiju T.

Given $6 sec^{2}\theta -5 sec\theta +1=0$ Therefore $sec\theta =\frac{5+\sqrt{25-4.61}}{2.6}$ = 1/2 or 1/3 But $sec\theta$ cannot lie between -1 and +1 Hence $sec\theta$ is not equal to 1/2 or 1/3

### Subject:Calculus

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Question:

If k be a positive real number such that f(x+k)+f(x) = 0 for all x belongs to R( real number system). Prove that f(x) is a periodic function with period 2k.

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Baiju T.

We have f(x+k)+f(x) = 0 for all x belongs to R i.e. f(x+k) = -f(x) for all x belongs to R f(x+2k) = -f(x+k) = f(x) for all x belongs to R f(x+2k) = f(x) for all x belongs to R Hence f(x) is periodic with period 2k

### Subject:Algebra

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Question:

John is 47inches tall and is growing at the rate of 1/12 inch per month. Jerry is 43 inches tall and is growing at the rate of 1/3 inches per month. If they each continue to grow at these rates for the next four years, in how many months will they be the same height?

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Baiju T.

We can express the heights of John and Jerry as two linear equations as follows: John's height in inches as H1=47+(1/12)m, where m is the number of months from now. Jerry's height as H2 = 43+(1/3)m. The question means that we have to find the value of m such that H1=H2. i.e. 47+(1/12)m = 43+(1/3)m 47-43 = (1/3 - 1/12)m 4 = (4/12 - 1/12)m 4 = (3/12)m m = 16. so the answer is 16 months.

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