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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.
Answer:

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.
Answer:

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?

Inactive
Baiju T.
Answer:

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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