Question

Liala and Vinay both have to travel to various locations for advertising their company's products. The company reimburses their expenses such as accommodation, food etc. The company also blacklists an employee whenever the employee's expenditure in a given month exceeds ₹ 12000. The accounts department fits the data of monthly expenditure to the polynomial E_l(x)El​(x) and E_v(x)Ev​(x) (in ₹ ) for Liala and Vinay respectively, where xx is the number of months since they joined the company (i.e., x = 1x=1 represents the completion of one month). The polynomial fit is known to be applicable for a period of 33 months (i.e., x \leq 33x≤33). If E_l(x) - 12,000 = b(x-5.1)(x-14)(x-23), ~b>0El​(x)−12,000=b(x−5.1)(x−14)(x−23), b>0 and E_v(x) - 12,000 = a(x-1.5)^2(x-5.1)(x-24), ~a>0Ev​(x)−12,000=a(x−1.5)2(x−5.1)(x−24), a>0. If Vinay and Liala have been blacklisted together for atleast NN times in 33 months, then find the value of N.​

Answer 1

Answer:

24 and ,51

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Answer 2

Given polynomials used for estimation of monthly expenditure of Liala and Vinay respectively are

$E_l(x)-12000=b(x-5.1)(x-14)(x-23), 0\leq x\leq 33\\ \\E _l(x)-12000 > 0:x\epsilon(5.1,14)U(23,33]\\ \\E_v(x)-12000=a(x-1.5)^2(x-5.1)(x-24), 0\leq x\leq 33\\ \\E_v(x)-12000 > 0:x\epsilon[0,2.5)U(1.5,5.1)U(24,33]$

If Liala and Vinay are blacklisted together, then

$E_l(x)-12000 > 0\\E_v(x)-12000 > 0\\ and\\ x\epsilon(26,33]$

As 'x' is greater integer

$x=27,28,29,30,31,32,33$

By observing we can say that there are $7$ different values of $x$.

∴Vinay and Liala have been blacklisted for at least $7$ times in $33$ months.

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