Limit Continuity and Differentiability 7 Question 31

32. Let f,g:[1,2]R be continuous functions which are twice differentiable on the interval (1,2). Let the values of f and g at the points 1,0 and 2 be as given in the following table:

x=1 x=0 x=2
f(x) 3 6 0
g(x) 0 1 -1

In each of the intervals (1,0) and (0,2), the function (f3g) never vanishes. Then, the correct statement(s) is/are

(2015 Adv.) (a) f(x)3g(x)=0 has exactly three solutions in (1,0)(0,2)

(b) f(x)3g(x)=0 has exactly one solution in (1,0)

(c) f(x)3g(x)=0 has exactly one solution in (0,2)

(d) f(x)3g(x)=0 has exactly two solutions in (1,0) and exactly two solutions in (0,2)

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Answer:

Correct Answer: 32. f(0+)=0,f(0)=1

Solution:

  1. Given, y=(logcosxsinx)(logsinxcosx)1+sin12x1+x2

y=loge(sinx)loge(cosx)2+sin12x1+x2(loge(cosx)cotxdydx=2loge(sinx)loge(cosx)+loge(inx)tanx)loge(cosx)2+21+x2

dydxx=π4=212log12log12+21+π216=8loge2+3216+π2



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