Question

# In a triangle ABC, D,E,F are taken on the sides BC,CA,AB respectively. If BDDC=CEEA=AFFB=n and  Area (△DEF)=f(n)× Area (△ABC), then the correct statements is/aref(n) is monotonic function.f(n) is increasing in [1,∞)1∫0f(n) dn=52−3ln21∫0f(n) dn=−12+3ln2

Solution

## The correct options are B f(n) is increasing in [1,∞) C 1∫0f(n) dn=52−3ln2Assuming A to be origin and the position vectors of B and C be →b and →c respectively. Position vectors of D:n→c+→bn+1E:→cn+1       F:n→bn+1 Now, −−→FD=−−→AD−−−→AF=n→c+(1−n)→bn+1−−→FE=−−→AE−−−→AF=→c−n→bn+1 We know that, Area (△ABC)=12|→b×→c| So, Area (△DEF)=12|−−→FD×−−→FE|=12(n+1)2∣∣∣(n→c+(1−n)→b)×(→c−n→b)∣∣∣=12(n+1)2∣∣∣n2(→b×→c)+(n−1)→c×→b∣∣∣=n2−n+1(n+1)2×12∣∣∣(→b×→c)∣∣∣ Therefore,  f(n)=n2−n+1(n+1)2⇒f(n)=1−3n(n+1)2⇒f′(n)=−3[(n+1)2−2n(n+1)(n+1)4]⇒f′(n)=3(n−1)(n+1)3 So, f(n) is increasing when n≥1 Now let I=1∫0f(n) dn⇒I=1∫01−3n(n+1)2 dn⇒I=1−31∫0[n(n+1)2] dn⇒I=1−31∫0[1(n+1)−1(n+1)2] dn⇒I=1−3[ln(n+1)+1(n+1)]10⇒I=1−3ln2+32⇒1∫0f(n) dn=52−3ln2

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