Mathematics · Application of Derivatives

JEE Advanced 2021 — Paper 2 — Question 46

Let f1:(0,∞)→Rf_{1}:(0, \infty) \rightarrow R and f2:(0,∞)→Rf_{2}:(0, \infty) \rightarrow R be defined by f1(x)=∫0x∏j=121(t−j)jdt,x>0f_{1}(x)=\int_{0}^{x} \prod_{j=1}^{21}(t-j)^{j} d t, x>0 and f2(x)=98(x−1)50−600(x−1)49+2450,x>0f_{2}(x)=98(x-1)^{50}-600(x-1)^{49}+2450, x>0, where, for any positive integer nn and real numbers a1,a2a_{1}, a_{2}, ……,an,∏i=1nai\ldots \ldots, a_{n}, \prod_{i=1}^{n} a_{i} denotes the product of a1,a2,…..,ana_{1}, a_{2}, \ldots . ., a_{n}. Let mim_{i} and nin_{i}, respectively, denote the number of points of local minima and the number of points of local maxima of function fi,i=1,2f_{i}, i=1,2, in the interval ( 0 , ∞)\infty).

The value of 6m2+4n2+8m2n26 m_{2}+4 n_{2}+8 m_{2} n_{2} is \qquad

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

f2′(x)=(98)(50)(x−1)49−(600)(49)(x−1)48\mathrm{f}_{2}{ }^{\prime}(\mathrm{x})=(98)(50)(\mathrm{x}-1)^{49}-(600)(49)(\mathrm{x}-1)^{48} =(49)(100)(x−1)48(x−1−6)=(49)(100)(x-1)^{48}(x-1-6) m2=1,n2=0\mathrm{m}_{2}=1, \mathrm{n}_{2}=0 6m2+4n2+8m2n26 m_{2}+4 n_{2}+8 m_{2} n_{2} 6(1)+4(0)+8(0)=66(1)+4(0)+8(0)=6

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives
Topic
Local, Global extremum
Let f 1 :(0, ∞) rightarrow R and f 2 :(0, ∞) rightarrow R be defined… | JEE Advanced 2021 PYQ with Solution · DhiX AI