Mathematics · Application of Derivatives

JEE Advanced 2021 — Paper 2 — Question 45

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 2m1+3n1+m1n12 m_{1}+3 n_{1}+m_{1} n_{1} is \qquad

Answer: 57

Numerical answer — enter this value.

Step-by-step solution

f1(x)=∫0x∏j=121(t−j)jdtf_{1}(x)=\int_{0}^{x} \prod_{j=1}^{21}(t-j)^{j} d t ⇒f1′(x)=∏j=121(x−j)j\Rightarrow \quad \mathrm{f}_{1}^{\prime}(\mathrm{x})=\prod_{\mathrm{j}=1}^{21}(\mathrm{x}-\mathrm{j})^{\mathrm{j}} Therefore m1=6,n1=5\mathrm{m}_{1}=6, \mathrm{n}_{1}=5 2 m1+3n1+m1n12 \mathrm{~m}_{1}+3 \mathrm{n}_{1}+\mathrm{m}_{1} \mathrm{n}_{1}

=2(6)+3(5)+30=2(6)+3(5)+30 =12+15+30=57=12+15+30=57

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