Q29 / 57 · JEE Advanced 2021 (Paper 2) · Mathematics

Let $f_{1}:(0, \infty) \rightarrow \mathbb{R}$ and $f_{2}:(0, \infty) \rightarrow \mathbb{R}$ be defined by $$ f_{1}(x)=\int_{0}^{x} \prod_{j=1}^{21}(t-j)^{j} d t, \quad x > 0 $ and $$ f_{2}(x)=98(x-1)^{50}-600(x-1)^{49}+2450, \quad x > 0 $ where, for any positive integer $n$ and real numbers $a_{1}, a_{2}, \ldots, a_{n}, \prod_{i=1}^{n} a_{i}$ denotes the product of $a_{1}, a_{2}, \ldots, a_{n} .$ Let $m_{i}$ and $n_{i}$, respectively, denote the number of points of local minima and the number of points of local maxima of function $f_{i}, i=1,2$, in the interval $(0, \infty)$. The value of 6m2+4n2+8m2n2 is ____.

Correct answer: 6

Solution

f2x=98x-150-600x-149+2450,  x>0

f2'x=98×50x−149−600×49x−148

=4900x−148x−1−6

=4900x−148x−7

So extrema is at x=7 only, which is minima

m2=1,n2=0

Hence 6m2+4n2+8m2n2=6

Hence, the correct answer is 6.

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