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← JEE Main 2022 · 25 Jul · Shift 1
MathematicsInteger+4 / -1Question 23 of 90

JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 23

If the maximum value of the term independent of tt in the expansion of (t2x15+(1−x)110t)15\left(t^2x^{\frac15}+\dfrac{(1-x)^{\frac1{10}}}t\right)^{15}, x≥0x\ge0, is K, then 8 K is equal to ______.

Answer and worked solution

Official answer: 6006

Constant term. 15C10 x(1−x){}^{15}C_{10}\,x(1-x) (from r=10r=10). Its maximum at x=12x=\frac12 is 30034\frac{3003}4.

8K =6006=6006.

Answer: 6006