
Grade 11IIT JEE Entrance Exam
dx÷root of 2ax-x2=ansin-1[x÷a-1].How can i solve this by using dimensions??
dx÷root of 2ax-x2=ansin-1[x÷a-1].How can i solve this by using dimensions??

Let S={x:x belongs to R(√3+√2)^(x^2-4) - (√3-√2)^(x^2-4)=10}. Then n(S) is equal to-
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