JEE Advanced level problem asked in JEE Main 2024

[@dumb_tan2539]

Here there's an easier way out. So concept is f(x)=g(x) has same no. of solutions as f(x)-g(x)=0 In our case, lets analyse 2^x=x². Obviously in second quadrant there will be one cut (and no more as d/d(x) of 2^x is +ve and x² is -ve and this means that after intersections if we move left the curves will move away from each other so never intersecting) Now in 1st quadrant it will either have 0,1 or 2 cuts possible. Check for x=1, 2^1>1² fir x=2, 2²=2² gotcha! now check if the slops are same, ln2(2^2) < 2•x [comparing their derivatives at x=2]. Since slops be different, it means x² will dominate for some time but as we know power functions increase much faster than exponentials, (to verify 2^10=1024>10^`=1000) so the curves cut each other twice in first quadrant. So m=1+2=3 By same logic check for ln2•2^x=2x. Here at x=0, lhs>rhs For x=1, lhs

[@RajdeepMishra7]

It gave much trouble to the candidates sir.

[@SuvrajitMandal-xe2ys]

Very tough problem, but you solved easily , thank you sir❤

[@Trishanjit860]

Very much helpful lecture,, thank you sir❤❤❤

[@sortsofeverythingakash4799]

2:08 It is being highlighted here that f"(x) would have been 2^x(ln 2)^ 2 - 2.

[@sortsofeverythingakash4799]

Very nice explanation Sir

[@mekohai4458]

this question is actually and advanced question....i mean a similar question was asked in jee advanced

[@VibswithPreetam]

it was tough...........

[@BibhasRoy-wf5pf]

Good problem

[@shouvik10]

When original polynomial cutting x axis at 3 points, how is 2,4 the solution of it. There must be minimum 3 solutions

[@shivankshrivastav]

Sir in which chapter of calculus these type of questions are there, i want to revise them.

[@MATHXINFINITY]

This is not a problem of actually calculus. But using calculus you can make problems easier.

[@shivx3295]

I did it