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Good morning sir, Have you left Akash ? ~ C D A DAV

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Excellent stuffs and the way of explaining

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thank you sir for practice of such questions

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Ans : π/2- tan-¹(1/2)

Let Tn = arccot(n^2 + 3/4) or Tn = arccot[(4n^2 + 3)/4] Now, n^2 ≥ 0 or 4n^2 + 3 > 0 Therefore arccot[(4n^2 + 3)/4] = arctan[4/(4n^2 + 3)] Now we have to simplify this to arctan(a) - arctan(b), so 4/(4n^2 + 3), we have to denominator of form 1 + ab 4n^2 + 3 = 4 + 4n^2-1 = 4 + (2n+1)(2n-1) Thus, Tn = arctan[4/{4 + (2n+1)(2n-1)}] Dividing LHS/RHS by 4 Tn =arctan[1/{1+ ((2n+1)/2)((2n-1)/2)}] Tn = arctan[{(2n+1)/2 - (2n-1)/2}/{1+ ((2n+1)/2)((2n-1)/2)}] Tn = arctan[(2n+1)/2] - arctan[(2n-1)/2] Putting the value of n from 1 to ∞ S = π/2 - arctan(1/2)

(-1,-1/²√2) U (1/²√2,1)

So the overall answer should be 12π-33..🙏🙏

Respected sir, at 44:15 sec-¹(sec9) it should be x-2π rather than x-3π because sec-¹(secx) graph only has x-evenπ or evenπ-x, no odd terms so...... it should be 9-2π.

1) [2nπ + 3π/2 , 2nπ + 5π/2] 2) x ={6, -6}

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Let y = f(x) => x + 1/x = y => x²+1-yx=0 => x² - yx +1= 0 By Shreedharacharya method- x = [y ± √(y² -4)]/2 Now the domain of f(x) is +ve so we know that for +ve nos. x+1/x is always greater than or equal to 2. Thus value of y is always greater than or equal to 2 If we take x = [y - √(y² -4)]/2 => x = greater than 1 - something => x can be less than 1 This goes out of domain hence not of our use. If x = [y + √(y² -4)]/2 => x = 1 + something => x is inside the domain Hence Inverse of f(x) = [x + √(x²-4)]/2.

F'(x) = x+√(x²-4)/2

sir last ques me agar ham zero dale then to same y aega?

Yes sir Because of that 1 I had problem for the homework question.

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We can see here that 2 functions are involved- 1) sin²x 2) √x Now in g(f(x)), we have √ over sin²x and in f(g(x)) we have √ over x and then sin² as function. So we can conclude that: f(x) = sin²x g(x) = √x

Bdhiya hai sir

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@23:14 & @23:19 equivalence class of 4 is {4,1} & for 5 is {5,2}

done

Even function Because when x ≤ (-1), |x| will open (-x) so f(x) = -x² When x ≥ 1 |x| will open positive so in this case too f(x) = -x² Now, when -1 < x < 1, we have to make 3 cases: (-1) <x<0 -- [x+1] + [1-x] = 1 0<x<1 -- [x+1] + [1-x] = 1 x = 0 --- [x+1] + [1-x] = 2

done sorry sir mene lectures backlog me dal diye the 😅

done thank u for the video sir

done

Perhaps the best quality free content in TH-cam I will suggest this channel over anyother (yahan bas kaam ki baat hin hoti hai )

Thank you sir very much for such great explanation as usual🙏 Really grateful🙏

Thank you sir

Thank you sir for the video🙏

Thank you sir for this video🙏

@42:26 (n^2-n)/2 elements for any triangle.

Thank u for the video sir