Class 11th

Kindly consider the following Image 

lim (x→0) [a e? - b cos (x) + c e? ] / (x sin (x) = 2
Using Taylor expansions around x=0:
lim (x→0) [a (1+x+x²/2!+.) - b (1-x²/2!+.) + c (1-x+x²/2!+.)] / (x * x) = 2
lim (x→0) [ (a-b+c) + x (a-c) + x² (a/2+b/2+c/2) + O (x³)] / x² = 2
For the limit to exist, the coefficients of lower powers of x in the numerator must be zero.
a - b + c = 0
a - c = 0 ⇒ a = c
Substituting a=c into the first equation: 2a - b = 0 ⇒ b = 2a.
The limit becomes: lim (x→0) [x² (a/2 + b/2 + c/2)] / x² = (a+b+c)/2
(a + b + c) / 2 = 2 ⇒ a + b + c = 4.

|z+i|/|z-3i| = 1 ⇒ |z+i| = |z-3i|. This means z is on the perpendicular bisector of the segment from -i to 3i. The midpoint is i, so z = x+i.
w = z? - 2z + 2. Let z = x + iy.
w = (x² + y²) - 2 (x + iy) + 2 = (x² - 2x + 2 + y²) - 2iy.
Re (w) = x² - 2x + 2 + y² = (x - 1)² + 1 + y².
From the first condition, y=1. Re (w) = (x - 1)² + 1 + 1 = (x - 1)² + 2.
Re (w) is minimum for x = 1.
The common z is z = 1 + i.
w = (1+i) (1-i) - 2 (1+i) + 2 = 2 - 2 - 2i + 2 = 2 - 2i.
w² = (2 - 2i)² = 4 (1 - 2i - 1) = -8i.
w? = (-8i)² = -64 ∈ R.
∴ least n ∈ N for which w? ∈ R is n=4.

Var (1, 2, ., n) = (Σn²/n) - (Σn/n)² = 10.
(n (n+1) (2n+1)/6n) - (n (n+1)/2n)² = 10.
(n+1) (2n+1)/6 - (n+1)/2)² = 10.
(n+1)/12 * [2 (2n+1) - 3 (n+1)] = 10.
(n+1)/12 * (4n+2 - 3n-3) = 10.
(n+1) (n-1)/12 = 10.
n² - 1 = 120 ⇒ n² = 121 ⇒ n = 11.

Var (2, 4, ., 2m) = Var (2* (1, 2, ., m) = 2² * Var (1, 2, ., m) = 16.
4 * Var (1, 2, ., m) = 16.
Var (1, 2, ., m) = 4.
Using the formula from above: (m²-1)/12 = 4.
m² - 1 = 48 ⇒ m² = 49 ⇒ m = 7.
m + n = 7 + 11 = 18.

Digits are 1, 3, 5, 7, 9. We need to form a 6-digit number where exactly one digit is repeated.

Choose the digit to be repeated:? C? ways.

Choose the positions for these two repeated digits:? C? ways.

Arrange the remaining 4 distinct digits in the remaining 4 places:? P? = 4! ways.
Total numbers =? C? *? C? * 4! = 5 * 15 * 24 = 1800.
The solution in the image 5/2 (6!) seems to follow a different logic which is unclear. 5 * (6!/2) = 5 * 360 = 1800. This logic is: choose one of 5 digits to repeat. Arrange the 6 digits, and since two are identical, divide by 2!

Expression = (49)¹²? - 1) / 48
This uses the sum of a geometric series or a? - b? factorization.
(x? - 1) / (x - 1) = 1 + x + x² + . + x? ¹.
Let x = 49. (49¹²? - 1)/48 is an integer.
The solution shows (49? ³-1) (49? ³+1) / 48. This is correct factorization. Since 49 is odd, 49? ³ is odd. So 49? ³-1 and 49? ³+1 are consecutive even numbers. One is divisible by 2, the other by 4, so their product is divisible by 8. Also, 49 ≡ 1 (mod 3), so 49? ³-1 is divisible by 3. Hence the numerator is divisible by 24. It is also divisible by 48.