Class 11th

For the quadratic equation (k+1)tan²x - √2λ tanx + (k-1) = 0, the roots are tanα and tanβ.
Sum of roots: tanα + tanβ = √2λ / (k+1).
Product of roots: tanα tanβ = (k-1) / (k+1).
tan (α + β) = (tanα + tanβ) / (1 - tanα tanβ)
tan (α + β) = [√2λ / (k+1)] / [1 - (k-1)/ (k+1)]
tan (α + β) = [√2λ / (k+1)] / [ (k+1 - k + 1)/ (k+1)] = (√2λ) / 2 = λ/√2.
Given tan² (α+β) = 50.
(λ/√2)² = 50
λ²/2 = 50 ⇒ λ² = 100 ⇒ λ = ±10.

Truth table for (p → q) ∧ (q → ~p).
| p | q | p → q | ~p | q → ~p | (p → q) ∧ (q → ~p) |
|-|-|-|-|-|-|
| T | T | T | F | F | F |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | T | T | T |
The final column is F, T, which is the truth table for ~p.
Therefore, (p → q) ∧ (q → ~p) is equivalent to ~p.

The tangent to the parabola y² = 4ax is y = mx + a/m.

For y² = 4x, a=1. So, the tangent is y = mx + 1/m.
The given line is y = mx + 4.
Comparing the two, 1/m = 4 ⇒ m = 1/4.
The line is y = (1/4)x + 4.
This line is also tangent to x² = 2by.
Substitute y into the parabola equation:
x² = 2b (1/4)x + 4)
x² = ( b/2 )x + 8b
x² - ( b/2 )x - 8b = 0.
For tangency, the discriminant (D) is zero.
D = (-b/2)² - 4 (1) (-8b) = 0.
b²/4 + 32b = 0.
b ( b/4 + 32) = 0.
b = 0 (not possible) or b/4 = -32 ⇒ b = -128.

(3¹/? + 5¹/? )?
General term =? C? (3¹/? )? (5¹/? )? =? C? 3^ (60-r)/4) 5^ (r/8)
Terms are rational for r being a multiple of 8 and (60-r) being a multiple of 4.
If r is a multiple of 8, then 60-r is 60 - 8k. Since 60 is a multiple of 4, 60-8k is also a multiple of 4.
So, we just need r to be a multiple of 8.
r = 0, 8, 16, 24, 32, 40, 48, 56. (Total 8 rational terms)
Total terms are 61.
Number of irrational terms = 61 - 8 = 53 = n.
∴ n - 1 = 52.

g (f (x) = f² (x) + f (x) - 1.
g (f (5/4) = f² (5/4) + f (5/4) - 1.
Given g (f (5/4) = 5/4, let f (5/4) = y.
-5/4 = y² + y - 1 (There appears to be a typo in the image's solution)
y² + y - 1 + 5/4 = 0
y² + y + 1/4 = 0
(y + 1/2)² = 0
y = -1/2.
So, f (5/4) = -1/2.

y = √ (2cos²α / (sinα cosα) + 1/sin²α)
y = √ (2cotα + cosec²α)
y = √ (2cotα + 1 + cot²α) = √ (1 + cotα)²) = |1 + cotα|.
Given α is in a range where 1+cotα is negative, y = -1 - cotα.
dy/dα = - (-cosec²α) = cosec²α.
At α = 5π/6, dy/dα = cosec² (5π/6) = (1/sin (5π/6)² = (1/ (1/2)² = 2² = 4.

Given Re (z-1)/ (2z+i) = 1, where z = x + iy.
(z-1)/ (2z+i) = [ (x-1) + iy] / [2x + I (2y+1)]
To rationalize, multiply the numerator and denominator by the conjugate of the denominator [2x - I (2y+1)].
Numerator = [ (x-1) + iy] * [2x - I (2y+1)] = 2x (x-1) - I (x-1) (2y+1) + i2xy + y (2y+1)
Real part of the numerator = 2x (x-1) + y (2y+1).
Denominator = (2x)² + (2y+1)².
Re (z-1)/ (2z+i) = [2x (x-1) + y (2y+1)] / [ (2x)² + (2y+1)²] = 1.
2x² - 2x + 2y² + y = 4x² + 4y² + 4y + 1.
0 = 2x² + 2y² + 2x + 3y + 1.
So, 2x² + 2y² + 2x + 3y + 1 = 0.

. Let the terms in Arithmetic Progression be a – 2d, a – d, a, a + d, a + 2d.
Sum of terms: (a – 2d) + (a – d) + a + (a + d) + (a + 2d) = 5a.
5a = 25 ⇒ a = 5.
Product of terms: (5 – 2d) (5 – d) (5) (5 + d) (5 + 2d) = 2520.
5 (25 – 4d²) (25 – d²) = 2520.
(25 – 4d²) (25 – d²) = 504.
625 – 25d² – 100d² + 4d? = 504.
4d? – 125d² + 121 = 0.
Factoring the equation: (4d² - 121) (d² - 1) = 0.
So, d² = 1 or d² = 121/4.
d = ±1 or d = ±11/2.
If d = ±1, the terms are 3, 4, 5, 6, 7.
If d = ±11/2, the terms are -6, -1/2, 5, 21/2, 16.
The largest term is 5 + 2d = 5 + 2 (11/2) = 5 + 11 = 16.