Ncert Solutions Maths class 11th

Equation of chord of x² + y² = 25 with mid point (h, k) is xh + yk = h² + k².
Or, y = (-h/k)x + (h² + k²)/k.
If this touches the ellipse x²/9 + y²/16 = 1, then the condition for tangency c² = a²m² + b² must be satisfied.
Here, m = -h/k, c = (h²+k²)/k, a²=9, b²=16.
(h² + k²)/k)² = 9 (-h/k)² + 16
(h² + k²)²/k² = 9h²/k² + 16
⇒ (h² + k²)² = 9h² + 16k²
∴ Required locus (x² + y²)² = 9x² + 16y².

Kindly consider the following Image 

(81)^sin²x + (81)^cos²x = 30.
(81)^sin²x + (81)^ (1-sin²x) = 30.
Let y = 81^sin²x.
y + 81/y = 30
y² - 30y + 81 = 0
(y - 3) (y - 27) = 0
⇒ y = 3 or y = 27.
Either 81^sin²x = 3 ⇒ 3^ (4sin²x) = 3¹ ⇒ sin²x = 1/4 ⇒ sin x = ±1/2. x = π/6, 5π/6.
OR, 81^sin²x = 27 ⇒ 3^ (4sin²x) = 3³ ⇒ sin²x = 3/4 ⇒ sin x = ±√3/2. x = π/3, 2π/3.
(as 0 ≤ x ≤ π)
Total possible solutions = 4.

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.

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.