Maths

Applying Rolle's theorem in for function f (x), there exists c such that f' (c) = 0, c ∈ (0,1).
Again applying Rolle's theorem in [0, c] for function f' (x), there exists c? such that f' (c? ) = 0, c? ∈ (0, c).
Option A is correct.

Given equation is 2x (2x + 1) = 1 ⇒ 4x² + 2x - 1 = 0. Roots of the equation are α and β.
∴ α + β = -2/4 = -1/2 ⇒ β = -1/2 - α
and
4α² + 2α - 1 = 0 ⇒ α² = (1-2α)/4 = 1/4 - α/2
Now
α = 1/2 - 2α²
Substituting into the expression for β:
β = -1/2 - (1/2 - 2α²) = -1 + 2α²

f (x) = sin²x (λ + sinx)
f' (x) = 2sinxcosx (λ + sinx) + sin²x (cosx) = sinxcosx (2λ + 3sinx)
For extrema, f' (x) = 0
sinx = 0, cosx = 0, or sinx = -2λ/3
For more than 2 points of extrema in the interval, sinx = -2λ/3 must have solutions other than where sinx=0 or cosx=0.
-1 < -2/3 < 1 and -2/3 0
This gives λ ∈ (-3/2, 3/2) - {0}

Given curves are y = x² - 1 and y = 1 - x² so intersection points are (±1,0). Bounded area =
4∫? ¹ (1 - x²)dx = 4 [x - x³/3]? ¹
= 4 (1 - 1/3) = 8/3 sq. units

Let y = (ex)?
ln y = x ln (ex) = x [1 + ln x]
(1/y) (dy/dx) = (1) (1 + ln x) + x (1/x) = 2 + ln x
⇒ dy = (ex)? (2 + ln x)dx
∫? ² (ex)? (2 + log? x)dx = [ (ex)? ]? ² = (2e)² - (1e)¹ = 4e² - e

Clearly, ∫ [0 to n] {x}dx = n/2
∫ [0 to n] [x]dx = 1 + 2 + 3 . . . n − 1
= n (n-1)/2
∴ (n (n-1)/2)² = n/2 {10n (n-1)}
Solving, n = 21

x? = Σf? x? / Σf? = (10 + 15x + 50) / (4+x)
= (60+15x)/ (4+x) = 15
σ² = 50 = Σf? x? ²/Σf? - (x? )²
50 = (50+225x+1250)/ (4+x) - (15)²
50 = (1300+225x)/ (4+x) - 225
⇒ 275 (4+x) = 1300 + 225x
⇒ 50x = 200 ⇒ x = 4

Let P (2cosθ, 2sinθ)
∴ Q (-2cosθ, -2sinθ)
Given line x+y-2=0
∴ α = |2cosθ + 2sinθ – 2| / √2
β = |-2cosθ - 2sinθ – 2| / √2
∴ αβ = √2 (cosθ + sinθ – 1) · √2 (cosθ + sinθ + 1)
= 2|cos²θ + sin²θ + 2sinθcosθ – 1| = 2|sin2θ|
Max |sin2θ| = 1
∴ maximum αβ = 2.

Ways of selecting correct questions =? C? = 15
Ways of doing them correct = 1
Ways of doing remaining 2 questions incorrect = 3² = 9
∴ No. Of ways = 15 * 1 * 9 = 135