Class 12th

This is possible because these catalysts have a lower activation energy, which allows a comparatively higher number of particles to react at a faster rate. More number of particles will mean more frequent collisions which speeds up the process.

Probability (at most two machines will be out of service) = (3/4)³ . k
⇒ ?C?(1/4)?(3/4)? + ?C?(1/4)¹(3/4)? + ?C?(1/4)²(3/4)³ = (3/4)³ . k
⇒ 17/8 (3/4)³ = (3/4)³ . k
⇒ k = 17/8

f(x) = (3x²+ax-2-a)e?
f'(x) = (3x²+ax-2-a)e? + e?(6x+a)
= e?(3x²+(a+6)x-2)
∴ x=1 is a critical point
∴ f'(1)=0
∴ 3+a+6-2=0
a = -7
∴ f'(x) = e?(3x²-x-2)
= e?(3x+2)(x-1)

∴ maxima at x = -2/3
∴ minima at x = 1

A and B are independent events so P(A/B) = 1/3

y = (8²?-8?²?)/(8²?+8?²?) ⇒ (1+y)/(1-y) = 8? ⇒ 4x = log?((1+y)/(1-y))
x = (1/4)log?((1+y)/(1-y)), f?¹(x) = (1/4)log?((1+x)/(1-x))

sinx = t, cosxdx = dt
I = ∫dt/(t³(1+t?)^(2/3)) = ∫dt/(t?(1+1/t?)^(2/3))
Put 1+1/t? = r³ ⇒ -6/t? dt = 3r²dr
I = (-1/2)∫dr = -1/2 r + c = (-1/2)(1+1/sin?x)^(1/3) + c
f(x) = (-1/2)cosec²x(1+sin?x)^(1/3) and λ=3
⇒ f(π/3) = -2

Note D = |3, 4, 5; 1, 2, 3; 4, 4| = 0 (R? → R? - 2R? + 3R? )
Now let P? = 4x+4y+4z-δ=0. If the system has solutions it will have infinite solution, so P? = αP? + βP?
Hence 3α+β=4 and 4α+2β=4 ⇒ α=2 and β=-2
So for infinite solution 2µ-2=δ ⇒ for 2µ ≠ δ+2 System inconsistent

I = ∫ (cosθ / (2sin²θ + 7sinθ + 3) dθ
sinθ = t ⇒ cosθdθ = dt
= (1/2) ∫ (1 / (t² + (7/2)t + 3/2) dt
= (1/2) ∫ (1 / (t+7/4)² - (5/4)²) dt
= (1/5) ln | (2t+1)/ (t+3)| + c
= (1/5) ln | (2sinθ+1)/ (sinθ+3)| + C
So A = 1/5
B (θ) = 5 (2sinθ+1)/ (sinθ+3)

y = |x| (x-1)
= { 0,   0 ≤ x < 1
{ x-1, 1 ≤ x < 2
Area = ∫? ² 2√x dx - ∫? ² (1) (1)
= [ (4x³/²)/3]? ² - 1/2 = (8√2)/3 - 1/2

e? y'x? + 4x³e? + 2y' / (2√ (y+1) = 0 at (1,0)
y' + 4 + y' = 0 ⇒ y' = -2
equation of tangent at (1,0) is 2x + y - 2 = 0
So option (C) is correct.