Ncert Solutions Maths class 12th

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New answer posted•a year ago
S
a year ago

Let the equation of the plane passing through (1, 2, 3) be:
a (x - 1) + b (y - 2) + c (z - 3) = 0
The plane contains the y-axis, which has direction ratios (0, 1, 0).
Therefore, the normal to the plane must be perpendicular to the y-axis.
a (0) + b (1) + c (0) = 0 ⇒ b = 0
The equation becomes: a (x -

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New answer posted•a year ago
S
a year ago

P (at least 2 show 3 or 5) =? C? (2/6)² (4/6)² +? C? (2/6)³ (4/6)¹ +? C? (2/6)?
= (384+128+16)/6? = 11/27
n=27
∴ expectation of number of times = np
= 27 ⋅ (11/27) = 11

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New answer posted•a year ago
S
a year ago

We have, 1 - (probability of all shots result in failure) > 1/4
? 1 - (9/10)? > 1/4
? 3/4 > (9/10)? ? n? 3 > (9/10)? ? n? 3

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New answer posted•a year ago
S
a year ago

x? = Σf? x? / Σf?
Σf? =? C? +? C? +? C? + . +? C? = 2?
Σf? x? = (0 ×? C? ) + (2 ×? C? ) + (2² ×? C? ) + . + (2? ×? C? )
This sum is Σ? ? C? 2? = (Σ? ? C? 2? ) -? C?2? = (1+2)? - 1 = 3? - 1.
x? = (3? - 1)/2?
Given x? = 728 / (something that resolves to 2? ). Assuming it is 728/2?
(3? - 1)/2? = 728/2

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

P (A∪B∪C) = P (A) + P (B) + P (C) – P (A∩B) – P (B∩C) – P (C∩A) + P (A∩B∩C)
Given relations lead to: α = 1.4 – P (A∩B) – β ⇒ α + β = 1.4 - P (A∩B)
Again, from P (A∪B) = P (A) + P (B) – P (A∩B), and given values, it is found that P (A∩B) = 0.2.
From (1) and (2), α = 1.2 – β.
Now given 0.85 ≤ α ≤ 0.95
⇒ 0.

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New answer posted•a year ago
S
a year ago

Here D = | 2 -4 λ |
| 1 -6 1 | = (λ-3) (3λ+2)
| λ -10 4 |
D = 0 ⇒ λ = 3, -2/3

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New answer posted•a year ago
S
a year ago

Let A (α, 0,0), B (0, β, 0), C (0,0, γ), then the centroid is G (α/3, β/3, γ/3) = (1,1,2).
α = 3, β = 3, γ = 6
∴ Equation of plane is x/α + y/β + z/γ = 1
⇒ x/3 + y/3 + z/6 = 1
⇒ 2x + 2y + z = 6
∴ Required line passing through G (1,1,2) and normal to the plane is (x-1)/2 = (y-1)/2 = (z-2)/1.

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

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 < -2λ/3 < 1 and -2λ/3 ≠ 0
This gives λ ∈ (-3/2, 3/2) -

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