Ncert Solutions Maths class 12th

Kindly consider the following figure

Kindly consider the following figure

We need to evaluate lim (x→0? ) (cos? ¹ (x - [x]²) ⋅ sin? ¹ (x - [x]²) / (x - x²).
As x → 0? , the greatest integer [x] = 0.
So the expression becomes:
lim (x→0? ) (cos? ¹ (x) ⋅ sin? ¹ (x) / (x (1 - x²)
= lim (x→0? ) cos? ¹ (x) * lim (x→0? ) (sin? ¹ (x) / x) * lim (x→0? ) (1 / (1 - x²)
We know lim (x→0) (sin? ¹ (x) / x) = 1.
lim (x→0? ) cos? ¹ (x) = cos? ¹ (0) = π/2.
lim (x→0? ) (1 / (1 - x²) = 1 / (1 - 0) = 1.
So the limit is (π/2) * 1 * 1 = π/2.

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 - 1) + c (z - 3) = 0
ax + cz = a + 3c
The plane also passes through the origin (0,0,0) since it contains the y-axis.
a (0) + c (0) = a + 3c ⇒ a + 3c = 0 ⇒ a = -3c
Substitute a = -3c into the plane equation:
-3c (x - 1) + c (z - 3) = 0
-3 (x - 1) + (z - 3) = 0
-3x + 3 + z - 3 = 0
-3x + z = 0 ⇒ 3x - z = 0

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

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

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
⇒ 0.85 ≤ 1.2 – β ≤ 0.95
⇒ -0.35 ≤ -β ≤ -0.25
⇒ 0.25 ≤ β ≤ 0.35, so β ∈ [0.25, 0.35]

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