Ncert Solutions Maths class 11th

Evaluate the limit:
L = lim (x→0) [sin? ¹ (x) - tan? ¹ (x)] / 3x³

Using Taylor series expansions around x=0:
sin? ¹ (x) = x + x³/6 + O (x? )
tan? ¹ (x) = x - x³/3 + O (x? )

L = lim (x→0) [ (x + x³/6) - (x - x³/3) ] / 3x³
L = lim (x→0) [ x³/6 + x³/3 ] / 3x³
L = lim (x→0) [ (1/6 + 1/3)x³ ] / 3x³
L = (1/2) / 3 = 1/6

The solution shows 3L = 1/2, which is correct. And 6L = 1, also correct.
The final line 6L+1=2 implies 6L=1, confirming the result.

Given the determinant:
| α β γ |
| β γ α | = 0
| γ α β |

The expansion of this determinant is - (α³ + β³ + γ³ - 3αβγ) = 0.
This implies (α+β+γ) (α²+β²+γ²-αβ-βγ-γα) = 0.

From a cubic equation x³ + ax² + bx + c = 0 with roots α, β, γ:
α+β+γ = -a
αβ+βγ+γα = b
αβγ = -c

Substituting into the determinant condition:
(-a) ( (α+β+γ)² - 3 (αβ+βγ+γα) ) = 0
(-a) ( (-a)² - 3b ) = 0
-a (a² - 3b) = 0
a (a² - 3b) = 0

This implies a=0 or a²=3b. If a≠0, then a²=3b, so a²/b = 3.

Given the family of parabolas y² = 4a (x+a).
Differentiate with respect to x:
2y (dy/dx) = 4a
a = (y/2) (dy/dx)

Substitute a back into the original equation:
y² = 4 * (y/2) (dy/dx) * [x + (y/2) (dy/dx)]
y² = 2y (dy/dx) * [x + (y/2) (dy/dx)]
y = 2 (dy/dx) * [x + (y/2) (dy/dx)]
y = 2x (dy/dx) + y (dy/dx)²
y (dy/dx)² + 2x (dy/dx) - y = 0

The general equation of a circle is given by:
az z? + α? z + αz? + d = 0

This can be rewritten as:
z? + (α? /a)z + (α/a)z? + d/a = 0

From this, we can identify the centre and radius:
Centre = -α/a
Radius = √ (|-α/a|² - d/a)

For a real circle to exist, the term under the square root must be non-negative:
|-α/a|² - d/a ≥ 0
|α|²/|a|² - d/a ≥ 0
|α|² - ad ≥ 0, where a ∈ R - {0}.

Kindly consider the following figure

We need to find the remainder of (2021)³? ² when divided by 17.
First, find the remainder of 2021 divided by 17.
2021 = 17 * 118 + 15.
So, 2021 ≡ 15 (mod 17).
Also, 15 ≡ -2 (mod 17).
So, (2021)³? ² ≡ (-2)³? ² (mod 17).
(-2)³? ² = 2³? ² = (2? )? ⋅ 2² = 16? ⋅ 4.
Since 16 ≡ -1 (mod 17),
16? ⋅ 4 ≡ (-1)? ⋅ 4 (mod 17).
≡ 1 ⋅ 4 (mod 17)
≡ 4 (mod 17).
The remainder is 4.

Kindly consider the following figure

Given equation of tangent is 2x - y + 1 = 0
equation of normal is x + 2y = 12
Solving with x - 2y = 4 we get centre at (6,2) radius = √ (36 + 9) = √45 = 3√5.