Ncert Solutions Maths class 12th

Given the function:
f(x) = { x(2 - sin(1/x)), if x ≠ 0
{ 0, if x = 0

For x < 0: f(x) = x(2 - sin(1/x))

For x > 0: f(x) = x(2 - sin(1/x))

The derivative f'(x) for x ≠ 0 is:
f'(x) = 1*(2 - sin(1/x)) + x*(-cos(1/x))*(-1/x²) = 2 - sin(1/x) + (1/x)cos(1/x)

The text calculates the derivative differently:
For x < 0: f'(x) = -2 + sin(1/x) - (1/x)cos(1/x)
For x > 0: f'(x) = 2 - sin(1/x) + (1/x)cos(1/x)

To check if f'(0) is defined, we would need to use the limit definition of the derivative at x=0. As x approaches 0, the term (1/x)cos(1/x) oscillates and does not approach a finite limit. Therefore, f'(0) is undefined.

A function f (x) is continuous at x=1, so lim (x→1? ) f (x) = lim (x→1? ) f (x) = f (1).
Assuming a piecewise function like f (x) = { -x, x<1; ax+b, x1 } (structure inferred from derivative).
Continuity at x=1: f (1) = 1. a (1)+b = 1 => a+b=1.

The function is differentiable at x=1. The derivative of f (x) at x=1 from the left is -1. The derivative from the right is a.
So, a = -1. (The image has 2a = -1, which would imply a function like -x and ax²+b). Let's assume f' (x) = 2a for x>1.
2a = -1 => a = -1/2.

From a+b=1, b = 1 - a = 1 - (-1/2) = 3/2.
So, a = -1/2 and b = 3/2.

A line passes through (1,3). Its equation is y - 3 = m (x - 1) or y = mx + (3-m).
The angle θ between this line and the line y = 3√2x - 1 (with slope m? = 3√2) is given by tanθ = √2.

tanθ = | (m - m? )/ (1 + m*m? )|
√2 = | (m - 3√2) / (1 + m*3√2)|

This gives two cases:

Case 1 (+ve):
√2 = (m - 3√2) / (1 + 3√2m)
√2 (1 + 3√2m) = m - 3√2
√2 + 6m = m - 3√2
5m = -4√2
m = -4√2 / 5

Case 2 (-ve):
-√2 = (m - 3√2) / (1 + 3√2m)
-√2 (1 + 3√2m) = m - 3√2
-√2 - 6m = m - 3√2
7m = 2√2
m = 2√2 / 7

Consider the series:
1/ (3²-1) + 1/ (5²-1) + 1/ (7²-1) + . + 1/ (201)²-1)

The general term T? can be written for the r-th term starting with r=1 for 3, r=2 for 5.
T? = 1/ (2r+1)² - 1) = 1/ (2r+1-1) (2r+1+1) = 1/ (2r * (2r+2) = 1/4 * 1/ (r (r+1)
T? = 1/4 * (1/r - 1/ (r+1)

The sum of the first n terms is:
S? = Σ T? = 1/4 * Σ (1/r - 1/ (r+1) from r=1 to n
S? = 1/4 * [ (1 - 1/2) + (1/2 - 1/3) + . + (1/n - 1/ (n+1) ]
S? = 1/4 * (1 - 1/ (n+1)

The last term is (201)²-1, so 2r+1 = 201, which gives r = 100. So, n=100.
S? = 1/4 * (1 - 1/101) = 1/4 * (100/101) = 25/101.

Given the function f (x) = cosec? ¹ (x) / √ {x - [x]} where [x] is the greatest integer function.

The domain of cosec? ¹ (x) is (-∞, -1] U [1, ∞).
For the denominator to be defined, x - [x] ≠ 0, which means {x} ≠ 0 (the fractional part of x is not zero). This implies that x cannot be an integer (x ∉ I).

Combining these conditions, the domain is all non-integer numbers except for those in the interval (-1, 1).

f (x) = (cos (sin x) - cos x) / x? We need lim (x→0) f (x) = 1/k.
Using cos C - cos D = -2 sin (C+D)/2) sin (C-D)/2).
f (x) = -2 sin (sin x + x)/2) sin (sin x - x)/2) / x?
For small x, sin x ≈ x.
lim (x→0) f (x) = lim -2 * ( (sin x + x)/2 ) * ( (sin x - x)/2 ) / x?
Using series expansion: sin x = x - x³/3! + x? /5! - .
sin x + x = 2x - x³/6 + .
sin x - x = -x³/6 + x? /120 - .
f (x) ≈ -2 * ( (2x)/2 ) * ( (-x³/6)/2 ) / x?
≈ -2 * (x) * (-x³/12) / x?
≈ (2x? /12) / x? = 2/12 = 1/6.
So, 1/k = 1/6 ⇒ k = 6.