Ncert Solutions Maths class 12th
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Consider the function f:R→R defined by f(x) = { |x|(2 - sin(1/x)), if x ≠ 0; 0, if x = 0 }. Then f is:
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Consider the function f:R→R defined by f(x) = { |x|(2 - sin(1/x)), if x ≠ 0; 0, if x = 0 }. Then f is:
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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/
Upvote 
If f(x) = { 1/|x| ; |x| ≥ 1, ax² + b ; |x| < 1 } is differentiable at every point of the domain, then the values of a and b are respectively:
If f(x) = { 1/|x| ; |x| ≥ 1, ax² + b ; |x| < 1 } is differentiable at every point of the domain, then the values of a and b are respectively:
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, x≥1 } (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
Upvote The equation of one of the straight lines which passes through the point (1,3) and makes an angle tan?¹(√2) with the straight line, y + 1 = 3√2x is:
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The equation of one of the straight lines which passes through the point (1,3) and makes an angle tan?¹(√2) with the straight line, y + 1 = 3√2x is:
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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 +
Upvote 1/(3²-1) + 1/(5²-1) + 1/(7²-1) + . + 1/((201)²-1) is equal to:
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1/(3²-1) + 1/(5²-1) + 1/(7²-1) + . + 1/((201)²-1) is equal to:
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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
Upvote The real function f(x) = arccosec(x) / √(x - [x]), where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to:
The real function f(x) = arccosec(x) / √(x - [x]), where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to:
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
Upvote If the function f(x) = (cos(sin x) - cos x) / x4 is continuous at each point in its domain and f(0) = 1/k, then k is .
If the function f(x) = (cos(sin x) - cos x) / x4 is continuous at each point in its domain and f(0) = 1/k, then k is .
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 - x
Upvote If the equation of the plane passing through the line of intersection of the planes 2x - 7y + 4z - 3 = 0, 3x – 5y + 4z + 11 = 0 and the point (-2, 1, 3) is ax + by + cz - 7 = 0, then the value of 2a + b + c - 7 is.
If the equation of the plane passing through the line of intersection of the planes 2x - 7y + 4z - 3 = 0, 3x – 5y + 4z + 11 = 0 and the point (-2, 1, 3) is ax + by + cz - 7 = 0, then the value of 2a + b + c - 7 is.
The equation of the plane passing through the line of intersection is:
(2x - 7y + 4z - 3) + λ (3x - 5y + 4z + 11) = 0.
The plane passes through the point (-2, 1, 3).
(2 (-2) - 7 (1) + 4 (3) - 3) + λ (3 (-2) - 5 (1) + 4 (3) + 11) = 0
(-4 - 7 + 12 - 3) + λ (-6 - 5 + 12 + 11) = 0
(-2) + λ (12) = 0 ⇒ 1
Upvote If f(x) = sin(cos⁻¹((1 - 2²ˣ) / (1 + 2²ˣ))) and its first derivative with respect to x is -(a/b)logₑ2 when x = 1, where a and b are integers, then the minimum value of |a² - b²| is.
If f(x) = sin(cos⁻¹((1 - 2²ˣ) / (1 + 2²ˣ))) and its first derivative with respect to x is -(a/b)logₑ2 when x = 1, where a and b are integers, then the minimum value of |a² - b²| is.
Kindly consider the following figure
Upvote The area of the triangle vertices A(z), B(iz) and C(z + iz) is:
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The area of the triangle vertices A(z), B(iz) and C(z + iz) is:
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Kindly consider the following figure
Upvote The value of lim(x→0⁺) (cos⁻¹(x - [x]²) ⋅ sin⁻¹(x - [x]²)) / (x - x²), where [x] denotes the greatest integer ≤ x is:
The value of lim(x→0⁺) (cos⁻¹(x - [x]²) ⋅ sin⁻¹(x - [x]²)) / (x - x²), where [x] denotes the greatest integer ≤ x is:
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)
Upvote