Maths

Contributor-Level 10

The function f (x) is non-differentiable at x=1, 3, 5.
Σ f (f (x) = f (f (1) + f (f (3) + f (f (5).
Assuming f (x) is defined such that f (1)=1, f (3)=1, f (5)=1 (based on context of absolute value functions).
Then Σ f (f (x) = f (1) + f (1) + f (1) = 1 + 1 + 1 = 3.

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New question posted

2 months ago

Consider an arithmetic series and a geometric series having four initial terms from the set {11,8,21,16,26,32,4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to.

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New answer posted

2 months ago

Maths Class 12th

A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :

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Contributor-Level 9

Two drawn cards are spades. There are 50 cards left.
The missing card could be a spade or not a spade.
P (missing card is spade) = 11/50 (since 11 spades remain out of 50 cards).
P (missing card is not spade) = 1 - 11/50 = 39/50

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2 months ago

Total number of 6 - digit numbers in which only and all the five digits 1,3,5,7 and 9 appear, is

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Contributor-Level 10

Digits are 1, 3, 5, 7, 9. We need to form a 6-digit number where exactly one digit is repeated.

Choose the digit to be repeated:? C? ways.

Choose the positions for these two repeated digits:? C? ways.

Arrange the remaining 4 distinct digits in the remaining 4 places:? P? = 4! ways.
Total numbers =? C? *? C? * 4! = 5 * 15 * 24 = 1800.
The solution in the image 5/2 (6!) seems to follow a different logic which is unclear. 5 * (6!/2) = 5 * 360 = 1800. This logic is: choose one of 5 digits to repeat. Arrange the 6 digits, and since two are identical, divide by 2!

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2 months ago

Maths Class 12th

Let P be a plane lx + my + nz = 0 containing the line, (1 - x)/1 = (y + 4)/2 = (z + 2)/3. If plane P divides the line segment AB joining points A(-3, -6, 1) and B(2, 4, -3) in ratio k : 1 then the value of k is equal to:

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Contributor-Level 9

A (-3, -6,1), B (2,4, -3). Point P divides AB in ratio k:1.
P = [ (2k-3)/ (k+1), (4k-6)/ (k+1), (-3k+1)/ (k+1)]
P lies on the plane lx + my + nz = 0.
l (2k-3) + m (4k-6) + n (-3k+1) = 0
k (2l + 4m - 3n) = 3l + 6m - n
⇒ k = (3l + 6m - n) / (2l + 4m - 3n)
Plane contains the line (x-1)/-1 = (y+4)/2 = (z+2)/3.
The plane passes through (1, -4, -2) and its normal is perpendicular to the line's direction vector.
-l + 2m + 3n = 0
l (1) + m (-4) + n (-2) = 0 ⇒ l - 4m - 2n = 0
Solving these gives l/-8 = m/-1 = n/-2. Let l=8, m=1, n=2.
k = (3 (8) + 6 (1) - 2) / (2 (8) + 4 (1) - 3 (2) = (24+6-2)/ (16+4-6) = 28/14 = 2.

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New answer posted

2 months ago

Let S? = ∑(r=1 to k) tan?¹(6? / (2²?¹ + 3²?¹)). Then lim (k→∞) S? is equal to :

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Contributor-Level 9

S? = ∑ tan? ¹ (6? / (2²? ¹ + 3²? ¹) from r=1 to k. (Assuming n in image is r)
t? = tan? ¹ (6? / (2²? ¹ + 3²? ¹)
= tan? ¹ ( (3/2) * (3/2)^ (2r) / ( (9/4) + (3/2)^ (2r+2) (This seems overly complex. Let's re-examine the image's simplification).
t? = tan? ¹ (6? / (2 * 4? + 3 * 9? ). The image simplifies the denominator to 2²? ¹ + 3²? ¹, which is different. Following the image's next step:
t? = tan? ¹ [ 6? / ( 1 + (3/2)^ (2r+1) ] (This denominator is incorrect).
The image seems to simplify t? into:
t? = tan? ¹ (3/2)? ¹) - tan? ¹ (3/2)? )
S? = [tan? ¹ (3/2)²) - tan? ¹ (3/2)] + [tan? ¹ (3/2)³) - tan? ¹ (3/2)²)] + . + [t

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2 months ago

The greatest positive integer k, for which 49ᵏ + 1 is a factor of the sum 49¹²⁵ + 49¹²⁴ + . + 49² + 49 + 1, is:

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Maths Maths Binomial Theorem

Contributor-Level 10

Expression = (49)¹²? - 1) / 48
This uses the sum of a geometric series or a? - b? factorization.
(x? - 1) / (x - 1) = 1 + x + x² + . + x? ¹.
Let x = 49. (49¹²? - 1)/48 is an integer.
The solution shows (49? ³-1) (49? ³+1) / 48. This is correct factorization. Since 49 is odd, 49? ³ is odd. So 49? ³-1 and 49? ³+1 are consecutive even numbers. One is divisible by 2, the other by 4, so their product is divisible by 8. Also, 49 ≡ 1 (mod 3), so 49? ³-1 is divisible by 3. Hence the numerator is divisible by 24. It is also divisible by 48.

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2 months ago

Let [x] denote greatest integer less than or equal to x. If for n ∈ N, (1 - x + x²)? = ∑(j=0 to 3n) a?x?, then ∑(j=0 to [3n/2]) a? + 4∑(j=0 to [(3n-1)/2]) a? is equal to :

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Contributor-Level 9

(1 - x + x²)³? = ∑ a? x? (from j=0 to 3n)
= a? + a? x + a? x² + . + a? x³? (I)
Let A = a? + a? + a? + .
Let B = a? + a? + a? + .
In (I) put x = 1: (1 - 1 + 1)³? = 1.
1 = a? + a? + a? + a? + . (A + B = 1)
In (I) put x = -1: (1 - (-1) + (-1)²)³? = 3³?
3³? = a? - a? + a? - a? + . (A - B = 3³? )
(This seems incorrect based on the provided solution. Following the image:)
In (I) put x = -1, (1+1+1)^n = 1. (There must be a typo in the original problem, probably (1-x+x²)^n).
Assuming (1-x+x²)^n. Put x=-1 gives 3^n.
The provided text says putting x=-1 gives 1.
1 = a? - a? + a? - a? + .
Adding the two equations: 2 = 2 (a? + a? + a? + .) = 2A

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2 months ago

Maths Maths Matrices

Let A = [[-i, 1], [-i, i]], i = √-1. Then, the system of linear equation A?[x; y] = [8; 64] has :

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Contributor-Level 9

A = [i, -i], [-i, i]
A² = [-2, 2], [2, -2]
A? = [8, -8], [-8, 8]
A? = [-128, 128], [128, -128]
A? [x, y]? =?
-128x + 128y = 8 ⇒ -16x + 16y = 1 ⇒ x - y = -1/16 (I)
128x - 128y = 64 ⇒ 16x - 16y = 8 ⇒ x - y = 1/2 (II)
System is inconsistent hence No solution

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2 months ago

Maths Vector Algebra

Kindly consider the following

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