Maths

Let f: R → R be defined as
f(x) = { λ|x²-5x+6| / µ(5x-x²-6), x<2; e^(tan(x-2))/(x-[x]), x>2; µ, x=2 }
where [x] is the greatest integer less than or equal to x. If f is continuous at x = 2, then λ + µ is equal to:

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Contributor-Level 9

LHL = lim (x→2? ) λ|x²-5x+6| / µ (5x-x²-6) = lim (x→2? ) -λ/µ
RHL = lim (x→2? ) e^ (tan (x-2)/ (x- [x]) = e¹
f (2) = µ
For continuity, -λ/µ = e = µ.
⇒ µ=e, -λ=µ²=e², λ=-e²
∴ λ + µ = -e² + e = e (1-e)

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8 months ago

Maths Maths Matrices

The values of a and b, for which the system of equations

2x + 3y + 6z = 8
x + 2y + az = 5
3x + 5y + 9z = b
Has no solution, are:

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Maths Ncert Solutions Maths class 12th

Contributor-Level 9

System of equations can be written as
[2 3 6; 1 2 a; 3 5 9] [x; y; z] = [8; 5; b]
R? →R? -1/2R? , R? →R? -3/2R?
. the system will have no solution if 3-a=0 and b-13≠0.
i.e. for a=3 & b≠13.

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8 months ago

The number of real roots of the equation e? - e? - 2e³? - 12e²? + e? + 1 = 0 is:

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Contributor-Level 9

e? - e? - 2e³? - 12e²? + e? + 1 = 0
e³? - 2 + e? ³? - e? [e³? + 12e? - 1] = 0
Let y=e? y? -y? -2y³-12y²+y+1=0.
The solution breaks the equation into (e³? - 4e? - 1) (e³? - 1 + 3e? ) = 0
Either e³? = 4e? + 1 (One Solution)
OR e³? = 1 - 3e? (One Solution)
∴ the equation has total 2 solutions.

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8 months ago

Let 9 distinct balls be distributed among 4 boxes, B?, B?, B? and B?. If the probability that B? contains exactly 3 balls is k(3/4)? then k lies in the set:

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Contributor-Level 9

Three balls can be given to B? in? C? ways. Now remaining 6 balls can be distributed into 3 boxes in 3? ways.
Total no. of favourable events =? C? * 3? = 28 * 3?
Total no. of events = 9 balls distributed into 4 boxes in 4? ways.
probability = 28 * 3? /4? = 28/9 * (3/4)? ⇒ k = 28/9
k ∈ |x-3|<1

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8 months ago

The area (in sq. units) of the region, given by the set {(x,y) ∈ R*R | x ≥ 0, 2x² ≤ y ≤ 4 – 2x} is:

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Contributor-Level 9

{ (x, y) ∈ R*R, x ≥ 0, 2x² ≤ y ≤ 4 − 2x}.
Required area = ∫? ¹ (4 - 2x - 2x²)dx
= [4x - x² - (2/3)x³]? ¹ = 4-1-2/3 = 7/3

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New answer posted

9 months ago

Maths Sets

Let $X = {n \in N : 1 \leq n \leq 50}$ . If $A = {n \in X : n$ is a multiple of 2 $}$ and $B = {n \in X, n$ is a multiple of 7 $}$ , then the number of elements in the smallest subset of $X$ containing both $A$ and $B$ is

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Contributor-Level 10

A = {2,4,6,8, . . .50} ⇒ 25 element
A = {7,14,21, . . . .49} ⇒ 7 elements
A ∩ B = {14,28,42} = 3 elements
Required number of elements = 25 + 7 - 3 = 29

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9 months ago

If the foot of the perpendicular drawn from the point $(1,0, 3)$ on a line passing through $(α, 7,1)$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$ , then $α$ is equal to

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Contributor-Level 10

(5/3, 7/3, 17/3)
AD ⋅ PD = 0
((5/3-α)i?+(7/3-7)j?+(17/3-1)k?) . (2/3i?+7/3j?+8/3k?) = 0
(5/3-α)(2/3) + (-14/3)(7/3) + (14/3)(8/3) = 0
A(α, 7, 1) D(7/3, 7/3, 12/3)
⇒ 3α = 12
α = 4

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9 months ago

Maths Statistics

If the mean and variance of eight numbers $3,7, 9,12,13,20$ , $x$ and $y$ be 10 and 25 respectively, then $x y$ is equal to

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Contributor-Level 10

Mean = 10 = (3+7+9+12+13+20+x+y)/8
16 = x + y
Variance σ² = 25 = (Σx?²/8) - (mean)²
25 = (3²+7²+9²+12²+13²+20²+x²+y²)/8 = 100
x² + y² = 148
(x+y)² = x² + y² + 2xy
256 = 148 + 2xy
x . y = 54

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9 months ago

If the system of linear equations,

$x + y + z = 6$

$x + 2 y + 3 z = 10$

$3 x + 2 y + λ z = μ$

has more than two solutions, then $μ - λ^{2}$ is equal to

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Contributor-Level 10

Δ = |1 1 1| = 0
|1 2 3|
|3 2 λ|
⇒ 1(2λ - 6) - 1(λ - 9) + 1(-4) = 0
⇒ λ = 1
Δx = |6 1 1|
|10 2 3| = 0
|μ 2 λ|
⇒ 2λ + μ = 16
⇒ μ = 14
μ - λ² = 14 - 1 = 13

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9 months ago

Maths Class 11th

The $_{}^{}$ $s u m ? \sum_{k = 1}^{20} ? (1 + 2 + 3 + \dots . k)$ is(Series)

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