Maths

Consider functions f : A → B and g : B → C (A,B,C ⊂ R) such that (gof)⁻¹ exists, then:

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Maths Differential Equations

Contributor-Level 10

If (gof)? ¹ exist then gof is a bijective function. For gof to be bijective, f must be one-one and g must be onto.

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New answer posted

3 months ago

A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of any ellipse with eccentricity 1/3 and the distance of the nearer focus from this directrix is 8/√53, then the equation of the other directrix can be:

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Contributor-Level 10

dy/dx = e^ (3x+4y) = e³? e?
e? dy = e³? dx
∫e? dy = ∫e³? dx
-e? /4 = e³? /3 + C
y (0)=0 ⇒ -1/4 = 1/3 + C ⇒ C = -7/12.
-e? /4 = e³? /3 - 7/12
e? = (7 - 4e³? )/3
y = (-1/4)ln (7-4e³? )/3)
x = -2/3 ln2 = ln (2? ²/³) = ln (1/4¹/³)
e³? = e^ (ln (1/4) = 1/4.
y = (-1/4)ln (7-1)/3) = (-1/4)ln2.
α = -1/4.

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3 months ago

Let X be a random variable such that the probability function of a distribution is given by P(X=0) = 1/2, P(X = j) = (1/3)? (j = 1,2,3,.∞). Then the mean of the distribution and P(X is positive and even) respectively are:

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Contributor-Level 10

Mean = Σx? p? = 0 (1/2) + Σ? ∞ j (1/3)? = (1/3)/ (1-1/3)² = (1/3)/ (4/9) = 3/4
P (X is positive and even) = P (X=2) + P (X=4) + .
= (1/3)² + (1/3)? + . = (1/9)/ (1-1/9) = (1/9)/ (8/9) = 1/8

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3 months ago

Let y = y(x) be solution of the differential equation logₑ(dy/dx) = 3x + 4y, with y(0) = 0.
If y(-2/3 logₑ2) = α logₑ2, then the value of α is equal to:

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Contributor-Level 10

A = [, [-1, 4]. |A| = 2 - 1 = 1.
Characteristic equation: λ² - tr (A)λ + |A| = 0 ⇒ λ² - 3λ + 1 = 0.
By Cayley-Hamilton, A² - 3A + I = 0. A? ¹ (A² - 3A + I) = A - 3I + A? ¹ = 0.
A? ¹ = 3I - A.
Comparing with A? ¹ = αI + βA, we get α=3, β=-1.
4 (α - β) = 4 (3 - (-1) = 16.

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3 months ago

Maths Class 11th

The value of cot(π/24) is:

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Contributor-Level 10

cotθ = (1+cos2θ)/sin2θ
cot (π/24) = (1+cos (π/12)/sin (π/12)
cos (π/12) = cos (15°) = cos (45-30) = (√3+1)/2√2
sin (π/12) = sin (15°) = sin (45-30) = (√3-1)/2√2
cot (π/24) = (1+ (√3+1)/2√2)/ (√3-1)/2√2) = (2√2+√3+1)/ (√3-1)
= (2√2+√3+1) (√3+1)/2 = (2√6+2√2+3+√3+√3+1)/2
= √6 + √2 + √3 + 2

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New answer posted

3 months ago

Maths Class 11th

Let y = y(x) be solution of the differential equation logₑ(dy/dx) = 3x + 4y, with y(0) = 0.
If y(-2/3 logₑ2) = α logₑ2, then the value of α is equal to:

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Contributor-Level 10

S? : x² + y² - x - y - 1/2 = 0, C? : (1/2, 1/2), r? = √ (1/4)+ (1/4)+ (1/2) = 1.
S? : x² + y² - 4y + 7/4 = 0, C? : (0, 2), r? = √ (4 - 7/4) = 3/2.
S? : (x-2)² + (y-1)² ≤ r², C? : (2, 1).
A ∪ B ⊂ C means both circles S? and S? must be inside S?
Distance C? = √ (2-1/2)² + (1-1/2)²) = √ (9/4 + 1/4) = √10/2.
Condition: r ≥ C? + r? ⇒ r ≥ √10/2 + 1.
Distance C? = √ (2-0)² + (1-2)²) = √5.
Condition: r ≥ C? + r? ⇒ r ≥ √5 + 3/2.
√10/2 + 1 ≈ 1.58 + 1 = 2.58.
√5 + 3/2 ≈ 2.23 + 1.5 = 3.73.
So minimum r = (√10+2)/2 and (2√5+3)/2. We need the maximum of these two.
Let's recheck the question logic f

...more

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3 months ago

Maths Class 11th

The compound statement (P ∨ Q) ∧ (¬P) ⇒ Q is equivalent to:

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Maths Continuity and Differentiability

Contributor-Level 10

Truth table analysis shows that (P ∨ Q) ∧ (¬P) is equivalent to Q ∧ ¬P.
Then (Q ∧ ¬P) ⇒ Q. This is a tautology.
The provided solution seems to have an error.
Let's check the options. (P ∨ Q) is a tautology. (P ∧ ¬Q) is a contradiction.
~ (P ⇒ Q) ⇔ P ∧ ¬Q is true.

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3 months ago

Let f: [-π/4, π/4] → R be defined as
f(x) = { (1 + |sin x|)³ᵃ/|sin x|, -π/4 < x < 0; b, x = 0; e^(cot 4x / cot 2x), 0 < x < /4 }
If f is continuous at x = 0, then the value of 6a + b² is equal to:

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Contributor-Level 10

L.H.L = lim (x→0? ) (1 + |sin x|)³? /|sin x| = lim (h→0) (1 + sinh)³? /sinh = e³?
R.H.L = lim (x→0? ) e^ (cot 4x / cot 2x) = lim (x→0? ) e^ (tan 2x / tan 4x) = e¹/².
f (0) = b.
For continuity, e³? = e¹/² = b.
3a = 1/2 ⇒ a = 1/6. b = e¹/².
6a + b² = 6 (1/6) + (e¹/²)² = 1 + e

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3 months ago

Let the plane passing through the point (-1, 0, -2) and perpendicular to each of the planes 2x + y - z = 2 and x - y - z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to:

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Contributor-Level 10

Normal to the required plane is perpendicular to the normals of the given planes.
n = n? * n? = (2i + j - k) * (i - j - k) = -2i + j - 3k.
Equation of the plane is -2 (x+1) + 1 (y-0) - 3 (z+2) = 0
-2x - 2 + y - 3z - 6 = 0
-2x + y - 3z - 8 = 0
2x - y + 3z + 8 = 0
Comparing with ax + by + cz + 8 = 0, we get a=2, b=-1, c=3.
a+b+c = 2-1+3 = 4.

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New answer posted

3 months ago

If |a| = 2, |b| = 5 and |a * b| = 8, then a . b is equal to:

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