Maths

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2 months ago

The range of a ∈ R for which the function f(x) = (4a - 3)(x + log?5) + 2(a - 7)cot(x/2)sin²(x/2), x ≠ 2nπ, n ∈ N has critical points is:

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Contributor-Level 9

f (x) = (4a - 3) (x + ln5) + 2 (a - 7)cot (x/2)sin² (x/2)
= (4a - 3) (x + ln5) + (a - 7)sin (x)
f' (x) = (4a - 3) + (a - 7)cos (x)
For critical points f' (x) = 0
cos (x) = - (4a - 3) / (a - 7) = (3 - 4a) / (a - 7)
⇒ -1 ≤ (3 - 4a) / (a - 7) ≤ 1
Solving this inequality leads to:
⇒ a ∈ [4/3, 2]

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2 months ago

Maths Class 11th

The number of elements in the set {x ∈ R : (|x| - 3)|x + 4| = 6} is equal to :

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Contributor-Level 9

(|x| - 3)|x + 4| = 6

Case (i) x < -4
(-x - 3) (- (x + 4) = 6
(x + 3) (x + 4) = 6 ⇒ x² + 7x + 12 = 6 ⇒ x² + 7x + 6 = 0
(x + 1) (x + 6) = 0 ⇒ x = -6 (since x < -4)

Case (ii) -4 ≤ x < 0
(-x - 3) (x + 4) = 6
⇒ -x² - 7x - 12 = 6
⇒ x² + 7x + 18 = 0
The discriminant is D = 7² - 4 (1) (18) = 49 - 72 < 0, so no real solution.

Case (iii) x ≥ 0
(x - 3) (x + 4) = 6
⇒ x² + x - 12 = 6
⇒ x² + x - 18 = 0
x = [-1 ± √ (1² - 4 (1) (-18)] / 2 = [-1 ± √73] / 2
Since x ≥ 0 ⇒ x = (√73 - 1) / 2
Only two solutions.

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2 months ago

Maths Class 11th

Consider three observations a, b and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?

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Contributor-Level 9

Variance of a, b, c & a+2, b+2, c+2, are same.
Given: b = a + c (i)
d² = (1/3) (a² + b² + c²) - [ (a+b+c)/3]²
as a + c = b
d² = (1/3) (a² + b² + c²) - (2b/3)²
9d² = 3 (a² + b² + c²) - 4b²
⇒ b² = 3 (a² + c²) - 9d²

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New answer posted

2 months ago

Maths Class 11th

Let a complex number z, |z| ≥ 1, satisfy log?/√² (|z|+11)/(|z|-1)² ≤ 2. Then, the largest value of |z| is equal to…

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Maths Differential Equations

Contributor-Level 9

log? /? [ (|z|+11)/ (|z|-1)²] < 2
(|z|+11)/ (|z|-1)² > (1/2)²
(|z|+11)/ (|z|-1)² > 1/4
⇒ 4|z| + 44 > |z|² - 2|z| + 1
⇒ |z|² - 6|z| - 43 < 0

⇒ |z| - 7 ≤ 0
∴ |z|max = 7

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New answer posted

2 months ago

Let C₁ be the curve obtained by the solution of differential equation 2xy dy/dx = y² - x², x > 0. Let the curve C₂ be the solution of 2xy/(x²-y²) = dy/dx. If both the curves pass through (1, 1), then the area enclosed by the curves C₁ and C₂ is equal to :

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Contributor-Level 10

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New answer posted

2 months ago

Let f:S → S where S = (0,∞) be a twice differentiable function such that f(x+1) = xf(x). If g: S → R be defined as g(x) = logₑf(x), then the value of |g''(5)-g''(1)| is equal to :

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Contributor-Level 10

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New answer posted

2 months ago

Let P(x) = x² + bx + c be a quadratic polynomial with real coefficients such that ∫₀¹ P(x)dx = 1 and P(x) leaves remainder 5 when it is divided by (x - 2). Then the value of 9(b + c) is equal to:

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Maths Differential Equations

Contributor-Level 10

P (x) = x² + bx + c.
Given ∫? ¹ P (x) dx = 1.
∫? ¹ (x² + bx + c) dx = [x³/3 + bx²/2 + cx] from 0 to 1 = 1/3 + b/2 + c = 1.
2 + 3b + 6c = 6 => 3b + 6c = 4 - (i)
When P (x) is divided by (x-2), the remainder is 5. So, P (2) = 5.
(2)² + b (2) + c = 5 => 4 + 2b + c = 5 => 2b + c = 1 - (ii)
From (ii), c = 1 - 2b. Substitute into (i):
3b + 6 (1 - 2b) = 4
3b + 6 - 12b = 4
-9b = -2 => b = 2/9.
c = 1 - 2 (2/9) = 1 - 4/9 = 5/9.
We need to find 9 (b+c).
9 (2/9 + 5/9) = 9 (7/9) = 7.

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New answer posted

2 months ago

If y = y(x) is the solution of the differential equation dy/dx + (tanx)y = sinx, 0 ≤ x ≤ π/3, with y(0) = 0, then y(π/4) equal to:

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Contributor-Level 10

dy/dx + (tanx)y = sinx. This is a linear differential equation.
Integrating Factor (I.F.) = e^ (∫tanx dx) = e^ (ln|secx|) = secx.
The solution is y * I.F. = ∫ (sinx * I.F.) dx + C.
y * secx = ∫ (sinx * secx) dx = ∫tanx dx = ln|secx| + C.
Given y (0) = 0.
0 * sec (0) = ln|sec (0)| + C => 0 = ln (1) + C => C = 0.
So, y * secx = ln (secx).
y = cosx * ln (secx).
At x = π/4:
y = cos (π/4) * ln (sec (π/4) = (1/√2) * ln (√2) = (1/√2) * (1/2)ln (2) = ln (2) / (2√2).

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New answer posted

2 months ago

Let A denote the event that a 6-digit integer formed by 0,1,2,3,4,5,6 without repetitions, is divisible by 3. Then probability of event A is equal to :

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