Maths Applications of Derivatives

For the curve defined parametrically as y = 3sinθ.cosθ, x = e^θ.sinθ where θ∈[0,π], the tangent is parallel to x-axis when θ is :

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Contributor-Level 10

y = (3/2)sin (2θ)
x = e^θ sinθ
dy/dθ = 3cos (2θ)
dx/dθ = e^θ (cosθ + sinθ)
dy/dx = (3cos (2θ) / (e^θ (cosθ + sinθ) = (3 (cosθ - sinθ) / e^θ

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New answer posted

3 weeks ago

If the tangent to the curve y = x³ at the point P(t, t³) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is:

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Contributor-Level 10

y = x³

${\frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d y}{d x} |}_{(t, t^{3})} = 3 t^{2}$

Equation of tangent y – t³ = 3t² (x – t)

Let again meet the curve at $Q (t_{1}, t_{1}^{3})$

$\Rightarrow t_{1}^{3} - t^{3} = 3 t^{2} (t_{1} - t)$

$t_{1}^{2} + t t_{1} + t^{2} = 3 t^{2} [? t_{1} \neq t]$

$t_{1}^{2} + t t_{1} - 2 t^{2} = 0$

->t₁ = -2t

Required ordinate = $\frac{2 t^{3} + t_{1}^{3}}{3} = \frac{2 t^{3} - 8 t^{3}}{3} = - 2 t^{3}$

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New answer posted

3 weeks ago

Let f(x) be a polynomial function with negative coefficients and f(−x) = f(x) for all x. Then

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Contributor-Level 10

f' (0) = 0

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New answer posted

3 weeks ago

Let f:(a,b) →R be twice differentiable such that f(x) = ∫ₐˣ g(t)dt for a differentiable function g(x). If f(x) = 0 has exactly five distinct roots in (a, b), then g(x)g'(x) = 0 has at least :

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Contributor-Level 10

f' (x)=g (x)
f" (x)=g' (x)
⇒g (x).g' (x)=f' (x).f" (x)
f (x) has five roots
⇒f' (x) has atleast 4 roots.
And f" (x) has atleast of 3 roots
⇒g (x).g' (x)=0 has atleast 7 roots in (a, b)

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New answer posted

3 weeks ago

The lengths of the sides of a triangle are 10 + x², 10 + x² and 20 – 2x². If for x = k, the area of the triangle is maximum, then 3k² is equal to:

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Contributor-Level 10

CD = √ (10+x²)² – (10–x²)² = 2√10|x|
Area
= 1/2 * CD * AB = 1/2 * 2√10|x| (20–2x²)
=> 10 – x² = 2x
3x² = 10
x = k
3k² = 10

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New answer posted

4 weeks ago

Let the function, f: [-7,0] → R be continuous on [−7,0] and differentiable on (-7,0). If f(-7) = -3 and f'(x) ≤ 2, for all x ∈ (−7,0), then for all such functions f, f(−1) + f(0) lies in the interval:

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Contributor-Level 10

Using Lagrange's Mean Value Theorem (LMVT) for x ∈ [−7, -1].
[f (-1) - f (-7)] / [-1 - (-7)] ≤ 2
[f (-1) - (-3)] / 6 ≤ 2
f (-1) + 3 ≤ 12
f (-1) ≤ 9

Using LMVT for x ∈ [−7, 0].
[f (0) - f (-7)] / [0 - (-7)] ≤ 2
[f (0) - (-3)] / 7 ≤ 2
f (0) + 3 ≤ 14
f (0) ≤ 11

Therefore, f (0) + f (-1) ≤ 11 + 9 = 20.

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New answer posted

4 weeks 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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Raj Pandey

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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New answer posted

a month ago

Let P(x) be a real polynomial of degree 3 which vanishes at x = -3. Let P(x) have local minima at x = 1, local maxima at x = -1 and ∫?¹ P(x)dx = 18, then the sum of all the coefficients of the polynomial P(x) is equal to.

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Contributor-Level 10

P' (x) = a (x-1) (x+1) = a (x²-1).
P (x) = ∫ P' (x) dx = a (x³/3 - x) + b.
Given P (-3) = 0 ⇒ a (-9+3) + b = 0 ⇒ b = 6a.
Given ∫ P (x)dx = 18. Assuming the integration is over a symmetric interval like [-c, c] and using the fact that a (x³/3-x) is an odd function, ∫ (a (x³/3 - x)dx = 0. Then ∫ b dx = 18. If the interval is [-1, 1], this would be b (1 - (-1) = 2b = 18, so b=9.
With b=9, we find a = b/6 = 9/6 = 3/2.
So, P (x) = 3/2 (x³/3 - x) + 9 = x³/2 - 3x/2 + 9.
The sum of all coefficients is 1/2 - 3/2 + 9 = -1 + 9 = 8.

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New answer posted

a month ago

Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n number is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to ______.

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Contributor-Level 10

Let X1, ., X2n be the first 2n observations and Y1, ., Yn be the last n observations.
Given:
ΣXi / 2n = 6 => ΣXi = 12n (i)
ΣYi / n = 3 => ΣYi = 3n (ii)

Combined mean: (ΣXi + ΣYi) / 3n = 5 => ΣXi + ΣYi = 15n. This is consistent with (i) and (ii).

Combined variance: (ΣXi^2 + ΣYi^2) / 3n - (mean)^2 = 4
(ΣXi^2 + ΣYi^2) / 3n - 5^2 = 4
ΣXi^2 + ΣYi^2 = (4 + 25) * 3n = 87n.

New observations are Xi + 1 and Yi - 1.
New mean = (Σ (Xi + 1) + Σ (Yi - 1) / 3n = (ΣXi + 2n + ΣYi - n) / 3n = (15n + n) / 3n = 16n / 3n = 16/3.

New variance k:
k = (Σ (Xi + 1)^2 + Σ (Yi - 1)^2) / 3n - (new mean)^2
k = (Σ (Xi^2 + 2Xi + 1) + Σ (Yi^2 - 2Yi + 1) / 3

...more

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New answer posted

a month ago

Let the tangent to the circle x² + y² = 25 at the point R(3,4) meet x-axis and y-axis at points P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r² is equal to:

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