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Question: If $y = x^{n-1} \ln x$ satisfies $x^2 \frac{d^2y}{dx^2} + a_n x \frac{dy}{dx} + b_n y = 0$ and $f(x...

If y=xn−1ln⁡xy = x^{n-1} \ln x satisfies

x2d2ydx2+anxdydx+bny=0x^2 \frac{d^2y}{dx^2} + a_n x \frac{dy}{dx} + b_n y = 0 and f(x)=cot⁡−1x2−1+sec⁡−1xx2−1+csc⁡−1xf(x) = \cot^{-1} \sqrt{x^2 - 1} + \sec^{-1} \frac{x}{\sqrt{x^2 - 1}} + \csc^{-1} x,

then which of the following options is/are correct?

A

f′(2)=−32f'(2) = -\frac{\sqrt{3}}{2}

B

f′(−2)=−123f'(-2) = -\frac{1}{2\sqrt{3}}

C

a4+b4=4a_4 + b_4 = 4

D

a3−b3=7a_3 - b_3 = 7

Answer

(A), (B), (C)

Explanation

Solution

The question has two parts. First, we need to find the coefficients ana_n and bnb_n of a differential equation. Second, we need to evaluate the derivative of a function involving inverse trigonometric functions.

Part 1: Differential Equation

The given differential equation is x2d2ydx2+anxdydx+bny=0x^2 \frac{d^2y}{dx^2} + a_n x \frac{dy}{dx} + b_n y = 0, and y=xn−1ln⁡xy = x^{n-1} \ln x. We calculate the first and second derivatives of yy:

dydx=ddx(xn−1ln⁡x)=(n−1)xn−2ln⁡x+xn−1⋅1x=(n−1)xn−2ln⁡x+xn−2\frac{dy}{dx} = \frac{d}{dx}(x^{n-1} \ln x) = (n-1)x^{n-2} \ln x + x^{n-1} \cdot \frac{1}{x} = (n-1)x^{n-2} \ln x + x^{n-2}.

d2ydx2=ddx((n−1)xn−2ln⁡x+xn−2)=(n−1)((n−2)xn−3ln⁡x+xn−2⋅1x)+(n−2)xn−3=(n−1)(n−2)xn−3ln⁡x+(n−1)xn−3+(n−2)xn−3=(n−1)(n−2)xn−3ln⁡x+(2n−3)xn−3\frac{d^2y}{dx^2} = \frac{d}{dx}((n-1)x^{n-2} \ln x + x^{n-2}) = (n-1)((n-2)x^{n-3} \ln x + x^{n-2} \cdot \frac{1}{x}) + (n-2)x^{n-3} = (n-1)(n-2)x^{n-3} \ln x + (n-1)x^{n-3} + (n-2)x^{n-3} = (n-1)(n-2)x^{n-3} \ln x + (2n-3)x^{n-3}.

Substitute yy, dydx\frac{dy}{dx}, and d2ydx2\frac{d^2y}{dx^2} into the differential equation:

x2[(n−1)(n−2)xn−3ln⁡x+(2n−3)xn−3]+anx[(n−1)xn−2ln⁡x+xn−2]+bn[xn−1ln⁡x]=0x^2 [(n-1)(n-2)x^{n-3} \ln x + (2n-3)x^{n-3}] + a_n x [(n-1)x^{n-2} \ln x + x^{n-2}] + b_n [x^{n-1} \ln x] = 0

Multiply by powers of xx:

(n−1)(n−2)xn−1ln⁡x+(2n−3)xn−1+an(n−1)xn−1ln⁡x+anxn−1+bnxn−1ln⁡x=0(n-1)(n-2)x^{n-1} \ln x + (2n-3)x^{n-1} + a_n (n-1)x^{n-1} \ln x + a_n x^{n-1} + b_n x^{n-1} \ln x = 0

Assuming x≠0x \neq 0, divide by xn−1x^{n-1}:

(n−1)(n−2)ln⁡x+(2n−3)+an(n−1)ln⁡x+an+bnln⁡x=0(n-1)(n-2) \ln x + (2n-3) + a_n (n-1) \ln x + a_n + b_n \ln x = 0

Group terms by ln⁡x\ln x:

[(n−1)(n−2)+an(n−1)+bn]ln⁡x+[(2n−3)+an]=0[(n-1)(n-2) + a_n(n-1) + b_n] \ln x + [(2n-3) + a_n] = 0

This equation must hold for all xx in the domain of ln⁡x\ln x (x>0x > 0). Thus, the coefficients of ln⁡x\ln x and the constant term must be zero.

Coefficient of ln⁡x\ln x: (n−1)(n−2)+an(n−1)+bn=0  ⟹  n2−3n+2+an(n−1)+bn=0(n-1)(n-2) + a_n(n-1) + b_n = 0 \implies n^2 - 3n + 2 + a_n(n-1) + b_n = 0.

Constant term: 2n−3+an=0  ⟹  an=3−2n2n-3 + a_n = 0 \implies a_n = 3 - 2n.

Substitute ana_n into the first equation:

n2−3n+2+(3−2n)(n−1)+bn=0n^2 - 3n + 2 + (3 - 2n)(n-1) + b_n = 0

n2−3n+2+(3n−3−2n2+2n)+bn=0n^2 - 3n + 2 + (3n - 3 - 2n^2 + 2n) + b_n = 0

n2−3n+2−2n2+5n−3+bn=0n^2 - 3n + 2 - 2n^2 + 5n - 3 + b_n = 0

−n2+2n−1+bn=0  ⟹  bn=n2−2n+1=(n−1)2-n^2 + 2n - 1 + b_n = 0 \implies b_n = n^2 - 2n + 1 = (n-1)^2.

So, an=3−2na_n = 3 - 2n and bn=(n−1)2b_n = (n-1)^2.

Check options (C) and (D):

(C) a4+b4=4a_4 + b_4 = 4. For n=4n=4, a4=3−2(4)=−5a_4 = 3 - 2(4) = -5, b4=(4−1)2=9b_4 = (4-1)^2 = 9. a4+b4=−5+9=4a_4 + b_4 = -5 + 9 = 4. Option (C) is correct.

(D) a3−b3=7a_3 - b_3 = 7. For n=3n=3, a3=3−2(3)=−3a_3 = 3 - 2(3) = -3, b3=(3−1)2=4b_3 = (3-1)^2 = 4. a3−b3=−3−4=−7a_3 - b_3 = -3 - 4 = -7. Option (D) is incorrect.

Part 2: Derivative of f(x)f(x)

The function is f(x)=cot⁡−1x2−1+sec⁡−1xx2−1+csc⁡−1xf(x) = \cot^{-1} \sqrt{x^2 - 1} + \sec^{-1} \frac{x}{\sqrt{x^2 - 1}} + \csc^{-1} x. The domain of f(x)f(x) requires x2−1≥0x^2 - 1 \ge 0 and ∣x∣≥1|x| \ge 1, which is ∣x∣≥1|x| \ge 1. The derivative f′(x)f'(x) is defined for ∣x∣>1|x| > 1.

Consider x>1x > 1. Let x=sec⁡θx = \sec \theta, where θ∈(0,π/2)\theta \in (0, \pi/2).

x2−1=sec⁡2θ−1=∣tan⁡θ∣=tan⁡θ\sqrt{x^2 - 1} = \sqrt{\sec^2 \theta - 1} = |\tan \theta| = \tan \theta (since θ∈(0,π/2)\theta \in (0, \pi/2)).

cot⁡−1x2−1=cot⁡−1(tan⁡θ)=cot⁡−1(cot⁡(π/2−θ))=π/2−θ=π/2−sec⁡−1x\cot^{-1} \sqrt{x^2 - 1} = \cot^{-1} (\tan \theta) = \cot^{-1} (\cot (\pi/2 - \theta)) = \pi/2 - \theta = \pi/2 - \sec^{-1} x.

xx2−1=sec⁡θtan⁡θ=csc⁡θ\frac{x}{\sqrt{x^2 - 1}} = \frac{\sec \theta}{\tan \theta} = \csc \theta.

sec⁡−1xx2−1=sec⁡−1(csc⁡θ)=sec⁡−1(sec⁡(π/2−θ))=π/2−θ=π/2−sec⁡−1x\sec^{-1} \frac{x}{\sqrt{x^2 - 1}} = \sec^{-1} (\csc \theta) = \sec^{-1} (\sec (\pi/2 - \theta)) = \pi/2 - \theta = \pi/2 - \sec^{-1} x.

csc⁡−1x=csc⁡−1(sec⁡θ)=csc⁡−1(csc⁡(π/2−θ))=π/2−θ=π/2−sec⁡−1x\csc^{-1} x = \csc^{-1} (\sec \theta) = \csc^{-1} (\csc (\pi/2 - \theta)) = \pi/2 - \theta = \pi/2 - \sec^{-1} x.

For x>1x > 1, f(x)=(π/2−sec⁡−1x)+(π/2−sec⁡−1x)+(π/2−sec⁡−1x)=3π/2−3sec⁡−1xf(x) = (\pi/2 - \sec^{-1} x) + (\pi/2 - \sec^{-1} x) + (\pi/2 - \sec^{-1} x) = 3\pi/2 - 3\sec^{-1} x.

f′(x)=ddx(3π/2−3sec⁡−1x)=0−3⋅1∣x∣x2−1f'(x) = \frac{d}{dx}(3\pi/2 - 3\sec^{-1} x) = 0 - 3 \cdot \frac{1}{|x|\sqrt{x^2 - 1}}. For x>1x > 1, ∣x∣=x|x|=x, so f′(x)=−3xx2−1f'(x) = -\frac{3}{x\sqrt{x^2 - 1}}.

Check option (A): f′(2)=−32f'(2) = -\frac{\sqrt{3}}{2}. Since 2>12 > 1, we use f′(x)=−3xx2−1f'(x) = -\frac{3}{x\sqrt{x^2 - 1}}.

f′(2)=−3222−1=−323=−332⋅3=−32f'(2) = -\frac{3}{2\sqrt{2^2 - 1}} = -\frac{3}{2\sqrt{3}} = -\frac{3\sqrt{3}}{2 \cdot 3} = -\frac{\sqrt{3}}{2}. Option (A) is correct.

Consider x<−1x < -1. Let x=sec⁡θx = \sec \theta, where θ∈(π/2,π)\theta \in (\pi/2, \pi).

x2−1=sec⁡2θ−1=∣tan⁡θ∣=−tan⁡θ\sqrt{x^2 - 1} = \sqrt{\sec^2 \theta - 1} = |\tan \theta| = -\tan \theta (since θ∈(π/2,π)\theta \in (\pi/2, \pi)).

cot⁡−1x2−1=cot⁡−1(−tan⁡θ)=cot⁡−1(cot⁡(π−θ))=π−θ=π−sec⁡−1x\cot^{-1} \sqrt{x^2 - 1} = \cot^{-1} (-\tan \theta) = \cot^{-1} (\cot (\pi - \theta)) = \pi - \theta = \pi - \sec^{-1} x.

xx2−1=sec⁡θ−tan⁡θ=−csc⁡θ\frac{x}{\sqrt{x^2 - 1}} = \frac{\sec \theta}{-\tan \theta} = -\csc \theta.

Since x<−1x < -1, xx2−1<−1\frac{x}{\sqrt{x^2 - 1}} < -1. Let y=−csc⁡θy = -\csc \theta. Since θ∈(π/2,π)\theta \in (\pi/2, \pi), sin⁡θ∈(0,1]\sin \theta \in (0, 1], csc⁡θ∈[1,∞)\csc \theta \in [1, \infty). So y∈(−∞,−1]y \in (-\infty, -1].

f′(x)=ddxcot⁡−1x2−1+ddxsec⁡−1xx2−1+ddxcsc⁡−1xf'(x) = \frac{d}{dx} \cot^{-1} \sqrt{x^2 - 1} + \frac{d}{dx} \sec^{-1} \frac{x}{\sqrt{x^2 - 1}} + \frac{d}{dx} \csc^{-1} x.

For ∣x∣>1|x| > 1:

ddxcot⁡−1x2−1=−11+(x2−1)2⋅ddxx2−1=−11+x2−1⋅2x2x2−1=−1x2⋅xx2−1=−1xx2−1\frac{d}{dx} \cot^{-1} \sqrt{x^2 - 1} = -\frac{1}{1 + (\sqrt{x^2 - 1})^2} \cdot \frac{d}{dx} \sqrt{x^2 - 1} = -\frac{1}{1 + x^2 - 1} \cdot \frac{2x}{2\sqrt{x^2 - 1}} = -\frac{1}{x^2} \cdot \frac{x}{\sqrt{x^2 - 1}} = -\frac{1}{x\sqrt{x^2 - 1}}.

ddxsec⁡−1u=1∣u∣u2−1dudx\frac{d}{dx} \sec^{-1} u = \frac{1}{|u|\sqrt{u^2 - 1}} \frac{du}{dx}, where u=xx2−1u = \frac{x}{\sqrt{x^2 - 1}}.

dudx=x2−1⋅1−x⋅2x2x2−1x2−1=x2−1−x2x2−1x2−1=−1(x2−1)x2−1\frac{du}{dx} = \frac{\sqrt{x^2 - 1} \cdot 1 - x \cdot \frac{2x}{2\sqrt{x^2 - 1}}}{x^2 - 1} = \frac{\frac{x^2 - 1 - x^2}{\sqrt{x^2 - 1}}}{x^2 - 1} = \frac{-1}{(x^2 - 1)\sqrt{x^2 - 1}}.

∣u∣=∣xx2−1∣|u| = |\frac{x}{\sqrt{x^2 - 1}}|. u2−1=x2x2−1−1=x2−(x2−1)x2−1=1x2−1u^2 - 1 = \frac{x^2}{x^2 - 1} - 1 = \frac{x^2 - (x^2 - 1)}{x^2 - 1} = \frac{1}{x^2 - 1}. u2−1=1x2−1=1x2−1\sqrt{u^2 - 1} = \sqrt{\frac{1}{x^2 - 1}} = \frac{1}{\sqrt{x^2 - 1}}.

ddxsec⁡−1xx2−1=1∣xx2−1∣1x2−1⋅−1(x2−1)x2−1=x2−1∣x∣⋅1x2−1⋅−1(x2−1)x2−1=x2−1∣x∣⋅−1(x2−1)x2−1=−1∣x∣x2−1\frac{d}{dx} \sec^{-1} \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{|\frac{x}{\sqrt{x^2 - 1}}|\sqrt{\frac{1}{x^2 - 1}}} \cdot \frac{-1}{(x^2 - 1)\sqrt{x^2 - 1}} = \frac{\sqrt{x^2 - 1}}{|x| \cdot \frac{1}{\sqrt{x^2 - 1}}} \cdot \frac{-1}{(x^2 - 1)\sqrt{x^2 - 1}} = \frac{x^2 - 1}{|x|} \cdot \frac{-1}{(x^2 - 1)\sqrt{x^2 - 1}} = -\frac{1}{|x|\sqrt{x^2 - 1}}.

ddxcsc⁡−1x=−1∣x∣x2−1\frac{d}{dx} \csc^{-1} x = -\frac{1}{|x|\sqrt{x^2 - 1}}.

So, for ∣x∣>1|x| > 1, f′(x)=−1xx2−1−1∣x∣x2−1−1∣x∣x2−1f'(x) = -\frac{1}{x\sqrt{x^2 - 1}} - \frac{1}{|x|\sqrt{x^2 - 1}} - \frac{1}{|x|\sqrt{x^2 - 1}}.

If x>1x > 1, ∣x∣=x|x| = x, so f′(x)=−1xx2−1−1xx2−1−1xx2−1=−3xx2−1f'(x) = -\frac{1}{x\sqrt{x^2 - 1}} - \frac{1}{x\sqrt{x^2 - 1}} - \frac{1}{x\sqrt{x^2 - 1}} = -\frac{3}{x\sqrt{x^2 - 1}}. This matches our previous result for x>1x > 1.

If x<−1x < -1, ∣x∣=−x|x| = -x, so f′(x)=−1xx2−1−1(−x)x2−1−1(−x)x2−1=−1xx2−1+1xx2−1+1xx2−1=1xx2−1f'(x) = -\frac{1}{x\sqrt{x^2 - 1}} - \frac{1}{(-x)\sqrt{x^2 - 1}} - \frac{1}{(-x)\sqrt{x^2 - 1}} = -\frac{1}{x\sqrt{x^2 - 1}} + \frac{1}{x\sqrt{x^2 - 1}} + \frac{1}{x\sqrt{x^2 - 1}} = \frac{1}{x\sqrt{x^2 - 1}}.

Check option (B): f′(−2)=−123f'(-2) = -\frac{1}{2\sqrt{3}}. Since −2<−1-2 < -1, we use f′(x)=1xx2−1f'(x) = \frac{1}{x\sqrt{x^2 - 1}}.

f′(−2)=1−2(−2)2−1=1−24−1=1−23=−123f'(-2) = \frac{1}{-2\sqrt{(-2)^2 - 1}} = \frac{1}{-2\sqrt{4 - 1}} = \frac{1}{-2\sqrt{3}} = -\frac{1}{2\sqrt{3}}. Option (B) is correct.