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Question: If f(x) satisfies the differential equation xf'(x) = f(x) + x² + 9(f(x))² and f(1) = 0, then which o...

If f(x) satisfies the differential equation xf'(x) = f(x) + x² + 9(f(x))² and f(1) = 0, then which of the following statements is/are true

A

lim⁡x→1+f(x)x−1=0\lim_{x\to 1^+} \frac{f(x)}{x-1} = 0

B

lim⁡x→1+f(x)x−1=1\lim_{x\to 1^+} \frac{f(x)}{x-1} = 1

C

lim⁡x→0+f(1+x)+f(1−x)x2=0\lim_{x\to 0^+} \frac{f(1+x)+f(1-x)}{x^2} = 0

D

lim⁡x→0−f(1+x)+f(1−x)x2=2\lim_{x\to 0^-} \frac{f(1+x)+f(1-x)}{x^2} = 2

Answer

B, D

Explanation

Solution

The given differential equation is xf′(x)=f(x)+x2+9(f(x))2xf'(x) = f(x) + x^2 + 9(f(x))^2. We are given the initial condition f(1)=0f(1) = 0.

Step 1: Transform the differential equation.

Rearrange the equation to isolate terms involving f(x)f(x) and f′(x)f'(x): xf′(x)−f(x)=x2+9(f(x))2xf'(x) - f(x) = x^2 + 9(f(x))^2

Divide both sides by x2x^2: xf′(x)−f(x)x2=1+9(f(x)x)2\frac{xf'(x) - f(x)}{x^2} = 1 + 9\left(\frac{f(x)}{x}\right)^2

Recognize that the left side is the derivative of f(x)x\frac{f(x)}{x}: ddx(f(x)x)=1+9(f(x)x)2\frac{d}{dx}\left(\frac{f(x)}{x}\right) = 1 + 9\left(\frac{f(x)}{x}\right)^2

Step 2: Solve the differential equation.

Let y=f(x)xy = \frac{f(x)}{x}. The equation becomes a separable differential equation: dydx=1+9y2\frac{dy}{dx} = 1 + 9y^2 dy1+9y2=dx\frac{dy}{1 + 9y^2} = dx

Integrate both sides: ∫dy1+(3y)2=∫dx\int \frac{dy}{1 + (3y)^2} = \int dx

To integrate the left side, let u=3yu = 3y, so du=3dy  ⟹  dy=du3du = 3dy \implies dy = \frac{du}{3}. ∫11+u2du3=∫dx\int \frac{1}{1 + u^2} \frac{du}{3} = \int dx 13arctan⁡(u)=x+C\frac{1}{3} \arctan(u) = x + C

Substitute back u=3yu = 3y: 13arctan⁡(3y)=x+C\frac{1}{3} \arctan(3y) = x + C

Step 3: Apply the initial condition to find C.

Given f(1)=0f(1) = 0. Since y=f(x)xy = \frac{f(x)}{x}, at x=1x=1, y(1)=f(1)1=01=0y(1) = \frac{f(1)}{1} = \frac{0}{1} = 0. Substitute x=1x=1 and y=0y=0 into the general solution: 13arctan⁡(3⋅0)=1+C\frac{1}{3} \arctan(3 \cdot 0) = 1 + C 13arctan⁡(0)=1+C\frac{1}{3} \arctan(0) = 1 + C 0=1+C  ⟹  C=−10 = 1 + C \implies C = -1

Step 4: Write the particular solution for f(x)f(x).

Substitute C=−1C = -1 back into the solution: 13arctan⁡(3y)=x−1\frac{1}{3} \arctan(3y) = x - 1 arctan⁡(3y)=3(x−1)\arctan(3y) = 3(x - 1) 3y=tan⁡(3(x−1))3y = \tan(3(x - 1)) y=13tan⁡(3(x−1))y = \frac{1}{3}\tan(3(x - 1))

Substitute back y=f(x)xy = \frac{f(x)}{x}: f(x)=x3tan⁡(3(x−1))f(x) = \frac{x}{3}\tan(3(x - 1))

Step 5: Evaluate the given limits.

(A) and (B): lim⁡x→1+f(x)x−1\lim_{x\to 1^+} \frac{f(x)}{x-1} Substitute the expression for f(x)f(x): lim⁡x→1+x3tan⁡(3(x−1))x−1\lim_{x\to 1^+} \frac{\frac{x}{3}\tan(3(x - 1))}{x-1} =lim⁡x→1+x3⋅tan⁡(3(x−1))x−1= \lim_{x\to 1^+} \frac{x}{3} \cdot \frac{\tan(3(x - 1))}{x-1}

Let h=x−1h = x - 1. As x→1+x \to 1^+, h→0+h \to 0^+. =lim⁡h→0+1+h3⋅tan⁡(3h)h= \lim_{h\to 0^+} \frac{1+h}{3} \cdot \frac{\tan(3h)}{h}

Using the standard limit lim⁡t→0tan⁡(kt)t=k\lim_{t\to 0} \frac{\tan(kt)}{t} = k: =1+03⋅3=13⋅3=1= \frac{1+0}{3} \cdot 3 = \frac{1}{3} \cdot 3 = 1 So, lim⁡x→1+f(x)x−1=1\lim_{x\to 1^+} \frac{f(x)}{x-1} = 1. Statement (A) is false. Statement (B) is true.

(C) and (D): lim⁡x→0+f(1+x)+f(1−x)x2\lim_{x\to 0^+} \frac{f(1+x)+f(1-x)}{x^2} and lim⁡x→0−f(1+x)+f(1−x)x2\lim_{x\to 0^-} \frac{f(1+x)+f(1-x)}{x^2} Let L=lim⁡x→0f(1+x)+f(1−x)x2L = \lim_{x\to 0} \frac{f(1+x)+f(1-x)}{x^2}. As x→0x \to 0, f(1+x)+f(1−x)→f(1)+f(1)=0+0=0f(1+x)+f(1-x) \to f(1)+f(1) = 0+0=0. The limit is of the form 00\frac{0}{0}. Apply L'Hopital's Rule: L=lim⁡x→0ddx(f(1+x)+f(1−x))ddx(x2)L = \lim_{x\to 0} \frac{\frac{d}{dx}(f(1+x)+f(1-x))}{\frac{d}{dx}(x^2)} L=lim⁡x→0f′(1+x)⋅1+f′(1−x)⋅(−1)2xL = \lim_{x\to 0} \frac{f'(1+x) \cdot 1 + f'(1-x) \cdot (-1)}{2x} L=lim⁡x→0f′(1+x)−f′(1−x)2xL = \lim_{x\to 0} \frac{f'(1+x) - f'(1-x)}{2x}

Again, as x→0x \to 0, f′(1+x)−f′(1−x)→f′(1)−f′(1)f'(1+x) - f'(1-x) \to f'(1) - f'(1). We need to find f′(1)f'(1). From the limit in (B), lim⁡x→1+f(x)−f(1)x−1=f′(1)\lim_{x\to 1^+} \frac{f(x)-f(1)}{x-1} = f'(1) (since f(1)=0f(1)=0). So f′(1)=1f'(1)=1. Thus, the limit is again of the form 00\frac{0}{0}. Apply L'Hopital's Rule again: L=lim⁡x→0ddx(f′(1+x)−f′(1−x))ddx(2x)L = \lim_{x\to 0} \frac{\frac{d}{dx}(f'(1+x) - f'(1-x))}{\frac{d}{dx}(2x)} L=lim⁡x→0f′′(1+x)⋅1−f′′(1−x)⋅(−1)2L = \lim_{x\to 0} \frac{f''(1+x) \cdot 1 - f''(1-x) \cdot (-1)}{2} L=lim⁡x→0f′′(1+x)+f′′(1−x)2L = \lim_{x\to 0} \frac{f''(1+x) + f''(1-x)}{2}

Since f′′(x)f''(x) is continuous around x=1x=1, we can substitute x=0x=0: L=f′′(1)+f′′(1)2=f′′(1)L = \frac{f''(1) + f''(1)}{2} = f''(1)

Now we need to calculate f′′(1)f''(1). First, find f′(x)f'(x): f(x)=x3tan⁡(3(x−1))f(x) = \frac{x}{3}\tan(3(x - 1)) Using the product rule and chain rule: f′(x)=13tan⁡(3(x−1))+x3⋅sec⁡2(3(x−1))⋅3f'(x) = \frac{1}{3}\tan(3(x - 1)) + \frac{x}{3} \cdot \sec^2(3(x - 1)) \cdot 3 f′(x)=13tan⁡(3(x−1))+xsec⁡2(3(x−1))f'(x) = \frac{1}{3}\tan(3(x - 1)) + x\sec^2(3(x - 1))

Now find f′′(x)f''(x): f′′(x)=13sec⁡2(3(x−1))⋅3+[1⋅sec⁡2(3(x−1))+x⋅2sec⁡(3(x−1))⋅(sec⁡(3(x−1))tan⁡(3(x−1))⋅3)]f''(x) = \frac{1}{3}\sec^2(3(x - 1)) \cdot 3 + [1 \cdot \sec^2(3(x - 1)) + x \cdot 2\sec(3(x - 1)) \cdot (\sec(3(x - 1))\tan(3(x - 1)) \cdot 3)] f′′(x)=sec⁡2(3(x−1))+sec⁡2(3(x−1))+6xsec⁡2(3(x−1))tan⁡(3(x−1))f''(x) = \sec^2(3(x - 1)) + \sec^2(3(x - 1)) + 6x\sec^2(3(x - 1))\tan(3(x - 1)) f′′(x)=2sec⁡2(3(x−1))+6xsec⁡2(3(x−1))tan⁡(3(x−1))f''(x) = 2\sec^2(3(x - 1)) + 6x\sec^2(3(x - 1))\tan(3(x - 1))

Evaluate f′′(1)f''(1): f′′(1)=2sec⁡2(3(1−1))+6(1)sec⁡2(3(1−1))tan⁡(3(1−1))f''(1) = 2\sec^2(3(1 - 1)) + 6(1)\sec^2(3(1 - 1))\tan(3(1 - 1)) f′′(1)=2sec⁡2(0)+6sec⁡2(0)tan⁡(0)f''(1) = 2\sec^2(0) + 6\sec^2(0)\tan(0) Since sec⁡(0)=1\sec(0) = 1 and tan⁡(0)=0\tan(0) = 0: f′′(1)=2(1)2+6(1)2(0)=2+0=2f''(1) = 2(1)^2 + 6(1)^2(0) = 2 + 0 = 2

So, L=f′′(1)=2L = f''(1) = 2. This means lim⁡x→0+f(1+x)+f(1−x)x2=2\lim_{x\to 0^+} \frac{f(1+x)+f(1-x)}{x^2} = 2 and lim⁡x→0−f(1+x)+f(1−x)x2=2\lim_{x\to 0^-} \frac{f(1+x)+f(1-x)}{x^2} = 2. Statement (C) is false. Statement (D) is true.

The true statements are (B) and (D).