Application of DerivativeshardFree

Critical Points and Higher Derivatives of an Integral with Limit x^2 | JEE

JEE Maths question with a full step-by-step solution.

Question
Let
f(x)=∫0x2(t−1)(t−4)(t−9) dt,f(x) = \int_{0}^{x^{2}} (t-1)(t-4)(t-9)\,dt ,
 then
Af′′(x)=0f''(x) = 0 has 44 distinct positive solutions
Bf′′′(x)=0f'''(x) = 0 has 22 distinct positive solutionscorrect
Cf′′′(x)=0f'''(x) = 0 has 33 distinct positive solutions
Df(x)f(x) has 66 critical points
Solution
Step 1: Differentiate, using the chain rule because the upper limit is x2x^{2} rather than xx.
f′(x)=(x2−1)(x2−4)(x2−9)⋅ddx(x2)=2x(x2−1)(x2−4)(x2−9).f'(x) = \left(x^{2}-1\right)\left(x^{2}-4\right)\left(x^{2}-9\right)\cdot\frac{d}{dx}\left(x^{2}\right) = 2x\left(x^{2}-1\right)\left(x^{2}-4\right)\left(x^{2}-9\right).
 Step 2: Factorise completely and list the zeros.
f′(x)=2x(x−1)(x+1)(x−2)(x+2)(x−3)(x+3),f'(x) = 2x(x-1)(x+1)(x-2)(x+2)(x-3)(x+3) ,
 so
f′(x)=0  at  x=0, ±1, ±2, ±3,f'(x) = 0 \ \text{ at } \ x = 0,\ \pm1,\ \pm2,\ \pm3 ,
 Question attachment which is 77 distinct points. Every one is a genuine critical point, so (4) - which claims 66 - is false. Step 3: Count the zeros of f′′f''. Between consecutive zeros of f′f', Rolle's theorem gives a zero of f′′f''; with 77 zeros there are 66 gaps, so f′′f'' has at least 66 zeros. Since f′f' is a polynomial of degree 77, f′′f'' has degree 66 and therefore exactly 66 zeros. Step 4: Split those 66 by sign. f′f' is an odd function (every factor pairs up as x(x2−a2)x\left(x^{2}-a^{2}\right)), so its zeros are symmetric about 00 and so are the zeros of f′′f'':
3 positive and 3 negative.3 \ \text{positive and } 3 \ \text{negative} .
 So (1), which claims 44 positive, is false. Step 5: Count the zeros of f′′′f'''. It has degree 55, and Rolle's theorem between the 66 zeros of f′′f'' gives 55 zeros - so exactly 55. Step 6: Split those by sign. f′f' odd makes f′′f'' even, which makes f′′′f''' odd, so f′′′(0)=0f'''(0) = 0 and the remaining four zeros pair off:
2 positive,2 negative,and x=0.2 \ \text{positive}, \quad 2 \ \text{negative}, \quad \text{and } x = 0 .
 Step 7: So f′′′(x)=0f'''(x) = 0 has exactly 22 distinct positive solutions - option (2) - and (3), claiming 33, is false. Answer: (2).
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Application of Derivatives · medium
Let a function f:R→Rf : \mathbb{R} \to \mathbb{R} satisfy the functional equation f(x+y)=f(x)⋅f(y),f(x+y) = f(x)\cdot f(y), where f(0)≠0f(0) \ne 0 and f′(0)=1f'(0) = 1. Given g(x)=3xf(x),g(x) = \frac{3x}{f(x)}, then
Application of Derivatives · hard
Consider the function, for x∈[−2,3]x \in [-2, 3], f(x)={x3−2x2−5x+6x−1if x≠1−6if x=1f(x) = \begin{cases} \dfrac{x^{3}-2x^{2}-5x+6}{x-1} & \text{if } x \ne 1 \\[2mm] -6 & \text{if } x = 1 \end{cases} then
Application of Derivatives · medium
For the functions f(θ)=αtan⁡2θ+βcot⁡2θf(\theta)=\alpha\tan^2\theta+\beta\cot^2\theta and g(θ)=αsin⁡2θ+βcos⁡2θg(\theta)=\alpha\sin^2\theta+\beta\cos^2\theta, α>β>0\alpha>\beta>0, let min⁡0<θ<π/2f(θ)=max⁡0<θ<πg(θ)\displaystyle\min_{0<\theta<\pi/2}f(\theta)=\max_{0<\theta<\pi}g(\theta). If the first term of a G.P. is (α2β)\left(\dfrac{\alpha}{2\beta}\right), its common ratio is (2βα)\left(\dfrac{2\beta}{\alpha}\right), and the sum of its first 1010 terms is mn\dfrac{m}{n} with gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to
Application of Derivatives · hard
If xx and yy are real numbers such that x2+y2=27x^{2}+y^{2} = 27, then the maximum possible value of x−yx-y is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath