Continuity & DifferentiabilitymediumFree

Continuity & Differentiability: Satisfies Equation Maximum Value Minimum Value Match List

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

Question
f:R→Rf:\mathbb R\to\mathbb R satisfies the equation (f(x))3−(f(x))2−x2f(x)+x2=0\left(f(x)\right)^3-\left(f(x)\right)^2-x^2f(x)+x^2 = 0. The maximum value of f(x)f(x) is 11 and the minimum value is 00. Match List-I with List-II.
List-I
If(x)f(x) is a continuous function
II f(x)x\dfrac{f(x)}x, (x≠0)\left(x \ne 0\right) is non-differentiable at only integral values of xx
III f(x)f(x) is discontinuous at x=n2+16n+1x = \dfrac{n^2+1}{6n+1} (n∈N)\left(n \in \mathbb N\right)
IV f(x)f(x) is discontinuous as well as an even function
List-II
Pf(x)f(x) is non-differentiable at maximum two points
Qf(x)f(x) is non-differentiable at exactly three points
Rf(x)f(x) may be discontinuous at infinitely many points
Sf(x)f(x) is non-differentiable at five points
Tf(x)f(x) is non-differentiable at 77 points
AI →\to (Q); II →\to (R); III →\to (S); IV →\to (P)
BI →\to (Q); II →\to (P); III →\to (S); IV →\to (R)correct
CI →\to (S); II →\to (P); III →\to (T); IV →\to (P)
DI →\to (S); II →\to (P); III →\to (T); IV →\to (R)
Solution
Question attachment Step 1:
f3−f2−x2f+x2=f2(f−1)−x2(f−1)=(f−1)(f2−x2)=(f−1)(f−x)(f+x)=0f^3-f^2-x^2f+x^2 = f^2\left(f-1\right)-x^2\left(f-1\right) = \left(f-1\right)\left(f^2-x^2\right) = \left(f-1\right)\left(f-x\right)\left(f+x\right) = 0
 So for each xx, f(x)∈{1, x, −x}f(x) \in \left\{1,\ x,\ -x\right\}: the graph of ff lies inside the union of the three lines y=1y = 1, y=xy = x, y=−xy = -x. Step 2: 0≤f(x)≤10 \le f(x) \le 1 for every xx. For ∣x∣>1\left|x\right|>1 neither xx nor −x-x lies in [0,1]\left[0,1\right], so f(x)=1f(x) = 1 there; for ∣x∣≤1\left|x\right| \le 1 the admissible values are 11 and ∣x∣\left|x\right|. Also f(x)=0f(x) = 0 can only happen at x=0x = 0, and the minimum 00 is attained, so
f(0)=0f(0) = 0
 Step 3 - (I): the branch y=∣x∣y = \left|x\right| and the branch y=1y = 1 meet only at x=±1x = \pm1, so a continuous ff can switch between them only there. Starting from f(0)=0f(0) = 0,
f(x)=min⁡(∣x∣,1)={1,∣x∣>1∣x∣,∣x∣≤1.f(x) = \min\left(\left|x\right|,1\right) = \begin{cases}1,&\left|x\right|>1\\ \left|x\right|,&\left|x\right| \le 1 .\end{cases}
 This has corners at x=−1x = -1, 00, 11 and is smooth elsewhere, exactly three points, so (I) →\to (Q). That eliminates options (3) and (4). Step 4 - (IV): an even ff may switch branches at any symmetric set of points, for instance at x=±1mx = \pm\tfrac1m for every m∈Nm \in \mathbb N. So it may be discontinuous at infinitely many points, and nothing bounds its non-differentiability by two. (IV) →\to (R), not (P). That eliminates option (1). Step 5 - (III):
n2+16n+1=27, 513, 1019, 1725, 2631, 1, 5043,…\frac{n^2+1}{6n+1} = \frac27,\ \frac5{13},\ \frac{10}{19},\ \frac{17}{25},\ \frac{26}{31},\ 1,\ \frac{50}{43},\dots
 n2+16n+1≥1  ⟺  n2+1≥6n+1  ⟺  n(n−6)≥0  ⟺  n≥6\dfrac{n^2+1}{6n+1} \ge 1 \iff n^2+1 \ge 6n+1 \iff n\left(n-6\right) \ge 0 \iff n \ge 6. For x>1x>1 the only admissible value is f(x)=1f(x) = 1, and at x=1x = 1 the two admissible values 11 and ∣x∣\left|x\right| are equal, with both branches tending to 11 as x→1x \to 1, so ff is continuous at these points. So a discontinuity is possible only for n=1,2,3,4,5n = 1,2,3,4,5: five points of discontinuity, hence at least five points of non-differentiability, which is what (S) records. Answer: (2)
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