৪৯তম বিসিএস ⎯ ফলিত গণিত [৫৬১]

 Continuity requires that the limit of the function at c exists and equals the function value at c. Differentiability is stronger than continuity, so D is wrong.

Differentiability implies continuity, but the converse is not always true (e.g. f(x) = ∣x∣ continuous at 0 but not differentiable).

Continuous at 0, but left derivative = - 1, right derivative = 1. So, not differentiable.

• f(x) = sinx : f(0) = 0,f(π) = 0.  Conditions satisfied.

• f(x) = cos⁡x : endpoints different.

• f(x) = x² : endpoints not equal. 

• f(x) = ex : endpoints not equal

• First check conditions:

  • Continuous? Yes (polynomial).

  • Differentiable? Yes.

  • $f(0)=1,f(2)=1$ → equal endpoints.

• Apply Rolle’s theorem: f′(x)=2x−2

• Solve f′(c)=0⟹2c−2=0⟹c=1.

Function: f(x)=∣x∣ on [−1,1].

Conditions for Rolle’s theorem:

f continuous on [a,b] → true.
f differentiable on (a,b) → fails at x=0.
f(a)=f(b)⟹f(−1)=f(1)=1 →  true.
Since differentiability fails at 0, Rolle’s theorem does not apply.

Correct: C

• Step 1: Check Rolle’s theorem:

  • Continuous on [0, π]? 

  • Differentiable on (0, π)? 

  • Endpoints equal? f(0)=f(π)=0

• Step 2: Find derivative: f′(x)=cosx

• Step 3: Solve f′(c)=0: cos⁡c=0  ⟹  c=π/2

Rolle’s Theorem is the case of MVT when f(a) = f(b).

Generalization of single - variable continuity to several variables.

Young’s theorem provides conditions for equality of mixed partials even when continuity is weaker than Schwarz’s theorem.

Young's theorem is essentially the same as Schwarz’s or Clairaut’s theorem for functions of two variables, stating that if f_xy and f_yx exist and are continuous in an open region, then they are equal at every point in that region. The key is the continuity of the second partials, which guarantees commutativity of differentiation.

• sin⁡x/x not defined at 0.

• ln⁡x undefined at $x\le 0$.

• e^x continuous everywhere

• tan⁡x has poles at odd multiples of π/2

• $|x|$ not differentiable at 0.

• x²sin⁡x smooth → differentiable everywhere

• x/∣x∣ undefined at 0.

That’s the geometric meaning of MVT.

$f$ is continuous on [0,4], differentiable on (0,4), so MVT applies. Rolle does not because f(0)≠f(4).

Schwarz’s theorem is the standard result for equality of mixed partial derivatives.