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Mathematics · Chapter

Continuity and Differentiability

This chapter introduces two core concepts of calculus: continuity and differentiability. Continuity formalizes the idea of a function's graph having no breaks, jumps, or holes. You will learn to use limits to test for continuity at a point and over an interval, and to classify different types of discontinuities. The main sub-topics are the definition of continuity, the algebra of continuous functions, and the Intermediate Value Theorem.

The second half, differentiability, is a stricter condition implying that a function is 'smooth' enough to have a well-defined tangent at every point. You will learn to find derivatives from first principles and master the rules of differentiation like the product, quotient, and chain rules. The chapter also covers methods for differentiating implicit, parametric, inverse trigonometric, and logarithmic functions. It typically concludes with Rolle's Theorem and the Lagrange Mean Value Theorem (LMVT), which are foundational for later topics in calculus.

The chapter's difficulty often arises when applying these concepts to piecewise functions or functions involving modulus, greatest integer, or fractional parts. A very strong understanding of the 'Limits' chapter is essential, as the definitions of both continuity and differentiability are based on limits.

Formulas for Continuity and Differentiability

Condition for Continuity at a point x = a
lim⁡x→a−f(x)=lim⁡x→a+f(x)=f(a)\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) The left-hand limit, right-hand limit, and the function's value must all exist and be equal.
Derivative from First Principle
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} This limit, if it exists, is the derivative of f(x).
Product Rule
(u⋅v)′=u′v+uv′(u \cdot v)' = u'v + uv' For differentiating the product of two functions u(x) and v(x).
Quotient Rule
(uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} For differentiating the ratio of two functions u(x) and v(x), where v(x) ≠ 0.
Chain Rule
dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} Used for differentiating composite functions of the form y = f(g(x)).
Derivative of arcsin(x)
ddx(sin⁡−1x)=11−x2\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}} Valid for x in (-1, 1).
Derivative of arctan(x)
ddx(tan⁡−1x)=11+x2\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2} Valid for all real x.
Logarithmic Differentiation
dydx=y⋅ddx(ln⁡y)\frac{dy}{dx} = y \cdot \frac{d}{dx}(\ln y) A technique for functions of the form y = [f(x)]^{g(x)}.
Derivative of Parametric Functions
dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt} For functions defined by x = f(t) and y = g(t), provided dx/dt ≠ 0.
Rolle's Theorem
f′(c)=0f'(c) = 0 For some c in (a, b), if f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b).
Lagrange's Mean Value Theorem (LMVT)
f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b) - f(a)}{b-a} For some c in (a, b), if f is continuous on [a, b] and differentiable on (a, b).

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JEE Advanced 2023

Let SS be the set of all twice differentiable functions ff from R\mathbb{R} to R\mathbb{R} such that d2fdx2(x)>0\frac{d^{2}f}{dx^{2}}(x)>0 for all x∈(−1,1)x\in(-1,1). For f∈Sf\in S, let XfX_{f} be the number of points x∈(−1,1)x\in(-1,1) for which f(x)=xf(x)=x. Then which of the following statements is(are) true?
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S={f: f′′(x)>0 ∀x∈(−1,1)}S=\{f:\ f''(x)>0\ \forall x\in(-1,1)\} f is strictly convex on (−1,1)f \text{ is strictly convex on } (-1,1) Xf=#{x∈(−1,1):f(x)=x}X_f=\#\{x\in(-1,1): f(x)=x\} (1) ∃f: Xf=0(2) ∀f: Xf≤2\text{(1) } \exists f:\ X_f=0 \qquad \text{(2) } \forall f:\ X_f\le 2 (3) ∃f: Xf=2(4) ∄f: Xf=1\text{(3) } \exists f:\ X_f=2 \qquad \text{(4) } \nexists f:\ X_f=1 1. set g(x)=f(x)−x1.\ \text{set } g(x)=f(x)-x 2. Xf=#{zeros of g in (−1,1)}2.\ X_f = \#\{\text{zeros of } g \text{ in } (-1,1)\} 3. check each statement with examples3.\ \text{check each statement with examples} g(x)=f(x)−x,f(x)=x  ⟺  g(x)=0g(x)=f(x)-x, \quad f(x)=x \iff g(x)=0 g′′(x)=f′′(x)>0⇒g strictly convex on (−1,1)g''(x)=f''(x)>0 \Rightarrow g \text{ strictly convex on } (-1,1) Claim: a strictly convex g has≤2 zeros\text{Claim: a strictly convex } g \text{ has} \le 2 \text{ zeros} a<b<c,g(a)=g(b)=g(c)=0a<b<c,\quad g(a)=g(b)=g(c)=0 convexity⇒g(b)<0, a contradiction\text{convexity} \Rightarrow g(b)<0, \text{ a contradiction} Xf≤2 for every f∈S⇒(2) TRUEX_f\le 2 \text{ for every } f\in S \Rightarrow \text{(2) TRUE} g(x)=x2+1>0⇒Xf=0g(x)=x^2+1>0 \Rightarrow X_f=0 f(x)=x+x2+1, f′′=2>0⇒(1) TRUEf(x)=x+x^2+1,\ f''=2>0 \Rightarrow \text{(1) TRUE} g(x)=x2−14=0⇒x=±12g(x)=x^2-\dfrac14=0 \Rightarrow x=\pm\dfrac12 f(x)=x+x2−14⇒Xf=2⇒(3) TRUEf(x)=x+x^2-\dfrac14 \Rightarrow X_f=2 \Rightarrow \text{(3) TRUE} g(x)=x2⇒g(x)=0 only at x=0g(x)=x^2 \Rightarrow g(x)=0 \text{ only at } x=0 f(x)=x+x2⇒Xf=1⇒(4) FALSEf(x)=x+x^2 \Rightarrow X_f=1 \Rightarrow \text{(4) FALSE} (1) TRUE,(2) TRUE,(3) TRUE,(4) FALSE\text{(1) TRUE},\quad \text{(2) TRUE},\quad \text{(3) TRUE},\quad \text{(4) FALSE} Answer: (1), (2), (3)\boxed{\text{Answer: (1), (2), (3)}}

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