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Official PYQ · 2026Set 1 · Q22Calculus: Limits, Continuity, DifferentiabilityNAT1 mark

GATE CS 2026 Set 1 Q22 — Calculus: Limits, Continuity, Differentiability

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Consider the function f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} defined as follows:
f(x)={c1ex−c2log⁡e(1x), if x>03 otherwise f(x)=\left\{\begin{array}{lr} c_{1} e^{x}-c_{2} \log _{\mathrm{e}}\left(\frac{1}{x}\right), & \text{ if } x>0 \\ 3 & \text{ otherwise } \end{array}\right.
where c1,c2∈Rc_{1}, c_{2} \in \mathbb{R}.

If ff is continuous at x=0x=0, then c1+c2=c_{1}+c_{2}= ____\_\_\_\_. (answer in integer)
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Answer: 03 : 03

For the function f(x)f(x) to be continuous at x=0x = 0, the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at x=0x = 0 must all be equal:

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

Step 1: Find the Left-Hand Limit and f(0)f(0)

From the definition of f(x)f(x) for x≤0x \le 0:

f(0)=3f(0) = 3 lim⁡x→0−f(x)=3\lim_{x \to 0^-} f(x) = 3

Step 2: Find the Right-Hand Limit

For x>0x > 0, the function is defined as:

f(x)=c1ex−c2log⁡e(1x)=c1ex+c2log⁡e(x)f(x) = c_1 e^x - c_2 \log_e\left(\frac{1}{x}\right) = c_1 e^x + c_2 \log_e(x)

Taking the limit as x→0+x \to 0^+:

lim⁡x→0+f(x)=lim⁡x→0+(c1ex+c2log⁡e(x))\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left( c_1 e^x + c_2 \log_e(x) \right)

We know that as x→0+x \to 0^+:

  • ex→1e^x \to 1
  • log⁡e(x)→−∞\log_e(x) \to -\infty

For the limit to exist and be a finite real number, the coefficient of the term that diverges to −∞-\infty must be zero. Thus, we must have:

c2=0c_2 = 0

Substituting c2=0c_2 = 0 back into the limit expression:

lim⁡x→0+f(x)=lim⁡x→0+c1ex=c1\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} c_1 e^x = c_1

Step 3: Solve for c1c_1 and c1+c2c_1 + c_2

For continuity at x=0x = 0:

lim⁡x→0+f(x)=f(0)  ⟹  c1=3\lim_{x \to 0^+} f(x) = f(0) \implies c_1 = 3

Therefore, the sum is:

c1+c2=3+0=3c_1 + c_2 = 3 + 0 = 3

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