JEE Main 2025 Apr 3 Shift 1, Mathematics Q19: Continuity of Function

JEE Main2025Apr 3, Shift 1Mathematics

Let $f (x) = ⎩ ⎨ ⎧ (1 + a x)^{1/ x}, x < 0 1 + b, x = 0 \frac{( x + 4 ) ^{1/2} - 2}{( x + c ) ^{1/3} - 2}, x > 0$ be continuous at $x = 0$ . Then $e^{a} b c$ is equal to

A
64
B
72
C
48
D
36

Solution

Left limit: $x \to 0^{-} lim (1 + a x)^{1/ x} = e^{a}$ (standard).

Right limit: The form is $\frac{0}{0}$ as $x \to 0^{+}$ only if $(x + c)^{1/3} \to 2$ , i.e. $c = 8.$

Apply L'Hôpital (or Taylor expand) at $x = 0$ with $c = 8$ :

$lim_{x \to 0^{+}} \frac{( x + 4 ) ^{1/2} - 2}{( x + 8 ) ^{1/3} - 2} = \frac{\frac{1}{2} ( 4 ) ^{- 1/2}}{\frac{1}{3} ( 8 ) ^{- 2/3}} = \frac{1/4}{1/12} = 3.$

Continuity at $0$ : $e^{a} = 1 + b = 3$ , so $a = ln 3$ and $b = 2.$

$e^{a} \cdot b \cdot c = 3 \cdot 2 \cdot 8 = 48.$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q19: Continuity of Function

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