For $\mathrm{a}, \mathrm{b}>0$, let $f(\mathrm{x})=\left\{\begin{array}{c}\frac{\tan ((\mathrm{a}+1) \mathrm{x})+\mathrm{b} \tan \mathrm{x}}{\mathrm{x}}, \mathrm{x} 0\end{array}\right.$ be a continous function at $x=0$. Then $\frac{b}{a}$ is equal to

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Continuity and Differentiability

For

a, b > 0

, let

f (x) = ⎩ ⎨ ⎧ \frac{t a n (( a + 1 ) x ) + b t a n x}{x}, x < 0 3 \frac{ax + b ^{2} x ^{2} - ax}{b a x x}, x > 0

be a continous function at

x = 0

. Then

\frac{b}{a}

is equal to

Answer: (D)

Continuity at

x = 0

requires

f (0^{-}) = f (0) = f (0^{+}) = 3

.
Left limit:

lim_{x \to 0^{-}} \frac{tan (( a + 1 ) x ) + b tan x}{x} = (a + 1) + b

. Setting this to 3:

a + b = 2

... (i)
Right limit: write

a x + b^{2} x^{2} = a x 1 + \frac{b ^{2} x}{a} \approx a x (1 + \frac{b ^{2} x}{2 a})

for small

x

, so

a x + b^{2} x^{2} - a x \approx a x \cdot \frac{b ^{2} x}{2 a} = \frac{b ^{2}}{2 a} x^{3/2}

x \to 0^{+} lim \frac{a x + b ^{2} x ^{2} - a x}{b a x ^{3/2}} = \frac{b ^{2} / ( 2 a )}{b a} = \frac{b}{2 a}

Setting this to 3:

b = 6 a

... (ii)
From (i) and (ii):

a + 6 a = 2 ⟹ a = \frac{2}{7}, b = \frac{12}{7}

, and directly from (ii),

\frac{b}{a} = 6

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option