If $\alpha=\lim _{x \rightarrow 0^{+}}\left(\frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)$ and $\beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $a x^{2}+b x-\sqrt{e}=0$, then 12 $\log _{e}(a+b)$ is equal to

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Limits

α = lim_{x \to 0^{+}} (\frac{e ^{t a n x} - e ^{x}}{t a n x - x})

and

β = lim_{x \to 0} (1 + sin x)^{\frac{1}{2} c o t x}

are the roots of the quadratic equation

a x^{2} + b x - e = 0

, then 12

lo g_{e} (a + b)

is equal to

Answer: 6

α = x \to 0^{+} lim \frac{e ^{t a n x} - e ^{x}}{tan x - x} = t \to 0 lim \frac{e ^{t a n x} - e ^{x}}{tan x - x} = e^{0} = 1

(this is a difference-quotient form of

\frac{d}{d u} e^{u}

u \to 0

, which

\to 1

β = x \to 0 lim (1 + sin x)^{\frac{1}{2} c o t x} = e^{l i m \frac{s i n x}{2 t a n x}} = e^{1/2} = e

These are roots of

a x^{2} + b x - e = 0

. Product of roots

α β = 1 \cdot e = - (- e) / a ⟹ a = - 1

. Sum of roots

α + β = 1 + e = - b / a = b ⟹ b = 1 + e

12 lo g_{e} (a + b) = 12 lo g_{e} (- 1 + 1 + e) = 12 lo g_{e} (e) = 12 \times \frac{1}{2} = 6

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Integer answer