Let $f: R \rightarrow R$ be defined as \[ f(x)=\left\{\begin{array}{cc} \frac{\lambda\left|x^{2}-5 x+6\right|}{\mu\left(5 x-x^{2}-6\right)} & , \quad x 2 \\ \mu & , \quad x=2 \end{array}\right. \] where $[x]$ is the greatest integer less than or equal to $x$. If $f$ is continuous at $x=2$, then $\lambda+\mu$ is equal to :

(A) $\mathrm{e}(-\mathrm{e}+1)$

(B) $\mathrm{e}(\mathrm{e}-2)$

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Continuity and Differentiability

Let

f : R \to R

be defined as

f (x) = ⎩ ⎨ ⎧ \frac{λ ∣ x ^{2} - 5 x + 6 ∣}{μ ( 5 x - x ^{2} - 6 )} e^{\frac{t a n ( x - 2 )}{x - [ x ]}} μ, x < 2, x > 2, x = 2

where

[x]

is the greatest integer less than or equal to

x

. If

f

is continuous at

x = 2

, then

λ + μ

is equal to :

Answer: (A)

For

x

a little below

2

both

x - 2

and

x - 3

are negative, so

x^{2} - 5 x + 6 = (x - 2) (x - 3)

, while

5 x - x^{2} - 6 = - (x - 2) (x - 3)

. Hence

x \to 2^{-} lim f (x) = x \to 2^{-} lim \frac{λ ( x - 2 ) ( x - 3 )}{- μ ( x - 2 ) ( x - 3 )} = - \frac{λ}{μ}

For

x

a little above

2

[x] = 2

, so

x - [x] = x - 2

and

x \to 2^{+} lim f (x) = x \to 2^{+} lim e^{\frac{t a n ( x - 2 )}{x - 2}} = e

Continuity at

x = 2

makes both one-sided limits equal

f (2) = μ

, so

μ = e

and

- \frac{λ}{μ} = e

, which gives

λ = - e^{2}

. Then

λ + μ = - e^{2} + e = e (- e + 1)

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option