Let f : R → R be defined as
f(x) = { (λ | x² - 5x + 6 |)/(μ (5x - x² - 6)),, x < 2; e^((tan (x - 2))/(x - [x])),, x > 2; μ ,, x = 2
where [x] is the greatest integer is than or equal to x. If f is continuous at x = 2, then λ + μ is equal to :

Solution

Maths sir · 4 steps

Near x = 2 the modulus and the bracket both settle down. Find each side's limit, then match them to μ.

1. Left of 2. Factor x² - 5x + 6 = (x-2)(x-3). For x < 2 both factors are negative, so the product is positive and |x² - 5x + 6| = x² - 5x + 6. The denominator is μ(5x - x² - 6) = -μ(x² - 5x + 6), so f(x) = -λ/μ and the left limit is -λ/μ.
2. Right of 2. For 2 < x < 3 we have [x] = 2, so x - [x] = x - 2. The exponent becomes tan(x-2)/(x-2), which tends to 1.

   lim(x → 2⁺) f(x) = e^(lim(h → 0⁺) (tan h)/h) = e¹ = e

3. Match all three. Continuity needs left limit = right limit = f(2). So μ = e, and -λ/μ = e gives λ = -e × e = -e².
4. So. λ + μ = -e² + e = e(-e + 1), which is option (a).

The idea to keep. Before taking a limit, strip the modulus and the greatest integer function using the sign and the integer part just next to the point. Then the limit is usually a standard one.

Answer from the NTA final answer key. Solution by PrepWiser.