15. For each x ∈ R, let [x] be the greatest integer less than or equal to x. Then lim_{x → 0⁻} (x([x] + |x|) sin[x]) / |x| is equal to (A) -sin 1 (B) 0 (C) 1 (D) sin 1 16. Let A = {a₁, a₂, a₃, a₄} and B = {b₁, b₂, b₃, b₄, b₅}. Then the number of injective functions f from A to B, such that f(a₁) ≠ b₄ and f(a₂) = b₁ is: (A) 18 (B) 6
Answer (1)
Let's tackle each problem one at a time.
15. We need to evaluate the limit:
x → 0 − lim ∣ x ∣ x ([ x ] + ∣ x ∣ ) sin [ x ]
To solve this, consider the behavior of the greatest integer function [ x ] as x approaches zero from the left. For x → 0 − , the greatest integer less than or equal to x is − 1 . Therefore, [ x ] = − 1 .
Since x is negative and approaches 0 from the left, ∣ x ∣ = − x .
Substituting these into the function, we have:
∣ x ∣ x ( − 1 + ∣ x ∣ ) sin ( − 1 ) = − x x ( − 1 + − x ) sin ( − 1 )
Simplifying, we get:
− x x ( − 1 − x ) ( − sin 1 )
Cancelling x and resolving the signs:
( 1 + x ) sin 1
As x → 0 − , the term x approaches 0, so:
x → 0 − lim ( 1 + x ) sin 1 = sin 1
Thus, the correct answer is (D) sin 1 .
16. We are tasked with finding the number of injective functions f : A → B such that f ( a 1 ) = b 4 and f ( a 2 ) = b 1 .
Assign f ( a 2 ) = b 1 . This is fixed.
For the remaining elements a 1 , a 3 , a 4 of set A , we need to select from the remaining elements of set B which are {b_2, b_3, b_4, b_5}. However, f ( a 1 ) = b 4 , so we must choose from {b_2, b_3, b_5}.
The number of ways to select an element for a 1 is 3 (because f ( a 1 ) = b 4 ).
After choosing for a 1 , there will be 3 elements left to choose for a 3 .
Then, 2 elements remain for a 4 .
The total number of injective function permutations given these selections is:
3 × 3 × 2 = 18
Therefore, the number of injective functions is (A) 18 .
