Q5. If f(x) = x2, find (f(1.1) − f(1))/(1.1 − 1) (NCERT Miscellaneous Exercise, Q2).
f(1.1) = 1.21 and f(1) = 1, so the value is 0.21/0.1 = 2.1.
Q6. A function f : Z → Z, given by f(x) = ax + b for integers a and b, contains the pairs (1, 1), (2, 3), (0, −1) and (−1, −3). Find a and b (adapted from NCERT Miscellaneous Exercise, Q8). These are four values of the function, not its entire graph.
f(0) = b = −1 and f(1) = a + b = 1, so a = 2. Check: 2 × 2 − 1 = 3 and 2 × (−1) − 1 = −3. So a = 2, b = −1.
Q7. A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Is f (i) a relation from A to B; (ii) a function from A to B? (NCERT Miscellaneous Exercise, Q10)
(i) Yes: every first entry is in A and every second entry is in B, so f ⊂ A × B. (ii) No: 2 has the two images 9 and 11.
Q8. A = {9, 10, 11, 12, 13} and f : A → N is defined by f(n) = the highest prime factor of n. Find the range of f (NCERT Miscellaneous Exercise, Q12).
f(9) = 3, f(10) = 5, f(11) = 11, f(12) = 3, f(13) = 13. The range is {3, 5, 11, 13}.
Q9. Find the domain of f(x) = 1/√(4 − x2).
The square root needs 4 − x2 ≥ 0 and the denominator needs it ≠ 0, so 4 − x2 > 0, that is −2 < x < 2. Domain (−2, 2).
Q10. Find [3.9], [−0.5] and [−4], and solve [x] = 2.
[3.9] = 3, [−0.5] = −1 and [−4] = −4. [x] = 2 exactly when 2 ≤ x < 3.