Questions and Answers

PART - A

QUESTIONS AND ANSWERS

1. Find the domain of f(x) =

Ans: Domain is the set of all real x for which f(x) is real

⸫ 1 - x2 ≥ 0 ⇒ x2 – 1 ≤ 0  ⇒ -1 ≤ x ≤ 1

⸫ Df = [-1, 1]

2. Find the domain of the function f(x) =

Ans: f(x) is real if x2 - 2x + 5 ≥ 0, since x2 + 1 ≠ 0   ∀ x∈ℝ

For this quadratic, discriminant D = 4 - 4.5 = -16 < 0

⸫ x2 - 2x + 5 ≥ 0 ∀ x∈ℝ

Hence, Df = ℝ

3. Find the domain of

Ans: f(x) is real if 1 – x > 0, 1 - x ≠ 1 and x + 2 ≥ 0

⇒ x - 1 < 0, x + 1 and x + 2 ≥ 0

⇒ x < 1 and x ≥ −2 ⇒ −2 ≤ x < 1, x ≠ 0

⸫ Df = [−2, 0) ∪ (0, 1)

4. If f(x) =   then find the domain of

Ans: f(x) = 3√x - 2

5. If f(x) = find the domain of fg and

Ans: f(x) = log(4 - x2)

⸫ Df is given by 4 − x2 > 0 ⇒ x2 − 4 < 0 ⇒ (x + 2)(x − 2) < 0

⇒                                                                       -2 < x <2

⸫ Df = (−2, 2) 

6.

Find the values of ƒ (0), ƒ (1), ƒ(1.5)

Ans: 

When x = 0, ƒ(0) = |0 +1| = 1

When x = −1, ƒ(−1) = |−1+1| = 0

and when x = 1.5, ƒ(1.5) = |1.5 - 1| = |+0.5| = 0.5

7. Is the function in the question no.6 even?

Ans:

Hence, ƒ is even.

8. If f(x) = − Test ƒ is even or odd

Ans:

Hence, ƒ is odd

9. If then ƒ is even - True or False.

Ans:

10. is even, True or False

Ans:

Hence, ƒ is odd

⸫ f(x) is even is false

11. If then find f(g(2)) and g(f(2))

Ans:

12. If f(x) = x + 5, g(x) = x2 + 3 find f(g(0)) and ƒ(ƒ(−5))

Ans:

13. If

then find g(f(3)) and g(f(2)).

Ans: 

Put x = 3, then g(f(3)) = 3 − 2 = 1, Put x = 2, then g(f(2)) = 1 - 2 = −1

14. If   where a > 0 and is a positive integer, show that

f(f(x)) = x.

Ans: Given f(x) =

⸫ f(f(x)) = x

15. If ƒ is an even function defined on [−5, 5] and if f(x) = find the number of solutions of this equation in [−5, 5].

Ans: Given f is an even function

⸫ There are 4 solutions in [-5, 5]

16. In the above problem if the interval is [−2, 2]. Find the number of solutions.

Ans: Only 3 solutions belong to [−2, 2]

17. If f(x) = then for what values of a, f(f(x)) = x?

Ans:

18. Find the inverse of f(x) = x3 +1, x ∈ ℝ

Ans: Given f(x) = x3 +1, x ∈ ℝ

Clearly if x1 ≠ x2, then f(x1) ≠  f(x2)

⸫ f is one-one

Let y = x3 + 1 ⇒ x3 = y – l ⇒ x = (y - 1)1/3

⸫ The inverse is ƒ-1(y) = (y − 1)1/3, y∈ℝ

or ƒ-1(x) = (x − 1)1/3, x∈ℝ

19. Find the inverse of f(x) =

Ans: 

Clearly f is one-one

20. Prove that the inverse of

Ans: 

21. If where [x] is the greatest integer ≤ x, then find the values of

Ans: Given f(x) = cos[π2]x + cos[-π2]x

We know π = 3.14 (approximately). 

⸫ π2 = 9.9 (approximately)

22. Find the vertical asymptote of the graph of the function

Ans: 

23. Find

Ans: 

24. is not true, why?

Ans: 

Hence, we can not write as difference of limits.

25. Evaluate

Ans: 

26. is not true. Why?

Ans: 

27. The equation x10 - 10x2 + 5 = 0 has a root in the interval (0, 2) True or false.

Ans: Let f(x) = x10 - 10x2 + 5. It is continuous in (0, 2). ƒ(0) = 5 > 0 and f(1) = 1 − 10 + 5 = −4 < 0

Since ƒ(0) and ƒ(1) have opposite signs, there is a root between 0 and 1. 

Hence there is a root in (0, 2) – True

28. Ifƒ'(r) exists, then True or false

Ans: True, since differentiability ⇒ continuity

29. Find the expression for the function whose graph is the line segment joining the points (-5, 10) and (7,−10)

Ans: Equation of the line joining the points(-5, 10) and (7, −10) is

30. Find means x degrees.

Ans: We know 180° is π radian

31. Find

Ans:

32. Find

Ans: 

Hence, f is continuous at x = 1.

32.

Ans: 

33. If

Ans: 

34. If ƒ and g are are continuous functions with f(3) = 5 and

Ans: Given ƒ and g are continuous

35.

Ans:

36. If ƒ is an odd function and if exists, find its value.

Ans: 

37.

Test the continuity of the function

Ans: 

Hence, ƒ is continuous at x = 0 and hence, ƒ is continuous ∀x.

38. is continuous at x = 2

Find the value of a

Ans: g is continuous at x = 2

39. If ƒ is continuous at x = 0, find K.

Ans: Given f is continuous at x = 0

40. If

Ans: 

41. Ifƒ'(2) = 4, g(3) = 6 and g′(3) = 5

find h'(3) if h(x) =

Ans: 

42. If f(x) = mx + c and ƒ(0) = ƒ'(0) = 1, then what is ƒ'(2) ?

Ans:

f(x) = mx + c

⸫ f'(x) = m

f'(0) = 1 ⇒ m = 1

Hence, f'(x) = 1 ⸫ ƒ'(2) = 1

43. At what point on the curve y = [In (x + 4)]2 is the tangent horizontal?

Ans: 

⸫ the point is (−3, 0)

44. true or false

Ans: Let ƒ(x) = |x2 + x|

45. If the line x + y = a is a tangent to the curve then find the value of a

Ans: The line x + y = a ⇒ y = a - x will touch if it intersects the ellipse in coincident points. The point of intersection are given by

46. If the normal to the curve y = f(x) at the point (3, 4) makes an angle with the positive x-axis, then find the value of ƒ’(3).

Ans: f'(3) is the slope of the tangent at (3, 4) on y = f(x)

Given normal at (3, 4) makes angle of 3π/4

⸫ Slope of the normal = tan 3π/4 = -1

47. If the tangent to the curve y = f(x) at the point (4, 3) passes through the point (0, 2), then find the value of ƒ' (4).

Ans: Slope of the tangent at (4, 3) on y = f(x) is ƒ'(4) since the tangent passes through (0, 2).

ƒ'(4) = slope the line joining the points (4, 3) and (0, 2).

48. At what point on the curve y = ex, the tangent line is parallel to y = 2x?

Ans: 

⸫ the point is (loge 2,2)

49.   Find the point at which f(x) is maximum.

Ans: 

Hence, f(x) changes from positive to negative as x increases in a ngd. Of 1/4

⸫ ƒ has max at x = 1 / 4

50. Find the absolute maximum of f(x) =

Ans: 

ƒ'(0) does not exist

⸫ x = 0 is a critical point

which is the absolute maximum at the end point x = 3.

Note that the absolute minimum is f(0) = 0, at the critical point x = 0

51. Determine the critical points of the function

Ans: 

g'(x) does not exist when x = 0 or 2; which are the end points and so not 1 critical points. So the only critical point is x = 1

[Note that initial points are interior points of the domain]

52. If an odd function ƒ has a local minimum at x = e, then what can you say about the value of ƒ at x = −e.

Ans: At x = -e, the function has an local maximum

53. Find the maximum value of the product of two numbers whose sum is 12.

Ans: Let x; y be the numbers x + y = 12 ⇒ y = 12 - x

P = xy = x(12 - x) = 12x - x2

⸫ Maximum value = 6⸱6 = 36