Page No 29.11:
Question 1:
Show that does not exist.
Answer:
Left hand limit:
Right hand limit:
Left hand limit ≠ Right hand limit
Page No 29.11:
Question 2:
Find k so that may exist, where
Answer:
Now, exists if the left hand limit is equal to the right hand limit.
⇒7 = 2 + k
k = 5
Page No 29.11:
Question 3:
Show that does not exist.
Answer:
Page No 29.11:
Question 4:
Let f(x) be a function defined by
Show that does not exist.
Answer:
Page No 29.11:
Question 5:
Let Prove that does not exist.
Answer:
Page No 29.11:
Question 6:
Let . Prove that does not exist.
Answer:
We have,
LHL of f(x) at x = 0
=
RHL of f(x) at x = 0
=
Clearly,
Hence, does not exist.
Page No 29.11:
Question 7:
Find , where
Answer:
We have,
LHL of f(x) at x = 3
=
RHL of f(x) at x = 3
=
Clearly,
Page No 29.11:
Question 8:
If Find and .
Answer:
(ii)
Page No 29.11:
Question 9:
Find , if
Answer:
Page No 29.11:
Question 10:
Evaluate , where
Answer:
Page No 29.11:
Question 11:
Let a1, a2, ..., an be fixed real numbers such that
f(x) = (x − a1) (x − a2) ... (x − an)
What is For a ≠ a1, a2, ..., an. Compute
Answer:
Page No 29.11:
Question 12:
Find
Answer:
Page No 29.11:
Question 13:
Evaluate the following one sided limits:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
Answer:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
Page No 29.12:
Question 14:
Show that does not exist.
Answer:
Page No 29.12:
Question 15:
Find:
Answer:
(i)
(ii)
(iii)
Page No 29.12:
Question 16:
Prove that for all a ∈ R. Also, prove that
Answer:
Page No 29.12:
Question 17:
Show that
Answer:
Page No 29.12:
Question 18:
Find Is it equal to
Answer:
Page No 29.12:
Question 19:
Find
Answer:
We know:
Page No 29.12:
Question 20:
Evaluate (if it exists), where
Answer:
Page No 29.12:
Question 21:
Show that does not exist.
Answer:
= – (An oscillating number that oscillates between –1 and 1)
Page No 29.12:
Question 22:
Let and if , find the value of k.
Answer:
We have,
It is given that,
Hence, the value of k is 6.
Page No 29.18:
Question 1:
Answer:
Page No 29.18:
Question 2:
Answer:
Page No 29.18:
Question 3:
Answer:
Page No 29.18:
Answer:
Page No 29.18:
Question 5:
Answer:
Page No 29.18:
Question 6:
Answer:
Page No 29.18:
Question 7:
Answer:
Page No 29.18:
Answer:
f(x) = 9 is a constant function.
Its value does not depend on x.
Page No 29.18:
Page No 29.18:
Question 10:
Answer:
Page No 29.18:
Question 11:
Answer:
Page No 29.18:
Question 12:
Answer:
Page No 29.18:
Question 13:
Answer:
Page No 29.18:
Question 14:
Answer:
Page No 29.23:
Question 1:
Answer:
Page No 29.23:
Question 2:
Answer:
Page No 29.23:
Question 3:
Answer:
Page No 29.23:
Question 4:
Answer:
Page No 29.23:
Question 5:
Answer:
Page No 29.23:
Question 6:
Answer:
Page No 29.23:
Question 7:
Answer:
Page No 29.23:
Question 8:
Answer:
Page No 29.23:
Question 9:
Answer:
Page No 29.23:
Question 10:
Answer:
Page No 29.23:
Question 11:
Answer:
Page No 29.23:
Question 12:
Answer:
Page No 29.23:
Question 13:
Answer:
Page No 29.23:
Question 14:
Answer:
Page No 29.23:
Question 15:
Answer:
Page No 29.23:
Question 16:
Answer:
Page No 29.23:
Question 17:
Answer:
Page No 29.23:
Question 18:
Answer:
Page No 29.23:
Question 19:
Answer:
Page No 29.23:
Question 20:
Answer:
Page No 29.23:
Question 21:
Answer:
Page No 29.23:
Question 22:
Answer:
Page No 29.23:
Question 23:
Answer:
Page No 29.23:
Question 24:
Answer:
Page No 29.23:
Question 25:
Answer:
p(x) = x4 3x3 + 2
p(1) = 1 3 + 2
= 0
Now, is a factor of p(x).
q(x) = x3 5x2 + 3x + 1
q(1) = 1 5 + 3 + 1
= 0
is a factor of q(x).
Page No 29.23:
Question 26:
Answer:
It is of the form
Let p(x) = x3 + 3x2 9x 2
p(2) = 8 + 12 18 2
= 0
Now, is a factor of p(x).
Let q(x) = x3 – x – 6
q(2) = 8 – 2 – 6
= 0
is a factor of q(x).
Page No 29.23:
Question 27:
Answer:
Page No 29.23:
Question 28:
Answer:
Page No 29.23:
Question 29:
Answer:
Let p(x) = x3 + x2 + 4x + 12
p(–2) = 0
Thus, x = –2 is the root of p(x).
Now, is a factor of p(x).
p(x) = x3 + x2 + 4x + 12
= (x + 2)(x2 – x + 6)
Let q(x) = x3 – 3x + 2
q = 8 + 6 + 2
= 0
Thus, x = 2 is the root of q(x).
Now, is a factor of q(x).
q(x) = (x + 2)(x2 – 2x + 1)
Page No 29.23:
Question 30:
Answer:
Let p(x) = x3 + 3x2 6x + 2
p(1) = 1 + 3 6 + 2
= 0
Now, is a factor of p(x).
p(x) = (x 1)(x2 + 4x 2)
q(x) = x3 + 3x2 3x + 2
q(1) = 1 + 3 3 1
= 0
is a factor of p(x).
Page No 29.24:
Question 31:
Answer:
Page No 29.24:
Question 32:
Answer:
Page No 29.24:
Question 33:
Answer:
Page No 29.24:
Question 34:
Evaluate the following limits:
Answer:
When x = 1, the expression assumes the form . So, (x − 1) is a factor of numerator and denominator.
Using long division method, we get
and
Page No 29.28:
Question 1:
Answer:
When x = 0, the expression takes the form .
Rationalising the numerator:
Page No 29.28:
Question 2:
Answer:
When x = 0, then the expression becomes .
Rationalising the denominator:
=
=
=
=
Page No 29.28:
Question 3:
Answer:
On putting x = 0 in the expression , it becomes
Rationalising the numerator:
​
=
=
=
Page No 29.28:
Question 4:
Answer:
It is of the form
Rationalising the numerator:
=
=
=
=
Page No 29.28:
Question 5:
Answer:
It is of the form
Rationalising the numerator:
Page No 29.28:
Question 6:
Answer:
It is of form
Rationalising the denominator:
=
=
=
=
=
=
Page No 29.28:
Question 7:
Answer:
It is of the form .
Rationalising the denominator:
=
=
=
=
= 1
Page No 29.28:
Question 8:
Answer:
It is of the form
Rationalising the numerator:
=
=
=
=
= 2
Page No 29.28:
Question 9:
Answer:
It is of the form .
Rationalising the denominator:
=
=
=
=
=
= 2
Page No 29.28:
Question 10:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
Page No 29.28:
Question 11:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
=
= 1
Page No 29.28:
Question 12:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
Page No 29.28:
Question 13:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
=
=
Page No 29.28:
Question 14:
Answer:
It is of the form .
⇒
=
=
=
Page No 29.28:
Question 15:
Answer:
It is of the form .
Rationalising the numerator and the denominator:
=
=
=
=
=
=
Page No 29.28:
Question 16:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
Page No 29.28:
Question 17:
Answer:
It is of the form .
Rationalising the denominator:
=
=
=
=
=
= 5
Page No 29.28:
Question 18:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
=
Page No 29.28:
Question 19:
Answer:
It is of the from .
Rationalising the numerator:
=
=
=
=
=
Page No 29.28:
Question 20:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
=
=
Page No 29.29:
Question 21:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
= 0
Page No 29.29:
Question 22:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
=
=
Page No 29.29:
Question 23:
Answer:
=
=
=
=
=
Page No 29.29:
Question 24:
Answer:
=
=
=
=
Page No 29.29:
Question 25:
Answer:
It is of the form .
Rationalising the numerator:
Page No 29.29:
Question 26:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
=
Page No 29.29:
Question 27:
Answer:
It is of the form .
Rationalising the numerator:
Page No 29.29:
Question 28:
Answer:
It is of the form .
⇒
=
=
=
=
=
=
=
Page No 29.29:
Question 29:
Answer:
It is of the form .
Rationalising the numerator and the denominator:
Page No 29.29:
Question 30:
Answer:
It is of the form .
=
=
=
= 1 (1 + 1 + 1)
= 3
Page No 29.29:
Question 31:
Answer:
It is of the form .
Rationalising the numerator:
=
=
=
=
Page No 29.29:
Question 32:
Answer:
=
=
=
Rationalising the numerator:
=
=
=
=
=
Page No 29.29:
Question 33:
Answer:
=
=
=
Rationalising the numerator:
=
=
=
=
=
=
=
=
Page No 29.29:
Question 34:
Answer:
=
=
=
Rationalising the numerator:
=
=
=
=
=
=
=
=
=
Page No 29.33:
Question 1:
Answer:
Let y = x + 2 and b = a + 2.
When x → a, then x + 2 ​→ a + 2.
⇒ y ​→ b
Page No 29.33:
Question 2:
Answer:
Let y = x + 2 and b = a + 2.
When x ​→ a and x + 2 ​→ a + 2.
y ​→ b
Page No 29.33:
Question 3:
Answer:
Let y = 1 + x
When x ​→ 0, then 1 + x → 1.
y ​→ 1
Page No 29.33:
Question 4:
Answer:
Page No 29.33:
Question 5:
Answer:
Page No 29.33:
Question 6:
Answer:
Let y = 2x
Page No 29.33:
Question 7:
Answer:
Page No 29.33:
Question 8:
Answer:
Page No 29.33:
Question 9:
Answer:
Page No 29.33:
Question 10:
Answer:
Page No 29.33:
Question 11:
Answer:
Page No 29.33:
Question 12:
If find the value of n.
Answer:
⇒ x(3)n – 1 = 108
⇒ x(3)n – 1 = 4 × 33
On comparing LHS and RHS, we observe that x is equal to 4.
Page No 29.33:
Question 13:
If find all possible values of a.
Answer:
Page No 29.33:
Question 14:
If find all possible values of a.
Answer:
Page No 29.33:
Question 15:
If find all possible values of a.
Answer:
Page No 29.33:
Question 16:
If find all possible values of a.
Answer:
Page No 29.38:
Question 1:
Answer:
Page No 29.38:
Question 2:
Answer:
Page No 29.38:
Question 3:
Answer:
Page No 29.38:
Question 4:
Answer:
It is of the form ∞ - ∞.
Rationalising the numerator:
Page No 29.38:
Answer:
It is of the form ∞–​​∞.
On rationalising, we get:
Page No 29.38:
Question 6:
Answer:
Page No 29.38:
Question 7:
Answer:
Page No 29.38:
Question 8:
Answer:
Page No 29.39:
Question 9:
Answer:
Page No 29.39:
Question 10:
Answer:
Page No 29.39:
Question 11:
Answer:
Dividing the numerator and the denominator by n:
Page No 29.39:
Question 12:
Answer:
Page No 29.39:
Question 13:
Answer:
Page No 29.39:
Question 14:
Answer:
Page No 29.39:
Question 15:
Answer:
Page No 29.39:
Question 16:
Answer:
Page No 29.39:
Question 17:
Answer:
Dividing the numerator and the denominator by n4:
Page No 29.39:
Question 18:
Answer:
Dividing the numerator and the denominator by :
Page No 29.39:
Question 19:
Answer:
Page No 29.39:
Question 20:
, where a is a non-zero real number.
Answer:
Dividing the numerator and the denominator by x4:
Page No 29.39:
Question 21:
and then prove that f(−2) = f(2) = 1
Answer:
Dividing the numerator and the denominator by x2:
Page No 29.39:
Question 22:
Show that
Answer:
Rationalising the numerator:
Page No 29.39:
Question 23:
Answer:
Let x =m
When n → – ∞, then m → ∞.
Dividing the numerator and the denominator by m:
Page No 29.39:
Question 24:
Answer:
Let x = –m
When x → –∞, then m → ∞.
Dividing the numerator and the denominator by m:
Page No 29.39:
Question 25:
Evaluate:
Answer:
Consider the identity
.....(1)
Putting k = 1, 2, 3,..., n in (1) and then adding the equations, we have
This expression on further simplification gives
Page No 29.39:
Question 26:
Evaluate:
Answer:
Page No 29.49:
Question 1:
Answer:
=
=
=
Page No 29.49:
Question 2:
Answer:
We know that .
Page No 29.49:
Question 3:
Answer:
Let
⇒
= 1
Page No 29.49:
Question 4:
Answer:
=
=
=
=
Page No 29.50:
Question 5:
Answer:
=
=
= 1 × 3
= 3
Page No 29.50:
Question 6:
Answer:
Page No 29.50:
Question 7:
Answer:
Page No 29.50:
Question 8:
Answer:
Page No 29.50:
Question 9:
Answer:
It is of the form .
Page No 29.50:
Question 10:
Answer:
It is of the form .
Dividing the numerator and the denominator by x:
Page No 29.50:
Question 11:
Answer:
It is of the form .
Page No 29.50:
Question 12:
Answer:
Page No 29.50:
Question 13:
Answer:
Page No 29.50:
Question 14:
Answer:
Page No 29.50:
Question 15:
Answer:
Page No 29.50:
Question 16:
Answer:
Page No 29.50:
Question 17:
Answer:
Page No 29.50:
Question 18:
Answer:
Page No 29.50:
Question 19:
Answer:
Dividing the numerator and the denominator by x, we get:
Page No 29.50:
Question 20:
Answer:
Dividing the numerator and the denominator by x, we get:
Page No 29.50:
Question 21:
Answer:
Dividing the numerator and the denominator by x:
Page No 29.50:
Question 22:
Answer:
Page No 29.50:
Question 23:
Answer:
Page No 29.50:
Question 24:
Answer:
Page No 29.50:
Question 25:
Answer:
Dividing the numerator and the denominator by x:
Page No 29.50:
Question 26:
Answer:
Page No 29.50:
Question 27:
Answer:
Page No 29.50:
Question 28:
Answer:
Page No 29.50:
Question 29:
Answer:
Page No 29.50:
Question 30:
Answer:
Page No 29.50:
Question 31:
Answer:
Page No 29.50:
Question 32:
Answer:
Page No 29.50:
Question 33:
Answer:
Page No 29.50:
Question 34:
Answer:
Page No 29.50:
Question 35:
Answer:
Page No 29.50:
Question 36:
Answer:
Page No 29.50:
Question 37:
Answer:
Page No 29.50:
Question 38:
Answer:
Page No 29.50:
Question 39:
Answer:
Page No 29.50:
Question 40:
Answer:
Page No 29.50:
Question 41:
Answer:
Page No 29.50:
Question 42:
Answer:
Page No 29.51:
Question 43:
Answer:
Page No 29.51:
Question 44:
Answer:
Page No 29.51:
Question 45:
Answer:
Page No 29.51:
Question 46:
Answer:
Page No 29.51:
Question 47:
Answer:
Page No 29.51:
Question 48:
Answer:
Page No 29.51:
Question 49:
Answer:
Page No 29.51:
Question 50:
Answer:
Page No 29.51:
Question 51:
Evaluate the following limits:
Answer:
Page No 29.51:
Question 52:
Answer:
Page No 29.51:
Question 53:
Answer:
Page No 29.51:
Question 54:
Answer:
Page No 29.51:
Question 55:
Answer:
Page No 29.51:
Question 56:
Answer:
Page No 29.51:
Question 57:
Answer:
Page No 29.51:
Question 58:
Answer:
Page No 29.51:
Question 59:
Answer:
Page No 29.51:
Question 60:
Evaluate the following limits:
Answer:
Disclaimer: There is some mistake or some misprinting in the question.
Page No 29.51:
Question 61:
Evaluate the following limits:
Answer:
Page No 29.51:
Question 62:
Evaluate the following limits:
Answer:
Page No 29.51:
Question 63:
If find k.
Answer:
Page No 29.62:
Question 1:
Answer:
Page No 29.62:
Question 2:
Answer:
Page No 29.62:
Question 3:
Answer:
Page No 29.62:
Question 4:
Evaluate the following limits:
Answer:
Page No 29.62:
Question 5:
Answer:
Page No 29.62:
Question 6:
Answer:
Page No 29.62:
Question 7:
Answer:
Page No 29.62:
Question 8:
Answer:
Page No 29.62:
Question 9:
Answer:
Page No 29.62:
Question 10:
Answer:
Page No 29.62:
Question 11:
Answer:
Page No 29.62:
Question 12:
Answer:
Page No 29.62:
Question 13:
Answer:
Page No 29.62:
Question 14:
Answer:
Page No 29.62:
Question 15:
Answer:
Page No 29.62:
Question 16:
Answer:
Page No 29.62:
Question 17:
Answer:
Page No 29.62:
Question 18:
Answer:
Page No 29.62:
Question 19:
where f(x) = sin 2x
Answer:
Page No 29.62:
Question 20:
Answer:
Page No 29.62:
Question 21:
Answer:
Page No 29.62:
Question 22:
Answer:
Page No 29.62:
Question 23:
Answer:
Let y = x – 1
If x → 1, then y → 0.
Page No 29.62:
Question 24:
Answer:
Page No 29.62:
Question 25:
Answer:
Page No 29.62:
Question 26:
Answer:
Page No 29.62:
Question 27:
Answer:
Page No 29.62:
Question 28:
Answer:
Page No 29.63:
Question 29:
Answer:
Page No 29.63:
Question 30:
Answer:
Rationalising the denominator, we get:
Page No 29.63:
Question 31:
Answer:
Page No 29.63:
Question 32:
Answer:
Page No 29.63:
Question 33:
Answer:
Page No 29.63:
Question 34:
Answer:
Page No 29.63:
Question 35:
Answer:
Page No 29.63:
Question 36:
Answer:
Page No 29.63:
Question 37:
Answer:
Page No 29.63:
Question 38:
Evaluate the following limits:
[NCERT EXEMPLAR]
Answer:
Put
When
Page No 29.65:
Question 1:
Answer:
Page No 29.65:
Question 2:
Answer:
Page No 29.65:
Question 3:
Answer:
Page No 29.65:
Question 4:
Answer:
Page No 29.65:
Question 5:
Answer:
Let x = π h
when x → π, then h → 0
Page No 29.65:
Question 6:
Answer:
Page No 29.71:
Question 1:
Answer:
Rationalising the denominator, we get:
Page No 29.71:
Question 2:
Answer:
Dividing the numerator and the denominator by x:
Page No 29.71:
Question 3:
Answer:
Page No 29.71:
Question 4:
Answer:
Page No 29.71:
Question 5:
Answer:
Page No 29.71:
Question 6:
Answer:
Page No 29.71:
Question 7:
Answer:
Page No 29.71:
Question 8:
Answer:
Page No 29.71:
Question 9:
Answer:
Page No 29.71:
Question 10:
Answer:
Let x = 2 + h
x → 2
∴ h → 0
Page No 29.71:
Question 11:
Answer:
Page No 29.71:
Question 12:
Answer:
Page No 29.71:
Question 13:
Answer:
Page No 29.71:
Question 14:
Answer:
Page No 29.71:
Question 15:
Answer:
Page No 29.71:
Question 16:
Answer:
Dividing the numerator and the denominator by x:
Page No 29.71:
Question 17:
Answer:
x → 0
∴ sin x → 0
Let y=sin x
x → 0
∴ y → 0
Page No 29.71:
Question 18:
Answer:
Page No 29.71:
Question 19:
Answer:
Page No 29.71:
Question 20:
Answer:
Page No 29.71:
Question 21:
Answer:
Page No 29.71:
Question 22:
Answer:
Page No 29.71:
Question 23:
Answer:
Page No 29.71:
Question 24:
Answer: