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 +
kk = 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:
Evaluate the following limits:
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 + 3
x2 9
x 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:
â
=
=
=
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 â→
a + 2.
⇒
y â→
b
Page No 29.33:
Question 2:
Answer:
Let
y =
x + 2 and
b =
a + 2.
When
x â→
a and
x + 2 â→
a + 2.
y â→
b
Page No 29.33:
Question 3:
Answer:
Let
y = 1 +
x
When
x â→ 0, then 1 +
x → 1.
y â→ 1
Page No 29.33:
Question 4:
Answer:
Page No 29.33:
Question 5:
Answer:
Page No 29.33:
Question 6:
Answer:
Let
y = 2
x
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 × 3
3
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 ∞–ââ∞.
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:
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 2
x
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
xx → 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:
Page No 29.71:
Question 25:
Answer:
Dividing the numerator and the denominator by
x2:
Page No 29.71:
Question 26:
Answer:
Rationalising the numerator:
Page No 29.71:
Question 27:
Answer:
Page No 29.71:
Question 28:
Answer:
Page No 29.71:
Question 29:
Answer:
Dividing numerator and the denominator by x, we get:
â
Left hand limit ≠ Right hand limit
Thus, limit does not exist.
Page No 29.71:
Question 30:
Answer:
Page No 29.72:
Question 31:
Answer:
Page No 29.72:
Question 32:
Answer:
Page No 29.72:
Question 33:
Answer:
Page No 29.72:
Question 34:
Answer:
Page No 29.72:
Question 35:
Answer:
Page No 29.72:
Question 36:
Answer:
If
x → 0, then tan
x → 0.
Let y = tan
x
Page No 29.72:
Question 37:
Answer:
Page No 29.72:
Question 38:
Answer:
Page No 29.72:
Question 39:
Answer:
Page No 29.72:
Question 40:
Answer:
Page No 29.72:
Question 41:
Answer:
Disclaimer: The answer given in the textbook is incorrect.
Page No 29.72:
Question 42:
Answer:
Dividing the numerator and the denominator by
x2:
Page No 29.72:
Question 43:
Answer:
Page No 29.76:
Answer:
Page No 29.76:
Question 2:
Answer:
Page No 29.76:
Question 3:
Answer:
Page No 29.76:
Question 4:
Answer:
Page No 29.77:
Question 5:
Answer:
Page No 29.77:
Question 6:
Answer:
Disclaimer: The solution given in the book is incorrect. However, the solution given here has been created according to the question given in the book.
Page No 29.77:
Question 7:
Answer:
Disclaimer: The solution given in the book is incorrect. However, the solution given here has been created according to the question given in the book.
Page No 29.77:
Question 8:
Answer:
Page No 29.77:
Question 9:
Answer:
Page No 29.77:
Question 10:
Answer:
Page No 29.77:
Question 1:
is equal to
(a) 1
(b) 1/2
(c) 1/3
(d) 0
Answer:
(c) 1/3
Page No 29.77:
Question 2:
is equal to
(a) 0
(b) 1
(c) 1/2
(d) 2
Answer:
(d) 2
Page No 29.77:
Question 3:
If then
(a) 1
(b) 0
(c) −1
(d) does not exist
Answer:
(b) 0
Page No 29.77:
Question 4:
(a) 0(b
) 1
(c
) 2
(d) 4
Answer:
(a) 0
Page No 29.77:
Question 5:
(a) 10/3(b
) 3/10(c
) 6/5(d) 5/6
Answer:
(a) 10/3
Page No 29.77:
Question 6:
(a) 0(b
) 1(c
) 4
(d) not defined
Answer:
(b) 1
Page No 29.77:
Question 7:
is equal to
(a) 0
(b
) −1/2(c
) 1/2(d) none of these
Answer:
(b)
Page No 29.77:
Question 8:
equals
(a) 0
(b
) ∞(c
) 1
(d) does not exist
Answer:
(a)
Page No 29.77:
Question 9:
is equal to
(a) 1
(b
) π(c
) x(d) π/180
Answer:
(d)
Page No 29.78:
Question 10:
is equal to
â(a) 1
(b
) −1
(c
) 0
(d) does not exist
Answer:
(d) does not exist
LHL ≠ RHL
Therefore, limit does not exist.
Page No 29.78:
Question 11:
is equal at
â(a) nan(b
) nan−1(c
) na(d) 1
Answer:
(b
) nan−1
Page No 29.78:
Question 12:
is equal to
(a) (b
) (c
) (d) 1
Answer:
(b) 1/2
The correct answer is B.
Page No 29.78:
Question 13:
is equal to
(a) 1
(b
) 0
(c
) −1
(d) 1/2
Answer:
(d) 1/2
Hence, the correct answer is d.
Page No 29.78:
Question 14:
is equal to
(a) 2/3
(b
) 4/3
(c
) (d) −4/3
Answer:
(d) 4/3
Page No 29.78:
Question 15:
(a) −1/12
(b
) −4/3(c
) −16/3(d) −1/48
Answer:
(d) −1/48
Page No 29.78:
Question 16:
is equal to
(a) 0
(b
) 1/2
(c
) 1/9
(d) 2
Answer:
(b) 1/2
Page No 29.78:
Question 17:
is equal to
(a) −π(b
) π(c
) (d)
Answer:
(a) −π
Page No 29.78:
Question 18:
If then n equal
(a) 10
(b) 100
(c) 150
(d) none of these
Answer:
(b
) 100
Page No 29.78:
Question 19:
The value of is
(a) −1
(b) 1
(c) 2
(d) none of these
Answer:
(c
) 2
Page No 29.78:
Question 20:
is equal to
(a) (b
) 2(c
) 0
(d) 1
Answer:
(a)
Page No 29.79:
Question 21:
is equal to
(a) (b
) (c
) (d)
Answer:
(c
)
Page No 29.79:
Question 22:
is real to
(a) (b
) (c
) (d)
Answer:
(b
)
This series is an AGP of the form given below:
S = 1 + 2
r + 3
r2........
nrn–1
Page No 29.79:
Question 23:
is equal to
(a) (b
) (c
) 1(d) −1
Answer:
(b
)
Dividing the numerator and the denominator by
n:
Page No 29.79:
Question 24:
If then equals
(a) 1
(b) 0
(c) −1
(d) none of these
Answer:
(b
) 0
LHL:
Let
x = 0 –
h, where
h → 0.
= 0 × The oscillating number between –1 and 1
= 0
RHL
Let
x = 0 +
h, where
h → 0.
= 0 × The oscillating number between –1 and 1
= 0
LHL = RHL = 0
Page No 29.79:
Question 25:
is equal to
(a) (b
) 0(c
) 2(d) 1
Answer:
(b) 0
Page No 29.79:
Question 26:
is equal to
(a) (b
) (c
) (d) none of these
Answer:
(a)
Page No 29.79:
Question 27:
is equal to
(a) (b
) (c
) (d)
Answer:
Disclaimer: The question in the book has some error, which has been resolved here and the solution is created accordingly.
Page No 29.79:
Question 28:
is equal to
(a) b(b
) a(c
) a log
eb(d) b log
ea
Answer:
(a) b
Page No 29.79:
Question 29:
is equal to
(a)
(b)
(c)
(d)
Answer:
(c)
Page No 29.79:
Question 30:
If α is a repeated root of ax2 + bx + c = 0, then is
(a) a
(b) b
(c) c
(d) 0
Answer:
(a)
Page No 29.80:
Question 31:
The value of is
(a) a
(b)
(c) −a
(d)
Answer:
(d)
Page No 29.80:
Question 32:
The value of is
(a) 1
(b) 2
(c) −1
(d) −2
Answer:
(b) 2
Page No 29.80:
Question 33:
is equal to
(a) 1
(b) −1
(c)
(d)
Answer:
(c)
Page No 29.80:
Question 34:
The value of is
(a) 2
(b) −1
(c) 1
(d) 0
Answer:
(d) 0
Page No 29.80:
Question 35:
The value of is
(a) 0
(b) 1
(c) −1
(d) none of these
Answer:
(a) 0
Page No 29.80:
Question 36:
The value of is
(a) 0
(b) −1
(c) 1
(d) none of these
Answer:
(c) 1
Page No 29.80:
Question 37:
The value of is
(a) 10
(b) 100
(c) 1010
(d) none of these
Answer:
(b) 100
Page No 29.80:
Question 38:
The value of is
(a) 1/2
(b) 1
(c) −1
(d) −1/2
Answer:
(d) −1/2
Page No 29.80:
Question 39:
, where [.] is the greatest integer function, is equal to
(a) 1 (b) 2 (c) 0 (d) does not exist
Answer:
We have,
(k − 1 < k − h < k ⇒ [k − h] = k − 1, k ∈ Z)
Also,
Thus, does not exist.
Hence, the correct answer is option (d).
Page No 29.80:
Question 40:
is equal to
(a) 1 (b) −1 (c) 0 (d) does not exist
Answer:
We know that,
Now, for all x ≥ 0 (however, x may large be), .
Hence, the correct answer is option (a).
Page No 29.80:
Question 41:
is
(a) 1 (b) −1 (c) 0 (d) does not exist
Answer:
We have,
Now,
Clearly,
∴ does not exist.
Hence, the correct answer is option (d).
Note: Here, a small interval close to 0 is only considered.
Page No 29.80:
Question 42:
If , where [.] denotes the greatest integer function, then is equal to
(a) 1 (b) 0 (c) −1 (d) does not exist
Answer:
We have,
Now,
Clearly,
Thus, does not exist.
Hence, the correct answer is option (d).
Page No 29.80:
Question 43:
(a) 1
(b) 2
(c) –1
(d) –2
Answer:
Page No 29.81:
Question 44:
(a) 2
(b)
(c)
(d) 1
Answer:
Page No 29.81:
Question 45:
(a) n
(b) 1
(c) –n
(d) 0
Answer:
Hence, the correct answer is option A.
Page No 29.81:
Question 46:
â(a) 1
(b)
(c)
(d)
Answer:
Hence, the correct answer is option B.
Page No 29.81:
Question 47:
â(a)
(b)
(c)
(d) –1
Answer:
Hence, the correct answer is option A.
Page No 29.81:
Question 48:
â(a)
(b) 1
(c)
(d) 1
Answer:
Hence,the correct answer is option C.
Page No 29.81:
Question 49:
â(a) 3
(b) 1
(c) 2
(d)
Answer:
Hence, the correct answer is option C.
Page No 29.81:
Question 50:
â(a) 2
(b) 0
(c) 1
(d) –1
Answer:
Hence, the correct answer is option C.
Page No 29.81:
Question 51:
(c) 1
(d) none of these
Answer:
Hence, the correct answer is option B.
Page No 29.82:
Question 1:
If f(x) f(x) = _____________________________.
Answer:
Page No 29.82:
Question 2:
Answer:
Page No 29.82:
Question 3:
Answer:
Page No 29.82:
Question 4:
Answer:
Page No 29.82:
Question 5:
is equal to ______________________.
Answer:
Page No 29.82:
Question 6:
âis equal to ______________________.
Answer:
Page No 29.82:
Question 7:
Answer:
Page No 29.82:
Question 8:
ââis equal to _____________________.
Answer:
Page No 29.82:
Question 9:
is equal to __________________.
Answer:
Page No 29.82:
Question 10:
Let f (x) = , the quadratic equation whose roots are f (x) and f (x) is __________________.
Answer:
Page No 29.82:
Question 11:
is equal to _____________________.
Answer:
Page No 29.82:
Question 12:
is equal to _______________________.
Answer:
Page No 29.82:
Question 13:
is equal to_________________________.
Answer:
Page No 29.82:
Question 14:
________________________.
Answer:
Page No 29.82:
Question 15:
Answer:
Page No 29.82:
Question 16:
is equal to _____________________________.
Answer:
Page No 29.82:
Question 17:
The value of
Answer:
Page No 29.83:
Question 1:
Write the value of
Answer:
Page No 29.83:
Question 2:
Write the value of
Answer:
Page No 29.83:
Question 3:
Write the value of
Answer:
Disclaimer: The solution given in the book is incorrect. However, the solution given here has been created according to the question given in the book.
Page No 29.83:
Question 4:
Write the value of
Answer:
Page No 29.83:
Question 5:
Write the value of
Answer:
Page No 29.83:
Question 6:
Write the value of
Answer:
Page No 29.83:
Question 7:
Write the value of
Answer:
Page No 29.83:
Question 8:
Write the value of
Answer:
Page No 29.83:
Question 9:
Write the value of
Answer:
Page No 29.83:
Question 10:
Write the value of
Answer:
Page No 29.83:
Question 11:
Write the value of
Answer:
Page No 29.83:
Question 12:
Write the value of
Answer:
Page No 29.83:
Question 13:
Write the value of
Answer:
Page No 29.83:
Question 14:
Write the value of
Answer:
Page No 29.83:
Question 15:
Write the value of
Answer:
View NCERT Solutions for all chapters of Class 13