RBSE Solutions for Class 11 Maths Chapter 10

Rajasthan Board RBSE Class 11 Maths Chapter 10 Limits and Derivatives Ex 10.3

Question 1.
Find the derivative of x² – 2 at x = 10.
Solution:
Let f(x) = x² – 2

Question 2.
Find the derivative of 49x at x = 50.
Solution:
Let f(x) = 49x

Question 3.
Find the derivative of the following function from first principle:

Solution:
(i) Let y = x³ – 16
Again, let y + δy = (x + δx)³ – 16
⇒ δy = (x + δx)³ – 16 – y
⇒ δy = (x + δx)³ – 16 – x³ + 16
⇒ δy = (x + δx)³ – x³

(ii) Let y = (x – 1) (x – 2) = x² – 3x + 2
Again, let y + δy = (x + δx)2 – 3(x + δx) + 2
⇒ δy = (x + δx)² – 3(x + δx) + 2 – x² + 3x – 2

Question 4.
For the function

Prove that f'(1) = 100 f'(0).
Solution:

Then, putting 1 and 0 in place of x.
f'(1)= (1⁹⁹ + 1⁹⁸ + … + 1)+ 1
= 1 + 1 + 1 + …+ 99 term + 1
= 99+ 1 = 100 and f'(0) = 1
Hence, f'(1)= 100
∵ f'(1) = 100 f'(0) Hence Proved.

Question 5.
For any constant real number a, find the derivative of:
xⁿ + ax^{n – 1} + a²x^{n – 2} + … + a^{n – 1} x + aⁿ
Solution:
Let y =f(x) = xⁿ + ax^{n – 1} + a²x^{n – 2} + …… + a^{n – 1}x + aⁿ
Then, derivative of f(x),

Question 6.
For some constant a and b, find the derivative of the following functions :

Solution:
(i) Let y = f(x) = (x – a) (x – b) or y = f(x) = x² – (a + b)x + ab
Then, derivative of given function

Hence, derivative of given function (x – a) (x – b)
= 2x – a – b
(ii) Let y = f(x) = (ax² + b)²
or y = f(x) = a²x⁴+ 2abx² + b²
Then, derivative of given function


= 4a²x³ + 4abx = 4ax(ax² + b)
Hence, derivative of given function (ax² + b²)²
= 4a²x³ + 4abx or 4ax(ax² + b)

We know that if any function is in the form of fraction, then its derivative

Question 7.
For any constant a, find the derivative of
.
Solution:

Question 8.
Find the derivative of the following 3

Solution:


(ii) Let y = f(x) = (5x³ + 3x – 1) (x – 1)
The given function is product of two function.
Then, derivative of product of two functions

= 20x³ – 15x² + 6x – 4
Hence, derivative of given function = 20x³ – 15x² + 6x – 4.

(iii) Let y = x⁵(3 – 6x^-9)
Then, derivative of given function

Hence, derivative of given function

We can also solve this equation by product rule of derivative.

Question 9.
Find the derivative of cos x by first principle.
Solution:
Let
f(x) = cos x, then f(x + h) = cos(x + h)
Then

Question 10.
Find the derivatives of the following :
(i) sin x cos x
(ii) sec x
(iii) cosec x
(iv) 3 cot x + 5 cosec x
(v) 5 sin x – 6 cos x + 7
Solution:
(i) Let f(x) = sin x. cos x, which is product of two functions.
So, formula of derivative of product of two functions.

= – sin² x + cos² x
= cos² x – sin² x
= cos 2x ( ∵ cos² x – sin² x = cos²x)
Hence, derivative of given function sin x cos x = cos 2x

(ii) Let f(x) = sec x

Hence, derivative of the given function sec x = sec x tan x

(iii) Let f(x) = cosec x
Then, derivative of f(x)

= – cosec x cot x
Hence, derivative of the given function cosec x
= – cosec x cot x

(iv) Let f(x) = 3 cot x + 5 cosec x

Hence, derivative of the given function 3 cot x + 5 cosec x is – 3 cosec² x – 5 cosec x cot x

(v) Let f(x) = 5 sin x – 6 cos x + 7

Hence, derivative of the given function 5 sin x – 6 cos x + 7 is 5 cos x + 6 sin x.

RBSE Solutions for Class 11 Maths

Rajasthan Board RBSE Class 11 Maths Chapter 10 Limits and Derivatives Ex 10.2

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

RBSE Solutions for Class 11 Maths

Rajasthan Board RBSE Class 11 Maths Chapter 10 Limits and Derivatives Ex 10.1

Question 1.
Show that left and right limits of function
at x = 1 are equal and their value is 1.
Solution:

From equation (i) and (ii)
L.H.L. = R.H.L.
⇒ f(1 – 0) = f(1 + 0) = 1
Hence, left and right limits of the given function at x = 1 are equal and their value is 1.
Hence Proved.

Question 2.
Is limit of the function at x = 0 ?
Solution:

It is clear from equation (i) and (ii),
L.H.L. ≠ R.H.L.
⇒ f(0 – 0) ≠ x(0 + 0)
Hence, limit of function does not exist at x = 0.

Question 3.
Prove that at x = 0, limits of function f(x) = | x | + | x – 1| exists.
Solution:
f(x) = | x | + | x – 1 |
Right limit at x – 0
R.H.L. = lim_x→0⁺ + f(x) = f(0 + 0)
= lim_h→0 f(0 + h)
= lim_h→0 |0 + h| + |0 + h – 1 | (h> 0)
= lim_h→0 | h | + | h – 1 |
= 0 + | 0 – 1 | = 1 …(i)
Left limit at x = 0
L.H.L. = lim_h→0^– + f(x) = f(0 – 0)
= lim_h→0 f(0 – h)
= lim_h→0 | 0 – h | + | 0 – h – 1 |
= lim_h→0 | 0 – h | + | -(h – 1) |
= | -0 | + | -(0+ 1) |
= 0 + 1 = 1 …(ii)
It is clear from equation (i) and (ii),
L.H.L. = R.H.L.
⇒ f(0 – 0) = f(0 + 0) = 1
Hence, limit of function exists at x = 0. Hence Proved.

Question 4.
Prove that at x = 2, limits of function does not exists.

Solution:

Right limit at x = 2
R.H.L. = lim_x→0 + f(x)
= f(2 + 0) = lim_h→0 f(2 + h)
= lim_h→0[(2 + h)² + (2 + h) + 1]
= lim_h→0 [4 + h² + 4h + 2 + h + 1]
= lim_h→0 [h² + 5h + 7]
= 0² + 5(0) + 7 = 7 ….(i)
Left limit at x = 2
L.H.L. = lim_x→0 ^– f(x)
= f(2 – 0) = lim_h→0 f(2 – h)
= lim_h→0 [2 – h]
= 2 – 0 = 2 .
From equation (i) and (ii),
L.H.L. ≠ R.H.L.
⇒ f(2 – 0) ≠ f(2 + 0)
Hence, limit of function does not exist at x = 2.

Question 5.
Find the left and right limit of function f(x) = x cos ( 1/x) at x = 0.
Solution:

RBSE Solutions for Class 11 Maths