RBSE Solutions for Class 11 Maths Chapter 10 Limits and Derivatives

RBSE Solutions for Class 11 Maths Chapter 10 Limits and Derivatives

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:

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:

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.

Rajasthan Board RBSE Class 11 Maths Chapter 10 Limits and Derivatives Miscellaneous Exercise

Question 1.
The value of
 is:
(A) 1/3
(B) -1/3
(C) 1
(D) – 1
Solution:

Hence, option (B) is correct.

Question 2.
The value of

(A) 0
(B) ∞
(C) 1
(D) – 1
Solution:

Hence, option (A) is correct.

Question 3.
The value of

(A) 2/3
(B) 1/3
(C) 1/2
(D) 3/2
Solution:

Hence, option (D) is correct.

Question 4.
The value of

(A) 3
(B) 2
(C) 1
(D) – 1
Solution:

Hence, option (C) is correct.

Question 5.
The value of 
(A) π/4
(B) π/2
(C) 0
(D) ∞
Solution:


Hence, option (A) is correct.

Question 6.
The value of 
(A) 0
(B) 1
(C) log_e (ab)
(D) log_e (a/b)
Solution:

Hence, option (D) is correct.

Question 7.
The value of 
(A) 0
(B) 1
(C) π/180
(D) π
Solution:

Hence, option (C) is correct.

Question 8.
The value of 
(A) 0
(B) 1/2
(C) -1/2
(D) -1
Solution:

Hence, option (B) is correct.

Question 9.
The value of 
(A) 0
(B) 81
(C) 4
(D) 1
Solution:

Hence, option (B) is correct.

Question 10.
The value of

(A) 0
(B) ∞
(C) – 1
(D) 1
Solution:


Hence, option (C) is correct.

Question 11.
If y is function of x, then derivative of y with respect to x is :

Solution:
y = ax² + bx + c (Let)

Hence, option (C) is correct.

Question 12.
Derivative of xⁿ is :
(A) x^{n – 1}
(B) (n – 1)x^{n – 2}
(C) nx^{n – 1}
(D) x^{n + 1}/n + 1
Solution:
Let y = xⁿ

Hence, option (C) is correct.

Question 13.
Derivative of 1√x is:

Solution:

Hence, option (B) is correct.

Question 14.
d/dx(5^x) is equal to :
(A) 5^x
(B) 10^x
(C) 10^x log_e 5
(D) 5^x log_e 5
Solution:
d/dx (5^x) = 5^x log_e5 ( ∵d/dx (a^x .log a)
Hence, option (D) is correct.

Question 15.
d/dx (log_a x) is equal to :

Solution:

Hence, option (A) is correct.

Question 16.
If f(x) = x³ + 6x² – 5 then f'(1) is equal to :
(A) 0
(B) 9
(C) 4
(D) 15
Solution:
f(x) = x³ + 6x² – 5
⇒ f'(x)= 3x² + 12x – 0
⇒ f'(x)= 3x² + 12x
⇒ f'(1)= 3(1)² + 12(1)
⇒ f'(1)= 3 + 12 = 15
Hence, option (D) is correct

Question 17.
Derivative of sec x° is :

Solution:

Hence, option (C) is correct.

Question 18.
Derivative of log_x a is :

Solution:


Hence, option (B) is correct.

Question 19.
 and f'(0) = 0, then value of c is :
(A) 0
(B) 1
(C) 2
(D) -2
Solution:

Hence, option (D) is correct.

Question 20.
Derivative of log_e√x is :

Solution:

Hence, option (A) is correct

Question 21.

then find the value of a, b and c.
Solution:

Question 22.
Evaluate

Solution:

Question 23.
Evaluate 
Solution:

Question 24.
Evaluate

Solution:

Question 25.
Evaluate 
Solution:

Question 26.
Evaluate 
Solution:

Question 27.
Evaluate
Solution:

Question 28.

Solution:

Question 29.
If y = x³. e^x sin x, then find dy/dx.
Solution:
Given, y = x³. e^x sin x
On differentiating

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