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NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE PDF Download

Differentiate the functions with respect to x in Exercises 1 to 8.
Q1: sin (x2 + 5)
Ans: 
Given y = sin(x2+5)
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
x2+5=u
du/dx = 2x
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE

Q2: cos (sin x)
Ans:
Let y = cos (sin x)
Differentiate both sides w.r.t x, we get
dy/dx = d/dx  (cos (sin x))=−sin(sinx)d/dx (sin x)(By chain rule)
= - sin (sin x) cos x = -cos x sin (sin x)

Q3: sin (ax + b)
Ans:
Let y=sin(ax+b)
Thus using chain rule,
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE

Q4: sec (tan(√x))
Ans:
Let y = sec (tan√x)
Differentiate both sides w.r.t. x, we get
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE

Q5: sin (ax+b)/cos (cx+d)
Ans: 
We have, f(x)=NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
Thus using quotient rule and chain rule simualtaneously,
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE

Q6: cos x3 . sin2 (x5)
Ans:
Let y = cos3 sin2 (x5)
Differentiate both sides w.r.t. x,
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
(Using chain rule)
= (cos x3)(2sin x5)(cos x5)(5x4)−sin2(x5)(sin x3)(3x2)

Q7: 2√cot(x2)  
Ans:

NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE 
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
(using sin 2θ=2 sin θ cos θ)

Q8: cos(√x)  
Ans: 
Compute the derivative:
As we know, the formula for the first principle,
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
We can write that
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
Hence the required derivative is NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE.

Q9: Prove that the function f given by f(x) = |x – 1|, x ∈ R is not differentiable at x = 1.
Ans: 
The given function is f(x) = |x – 1|, x ∈ R.
It is known that a function f is differentiable at point x=c in its domain if both
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE are finite and equal.
To check the differentiability of the function at x=1, 
Consider the left hand limit of f at  x = 1
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
Consider the right hand limit of f at x - 1
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
Since the left and right hand limits of f at x=1 are not equal, f is not differentiable at x=1. 

Q10: Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.
Ans:
f(x) = [x], 0<x<3
At x = 1
f(x) is differentiable at x = 1y if
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE 
∞ ≠ 0
∴ f(x) is not differentaible at x = 1
For x=2
NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE
∞ ≠ 0
∴ f(x) is not differentaible at x = 2.

The document NCERT Solutions Exercise 5.2: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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FAQs on NCERT Solutions Exercise 5.2: Continuity & Differentiability - Mathematics (Maths) Class 12 - JEE

1. What is the definition of continuity of a function?
Ans. Continuity of a function at a point means that the function exists at that point, the limit of the function exists at that point, and the value of the function at that point is equal to the limit of the function at that point.
2. How is continuity related to differentiability?
Ans. A function is differentiable at a point if it is continuous at that point. However, a function can be continuous at a point but not differentiable at that point.
3. Can a function be continuous but not differentiable?
Ans. Yes, it is possible for a function to be continuous at a point but not differentiable at that point. A common example is the absolute value function at x = 0.
4. How is the continuity of a function tested?
Ans. The continuity of a function can be tested using the following criteria: 1) The function exists at the point in question. 2) The limit of the function as it approaches the point exists. 3) The value of the function at the point is equal to the limit.
5. What is the importance of continuity and differentiability in calculus?
Ans. Continuity and differentiability are crucial concepts in calculus as they help us understand the behavior of functions at specific points. They enable us to analyze the rate of change of a function, the existence of local extrema, and the behavior of the function near critical points.
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