NCERT Solutions Exercise 5.3: Continuity & Differentiability

NCERT Solutions Exercise 5.3: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE PDF Download

Find dy/dx in the following:
Q1: 2x + 3y = sin x
Ans:
2x+3y=sinx
Differentiating both sides w.r.t. x, we obtain

Q2: 2x + 3y = sin y
Ans:
2x+3y = siny

Q3: ax + by2 = cos y
Ans:
ax+by= cos y
Differentiating w.r.t. x, we have

Q4: xy + y2 = tan x + y
Ans:
Given, xy+y2 = tan x+y
Differentiating both sides w.r.t. x, we get

Q5: x+ xy + y2 = 100
Ans:
x+ xy + y2 = 100
Differentiating both sides w.r.t x we get,

Q6: x3 + x2y + xy2 + y3 = 81
Ans:
Given equation: x3 + x2y + xy2 + y3 = 81
Differentiating the above equation wrt. x

Q7: sin2 y + cos xy = κ
Ans:
We have, sin2y+cosxy = k
Differentiating both sides with respect to x, we obtain

Using chain rule, we obtain

From (1), (2) and (3), we obtain

Q8: sin2 x + cos2 y = 1
Ans:
We have sin2 x + cos2 y = 1
Differentiating both sides w.r.t. x, we obtain

Q9: y = sin–1 (2x/(1+x2))
Ans:
y = sin–1 (2x/(1+x2))
Differentiating above equation w.r.t. x, we have

Q10: y = tan–1 ((3x-x3)/(1 - 3x2)), -(1/√3) < x < (1/√3)
Ans:

Put x = tanθ

Q11: y = cos-1((1-x2)(1+x2)), 0 < x <1
Ans:

Put x=tanθ

= cos−1cos(2θ) = 2θ = 2tan−1x

Q12: y = sin-1((1-x2)/(1+x2)) 0 < x < 1.
Ans:

Q13: y = cos-1(2x/(1+x2)), -1 < x <1.
Ans:

Q14: y = sin-1(2x√1-x2), -(1/√2) < x < (1/√2)
Ans:
L.H.S = sin−1(2x√1−x2)
Substituting x = sinθ

Now,
= sin−1(2sinθcosθ)
= sin−1(sin2θ)

⇒ sin−1(sin2θ)
= 2θ = 2sin−1x
Hence proved.

Q15: y = sec-1(1/(2x2-1)), 0 < x < (1/√2).
Ans:

Put x = cosθ ⇒ θ = cos−1x
Then we have,
y = cos−1(2cos2θ−1) = cos−1(cos2θ) = 2θ = 2cos−1x
.

The document NCERT Solutions Exercise 5.3: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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Mathematics (Maths) Class 12

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FAQs on NCERT Solutions Exercise 5.3: Continuity & Differentiability - Mathematics (Maths) Class 12 - JEE

 1. What is the definition of continuity in calculus?
Ans. Continuity in calculus refers to a function where there are no holes, jumps, or breaks in its graph. A function is said to be continuous at a point if the limit of the function exists at that point and is equal to the value of the function at that point.
 2. How can we determine if a function is continuous at a given point?
Ans. To determine if a function is continuous at a given point, you need to check if the limit of the function as it approaches that point exists and is equal to the value of the function at that point. If both conditions are met, the function is continuous at that point.
 3. What is the difference between continuity and differentiability?
Ans. Continuity refers to the smoothness of a function at a particular point, while differentiability refers to the existence of the derivative of a function at that point. A function can be continuous but not differentiable, but if a function is differentiable at a point, it must also be continuous at that point.
 4. Can a function be differentiable but not continuous?
Ans. No, a function cannot be differentiable at a point where it is not continuous. Differentiability implies continuity, so if a function is differentiable at a point, it must also be continuous at that point.
 5. How can we determine if a function is differentiable at a given point?
Ans. To determine if a function is differentiable at a given point, you need to check if the derivative of the function exists at that point. The function is differentiable at a point if the limit of the difference quotient exists as it approaches that point.

Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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