NCERT Questions for Class 12 Maths Chapter 5 – Continuity and Differentiability

Important Questions with Solutions of Class 12 Maths Chapter 5 – Continuity and Differentiability

1) Find the value of k if

Ans – The limit at the break points of a continuous function must evaluate to the same value.

∴ The following condition must be met,

⇒ 3(2)² – k(2) + 5 = 1 – 3(2)

⇒ 12 – 2k + 5 = -5

⇒ 2 – 2k

k = 1

We know that the polynomials are continuous for their respective domains as the piecewise function is composed of polynomials. Hence, if the function is continuous, k must have a value of k = 1.

2) For what value of a and b,

Ans – In order for the function to be continuous, it is necessary that the following conditions are satisfied:

So the values will be a = 0 and b = -1.

3) Prove that f(x) = |x+1| is continuous, at x = -1 but not deriveable at x = -1.

Ans – Determine the limit of the breaking points to prove that the function is continuous at x = -1.

We conclude that the function is continuous since the above condition is met. We must now find the left-hand derivative and the right-hand derivative.

Since the above condition could not be met, the function is not derivable at x=-1.

4) Find dy/dx

Ans – Differentiating with respect to x,

5) Show that

Ans – Differentiating with respect to x,

6) Find the derivative of

Ans –

Let x = tan t

7) Differentiate w.r.t. x

Ans – Taking the natural logarithm on both sides and then simplifying further, it can be obtained as shown below

8) Find dy/dx, if ((cos x)^y) = ((cos y)^x)

Ans – Applying the natural logarithm on both sides,

⇒ ln((cos x)^y) = ln((cos y)^x)

⇒ y ln cos x = x ln cos y

Differentiating both sides w.r.t. x,

9) Find dy/dx

Ans – We have

Differentiating w.r.t. x,

10) Differetiate (ln x)^{ln x}, x>1 w.r.t. x.

Ans – Apply natural logarithm on both sides

ln y = ln x(ln(ln x))

Differentiating w.r.t. x,

Why are Continuity and Differentiability important?

Continuity and Differentiability are two wights of the central core of the concepts of calculus. Continuity would imply the smoothness of the function, and Differentiability at a point would imply that the derivative should exist at that point. In simpler words, a function is continuous when it does not have any breaks, holes, or jumps from one end to the other and is differentiable if it always has a well-defined slope at each point.

Knowing of continuity and differentiability is extremely useful in many fields today, including calculus. These notions play crucial issues and supplements in the analysis and deductions about functions, maximum and minimum values, and optimization and work as basic tools of further topics in calculus, including integrations and differential equations.