# For what value of λ is the function defined by f(x)={(λ(x^{2} − 2x), if x ≤ 0) (4x + 1, if x > 0) is continous at x = 0? What about continuity at x = 1?

**Solution:**

A function is said to be continuous when the graph of the function is a single unbroken curve.

The given function is

f(x) = {(λ(x^{2} − 2x), if x ≤ 0) (4x + 1, if x > 0)

If f is continuous at x = 0, then

lim_{x→0−} f(x) = lim_{x→0+} f(x) = f(0)

⇒ lim_{x→0−} λ(x^{2 }− 2x) = lim_{x→0+} (4x + 1)

= λ(0^{2 }− 2×0)

⇒ λ(0^{2 }− 2×0)

= 4(0) + 1= 0

⇒ 0= 1= 0 [which is not possible]

Therefore, there is no value of λ for which f is continuous at x = 0.

At x = 1

f(1) = 4x + 1 = 4(1) + 1 = 5

lim_{x→1} (4x + 1) = 4(1) + 1 = 5

⇒ lim_{x→1} f(x) = f(1)

Therefore, for any values of λ, f is continuous at x = 1

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.1 Question 18

## For what value of λ is the function defined by f(x)={(λ(x^{2} − 2x), if x ≤ 0) (4x + 1, if x > 0) is continous at x = 0? What about continuity at x = 1?

**Summary:**

For the function defined by f(x) = {(λ(x^{2} − 2x), if x ≤ 0) (4x + 1, if x > 0), there is no value of λ for which f is continuous at x = 0. For any values of λ, f is continuous at x = 1

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