What is the value of jump at x = 2, for f(x);
f(x) =

2

Open in App

Solution

The correct option is A 2
We can see that left hand limit at x = 2 is = 1 and Right hand limit at x = 2 is = 3
So, jump = $∣ ∣ ∣ l i m x \to 2^{-} f (x) - l i m x \to 2^{+} f (x) - ∣ ∣ ∣ = | 1 - 3 | = 2$

Suggest Corrections

Similar questions

R \to R

sin x cos y [f (2 x + 2 y) - f (2 x - 2 y)] = cos x sin y [f (2 x + 2 y) + f (2 x - 2 y)]

f^{'} (0) = \frac{1}{2}

f (x) = log (\frac{1 + x}{1 - x})

f (\frac{2 x}{1 + x^{2}})

\to

f (2 x + 3) + f (2 x + 7) = 2, \forall x ϵ R

f_{1} (x) = 2 x, f_{2} (x) = 3 sin x - x cos x,

x \in (0, \frac{π}{2})

Join BYJU'S Learning Program