Assertion (A): f(x) = x^2 + | x | is continuous at every real x. Reason (R): f(x) = x^2 + | x | is not differentiable at x = 0.

Question 73

MediumOriginal question

Assertion (A): $f(x) = x^{2} + \lvert x \rvert$ is continuous at every real $x$ .
Reason (R): $f(x) = x^{2} + \lvert x \rvert$ is not differentiable at $x = 0$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

A is true: $x^{2}$ and $\lvert x \rvert$ are continuous everywhere, and so is their sum.
R is true as well: the slopes at 0 are $-1$ and 1. But a corner says nothing about continuity, so R does not explain A.

Worth noting

The continuity in A comes from the algebra of continuous functions, not from anything about slopes.

Concept: Recognising continuous functions
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsContinuous functions

Practise this chapter →

Ten questions from Continuity & Differentiability, at your own pace.

Assertion (A): f(x)=x2+∣x∣f(x) = x^{2} + \lvert x \rvertf(x)=x2+∣x∣ is continuous at every real xxx.Reason (R): f(x)=x2+∣x∣f(x) = x^{2} + \lvert x \rvertf(x)=x2+∣x∣ is not differentiable at x=0x = 0x=0.

Explanation

Assertion (A): $f(x) = x^{2} + \lvert x \rvert$ is continuous at every real $x$ .
Reason (R): $f(x) = x^{2} + \lvert x \rvert$ is not differentiable at $x = 0$ .