Assertion (A): f(x) = (1 - cos 2x)/x^2 for x ≠ 0, with f(0) = 1, is continuous at x = 0. Reason (R): 1 - cos 2x = 2sin^2 x for every real x.

Question 50

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Assertion (A): $f(x) = \frac{1 - \cos 2x}{x^{2}}$ for $x \ne 0$ , with $f(0) = 1$ , is continuous at $x = 0$ .
Reason (R): $1 - \cos 2x = 2\sin^{2} x$ for every real $x$ .

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 D

Correct answer: D

R is true: it is the double-angle formula $\cos 2x = 1 - 2\sin^{2} x$ .
A is false: by R, $f(x) = 2\left(\frac{\sin x}{x}\right)^{2} \to 2$ , not 1.

Worth noting

Treating $1 - \cos 2x$ as $x^{2}$ near 0 gives 1; the factor 2 in $2\sin^{2} x$ makes the limit 2.

Concept: A constant from a limit
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsChoosing a constant for continuity

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

Assertion (A): f(x)=1−cos⁡2xx2f(x) = \frac{1 - \cos 2x}{x^{2}}f(x)=x21−cos2x​ for x≠0x \ne 0x=0, with f(0)=1f(0) = 1f(0)=1, is continuous at x=0x = 0x=0.Reason (R): 1−cos⁡2x=2sin⁡2x1 - \cos 2x = 2\sin^{2} x1−cos2x=2sin2x for every real xxx.

Explanation

Assertion (A): $f(x) = \frac{1 - \cos 2x}{x^{2}}$ for $x \ne 0$ , with $f(0) = 1$ , is continuous at $x = 0$ .
Reason (R): $1 - \cos 2x = 2\sin^{2} x$ for every real $x$ .