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  • Let <img alt="f(x)=\min \{1,1+x \sin x\}, 0 \leq x \leq 2 \pi" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%5Cmin%20%5C%7B1%2C1&plus;x%20%5Csin%20x%5C%7D%2C%200%20%5Cleq%20x

Let f(x)=\min \{1,1+x \sin x\}, 0 \leq x \leq 2 \pi. If m is the number of points, where f is not differentiable and n is the number of points, where f is not continuous, then the ardered pair (m, n) is equal to

Option: 1

(2,0)


Option: 2

(1,0)


Option: 3

(1,1)


Option: 4

(2,1)


Answers (1)

best_answer

\mathrm{f(x)=\min \{1,1+x \sin x\} ; 0 \leq x \leq 2 \pi}

Non differentiable at \mathrm{x=\pi}
\mathrm{\therefore \quad m=1, n=0 \quad ;(m, n) \equiv(1,0) }

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Rishabh

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