हिंदी

Let f(x) = |x2sinxp-1|, where p is a constant. The value of p for which f'(0) = 1 is ______. - Mathematics

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प्रश्न

Let f(x) = `|(x^2, sin x),(p, -1)|,` where p is a constant. The value of p for which f'(0) = 1 is ______.

विकल्प

  • `RR`

  • 1

  • 0

  • −1

MCQ
रिक्त स्थान भरें

उत्तर

Let f(x) = `|(x^2, sin x),(p, -1)|,` where p is a constant. The value of p for which f'(0) = 1 is -1.

Explanation:

Given f(x) = `|(x^2, sin x),(p, -1)|`

⇒ f(x) = −x2 − p sin x

⇒ f'(x) = −2x − p cos x

Also, given that f' (0) = 1

⇒ −2(0) − p(cos 0) = 1

⇒ −p = 1 

⇒ p = −1 

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2023-2024 (February) Delhi Set - 3
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