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Continuity And Differentiability

Question
CBSEENMA12036108

The equation of the tangent to the curvestraight y equals straight x space plus space 4 over straight x squared, that is parallel to the x-axis, is

  • y = 0

  • y = 1

  • y = 3

  • y =2

Solution

C.

y = 3

space straight y space equals space straight x space plus space 4 over straight x squared
On differentiating w.r.t, we get
dy/dx = 1-8/x3
since the tangent is parallel to X-axis, therefore
⇒ x3 = 8
⇒ x = 2 and y = 3

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