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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

જો \({e^y} + xy = e\) હોય તો \(x = 0\)  આગળ ક્રમયુક્ત જોડ \(\left( {\frac{{dy}}{{dx}},\frac{{{d^2}y}}{{d{x^2}}}} \right)\) ની કિમંત મેળવો.

  1. A \(\left( {\frac{1}{e}, - \frac{1}{{{e^2}}}} \right)\)
  2. B \(\left( {\frac{1}{e},  \frac{1}{{{e^2}}}} \right)\)
  3. C \(\left( { - \frac{1}{e},\frac{1}{{{e^2}}}} \right)\)
  4. D \(\left( { - \frac{1}{e}, - \frac{1}{{{e^2}}}} \right)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left( { - \frac{1}{e},\frac{1}{{{e^2}}}} \right)\)

Step-by-step Solution

Detailed explanation

\({e^y} = xy = e\) differentiate w.r.t. \(x\) \({e^y}\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0\) \({e^y}\frac{{dy}}{{dx}} + x\frac{{dy}}{{dx}} + y = 0\)…
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