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Standard XII

Mathematics

Symmetric Matrix

If yn is th...

Question

If $y_{n}$ is the $n$ th derivative of $y$ then the least value of $n$ so that $y_{n} = y_{n + 1}$ where $y = x^{2} + e^{x}$ is

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Solution

The correct option is A $3$
$y = x^{2} + e^{x}$
$y_{1} = 2 x + e^{x}$
$y_{2} = 2 + e^{x}$
$y_{3} = e^{x}$
$y_{4} = e^{x}$
Therefore $y_{3} = y_{4}$
Comparing with the expression
$y_{n} = y_{n + 1}$ , we get the least value of $n$ is $3$ .

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Similar questions

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y = \frac{1}{x^{2} - a^{2}}

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and

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If yn is the nth derivative of y then the least value of n so that yn=yn+1 where y=x2+ex is

The correct option is A 3y=x2+exy1=2x+exy2=2+exy3=exy4=exTherefore y3=y4Comparing with the expressionyn=yn+1, we get the least value of n is 3.

If $y_{n}$ is the $n$ th derivative of $y$ then the least value of $n$ so that $y_{n} = y_{n + 1}$ where $y = x^{2} + e^{x}$ is

The correct option is A $3$
$y = x^{2} + e^{x}$
$y_{1} = 2 x + e^{x}$
$y_{2} = 2 + e^{x}$
$y_{3} = e^{x}$
$y_{4} = e^{x}$
Therefore $y_{3} = y_{4}$
Comparing with the expression
$y_{n} = y_{n + 1}$ , we get the least value of $n$ is $3$ .