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JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

If

f(x)=x^n+4

then the value of

f(1)+

\frac{f^{\prime}(1)}{1 !}+\frac{f^{\prime \prime}(1)}{2 !}+\ldots+\frac{f^n(1)}{n !}

is

View answer and explanation

Correct option: B

2^n+4

Option B (

2^n+4

) is correct because the expression is a Taylor series expansion of

f(x)

around

x=1

, which evaluates to

f(2) = 2^n + 4

.

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y

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x=k, k

being an integer, the left derivative of

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y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

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z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}