If f x -tan -1-log e - x 2-log ex 2-tan -1-3-2 log x -1-6 log x - then the value of f - x isA- x 2B- XC-D- 0

The correct option is D 0f(x)=tan^ 1[log(ex^2)log(ex^2)]+tan^ 1[3+2log x1 6log x] f(x)=tan^ 1[1 2log x1+2log x]+tan^ 1[3+2log x1 3·2log x] f(x)=tan^ 1(1) tan ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

If f x =tan -...

Question

If $f (x) = {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{log (\frac{e}{x^{2}})}{log (e x^{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ + {tan}^{- 1} [\frac{3 + 2 log x}{1 - 6 log x}]$ then the value of $f^{''} (x)$ is

x^{2}

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Solution

The correct option is D $0$
$f (x) = {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{log (\frac{e}{x^{2}})}{log (e x^{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ + {tan}^{- 1} [\frac{3 + 2 log x}{1 - 6 log x}] f (x) = {tan}^{- 1} [\frac{1 - 2 log x}{1 + 2 log x}] + {tan}^{- 1} [\frac{3 + 2 log x}{1 - 3 \cdot 2 log x}] f (x) = {tan}^{- 1} (1) - {tan}^{- 1} (2 log x) + {tan}^{- 1} (3) + {tan}^{- 1} (2 log x) f (x) = {tan}^{- 1} (1) + {tan}^{- 1} (3) ∴ f^{''} (x) = 0$

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Similar questions

Q. If

f (x) = {tan}^{- 1} ⎡ ⎢ ⎣ \begin{matrix} \frac{log (\frac{e}{x^{2}})}{log (e x^{2})} \end{matrix} ⎤ ⎥ ⎦ + {tan}^{- 1} [\begin{matrix} \frac{3 + 2 log x}{1 - 6 log x} \end{matrix}]

, then the value of

f^{''} (x)

Q. If

y = {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{log (\frac{e}{x^{2}})}{log (e x^{2})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ + {tan}^{- 1} (\frac{3 + 2 log x}{1 - 6 log x})

, then

\frac{d^{2} y}{d x^{2}}

y = {tan}^{- 1} ⎡ ⎢ ⎢ ⎣ \frac{log (\frac{e}{x^{2}})}{log (e x^{2})} ⎤ ⎥ ⎥ ⎦ + {tan}^{- 1} (\frac{3 + 2 log x}{1 - 6 log x})

, then

\frac{d^{2} y}{d x^{2}} =

Q. If

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \frac{log (\frac{e}{x^{2}})}{log (e x^{2})} ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ + {tan}^{- 1} (\frac{3 + 2 log x}{1 - 6 log x})

, then

\frac{d^{n} y}{d x^{n}}

Q. If

y = T a n^{- 1} (\frac{log (e / x^{2})}{log e x^{2}}) + T a n^{- 1} (\frac{3 + 2 log x}{1 - 6 log x})

then

{(\frac{d y}{d x})}_{x = 2} + {(\frac{d y}{d x})}_{x = 3}

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If f(x)=tan−1⎡⎢ ⎢ ⎢ ⎢⎣log(ex2)log(ex2)⎤⎥ ⎥ ⎥ ⎥⎦+tan−1[3+2logx1−6logx] then the value of f′′(x) is

The correct option is D 0f(x)=tan−1⎡⎢ ⎢ ⎢ ⎢⎣log(ex2)log(ex2)⎤⎥ ⎥ ⎥ ⎥⎦+tan−1[3+2logx1−6logx]f(x)=tan−1[1−2logx1+2logx]+tan−1[3+2logx1−3⋅2logx]f(x)=tan−1(1)−tan−1(2logx)+tan−1(3)+tan−1(2logx)f(x)=tan−1(1)+tan−1(3)∴f′′(x)=0

If $f (x) = {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{log (\frac{e}{x^{2}})}{log (e x^{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ + {tan}^{- 1} [\frac{3 + 2 log x}{1 - 6 log x}]$ then the value of $f^{''} (x)$ is