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Limits Continuity and Differentiability question

2021 · 16 Mar · Shift 2 · Q29

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Limits Continuity and Differentiability
2021 · 16 Mar · Shift 2 · Q29

Limits Continuity and Differentiability question

2021 · 16 Mar · Shift 2 · Q29

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

Let f : S

\to

S where S = (0,

\infty

) be a twice differentiable function such that f(x + 1) = xf(x). If g : S

\to

R be defined as g(x) = loge f(x), then the value of |g''(5)

-

g''(1)| is equal to :

A
1
B
${{187} \over {144}}$
C
${{197} \over {144}}$
D
${{205} \over {144}}$

View written solutionFree

Correct answer: D

Given functional equation

We have $f(x+1)=x f(x), \qquad x>0.$

Also, $g(x)=\ln f(x).$

Taking natural log on both sides: $g(x+1)=\ln f(x+1)=\ln(xf(x))=\ln x+g(x).$

So, $g(x+1)-g(x)=\ln x.$

Differentiate once

Differentiate both sides with respect to $x$ : $g'(x+1)-g'(x)=\frac{1}{x}.$

Differentiate twice

Differentiate again: $g''(x+1)-g''(x)=-\frac{1}{x^2}.$

This recurrence will help us connect $g''(5)$ and $g''(1)$ .

Compute step by step

Put successive values of $x$ :

For $x=1$ , $g''(2)-g''(1)=-1.$
For $x=2$ , $g''(3)-g''(2)=-\frac14.$
For $x=3$ , $g''(4)-g''(3)=-\frac19.$
For $x=4$ , $g''(5)-g''(4)=-\frac1{16}.$

Now add all these equations: $g''(5)-g''(1)= -\left(1+\frac14+\frac19+\frac1{16}\right).$

Therefore, $|g''(5)-g''(1)|=1+\frac14+\frac19+\frac1{16}.$

Add the fractions

LCM of $1,4,9,16$ is $144$ .

So, $1=\frac{144}{144},\qquad \frac14=\frac{36}{144},\qquad \frac19=\frac{16}{144},\qquad \frac1{16}=\frac{9}{144}.$

Hence, $1+\frac14+\frac19+\frac1{16}=\frac{144+36+16+9}{144}=\frac{205}{144}.$

Thus, $|g''(5)-g''(1)|=\frac{205}{144}.$

Match with options

The correct option is: $\boxed{\text{D } \frac{205}{144}}.$

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