Question

Let F : R → R be a continuous odd function , which vanishes exactly at one point and f(1)=1/2
Suppose that F(x)= $\sf\displaystyle \int^x _{-1}f(t)dt$ for all x∈[-1,2] and
$\sf G(x)=\displaystyle \int^2 _{-1} t|f\{f(t)\}|dt$ for all x∈[-1,2] . if $\sf\displaystyle \lim_{x\to 1}\dfrac{F(x)}{G(x)} =\dfrac{1}{14}$ . then the value of f(1/2) is

Answer 1

$\large\underline{\bold{Given \: correct\:Question - }}$

Let F : R → R be a continuous odd function , which vanishes exactly at one point and f(1)=1/2.

Suppose that

$\bf \: F(x)= \sf\displaystyle \int^x _{-1}f(t)dt \: \: \: for \: all \: x \: \in \: [-1,2]$

and

$\bf G(x)=\displaystyle \int^x _{-1} t|f\{f(t)\}|dt \: for \: all \: x \: \in \: [-1,2] .$

$\sf \: if \: \bf\displaystyle \lim_{x\to 1}\dfrac{F(x)}{G(x)} =\dfrac{1}{14} \: \bf then \: value \: of \: f\bigg(\dfrac{1}{2} \bigg)$

$\large\underline{\bold{Solution-}}$

Given that

$\bf \: F(x)= \sf\displaystyle \int^x _{-1} \sf \: f(t)dt \: \: \: for \: all \: x \: \in \: [-1,2]$

and

$\bf G(x)=\displaystyle \int^x _{-1} \sf \: t|f\{f(t)\}|dt \: for \: all \: x \: \in \: [-1,2] .$

Now,

According to statement,

$\bf\displaystyle \lim_{x\to 1} \bf \: \dfrac{F(x)}{G(x)} =\dfrac{1}{14}$

On substituting the values of F(x) and G(x), we have

$\bf\displaystyle \lim_{x\to 1}\dfrac{\sf\displaystyle \int^x _{-1} \sf \: f(t)dt \:}{\displaystyle \int^x _{-1} \sf \: t|f\{f(t)\}|dt \:} =\dfrac{1}{14}$

Using L - Hospital Rule and Leibniz Rule for differentiation under the integral sign, we get

$\bf\displaystyle \lim_{x\to 1}\dfrac{0 + \dfrac{d}{dx}(x) \: \times f(x) - \dfrac{d}{dx}( - 1) \times f( - 1) }{0 + \dfrac{d}{dx}(x) \times x |f \{f(x) \}| - \dfrac{d}{dx}( - 1) \times ( - 1) |f \{f( - 1) \}| } = \dfrac{1}{14}$

$\bf\displaystyle \lim_{x\to 1} \: \bf\dfrac{f(x)}{x |f \{f(x) \}|} = \dfrac{1}{14}$

$\sf \: \dfrac{f(1)}{1 \times |f \{f(1) \}|} = \dfrac{1}{14}$

$\sf \: \dfrac{\dfrac{1}{2} }{ f\bigg( \dfrac{1}{2} \bigg) } = \dfrac{1}{14} \: \: \: \: \: \: \: \: \: \bigg\{ \because \: \bf \: f(1) = \dfrac{1}{2} \bigg\}$

$\therefore \bf \: f\bigg( \dfrac{1}{2} \bigg) = 14 \times \dfrac{1}{2} = 7$

Note :-

1. L' Hospital's rule can only be applied in the case where direct substitution yields an indeterminate form, such as 0/0 or ±∞/±∞. So if f and g are defined, L' Hospital would be applicable only if the value of both f and g is 0 or ∞.

2. Leibniz Rule for differentiation under the integral sign :-

$\sf \: \dfrac{d}{dx}\sf\displaystyle \int^{a(x)} _{{b(x)}} \sf \: f(t)dt \: = \dfrac{d}{dx}a(x) \times f\bigg( a(x)\bigg) - \dfrac{d}{dx}b(x) \times f\bigg( b(x)\bigg)$

Answer 2

Correct Question:-

Let $\displaystyle\sf {f:\mathbb{R}\to\mathbb{R}}$ be a continuous odd function, which vanishes exactly at one point and $\displaystyle\sf {f(1)=\dfrac{1}{2}.}$ Suppose that $\displaystyle\sf {F(x)=\int\limits_{-1}^xf(t)\ dt}$ for all $\displaystyle\sf {x\in[-1,\ 2]}$ and $\displaystyle\sf {G(x)=\int\limits_{-1}^xt|f(f(t))|\ dt}$ for all $\displaystyle\sf {x\in[-1,\ 2].}$ If $\displaystyle\sf {\lim_{x\to1}\dfrac{F(x)}{G(x)}=\dfrac{1}{14},}$ then find the value of $\displaystyle\sf {f\left(\dfrac{1}{2}\right).}$

Solution:-

Since $\displaystyle\sf {f}$ is an odd function, we have,

• $\displaystyle\sf{f(-x)=-f(x)}$

So,

• $\displaystyle\sf {f(-1)=-f(1)=-\dfrac{1}{2}}$

and,

• $\displaystyle\sf {f(-0)=f(0)=-f(0)\ \implies\ f(0)=0}$

also for some $\displaystyle\sf {a\in\mathbb{R},}$

• $\displaystyle\sf {\int\limits_{-a}^af(t)\ dt=0\quad\quad\dots(1)}$

Given,

$\displaystyle\longrightarrow\sf {F(x)=\int\limits_{-1}^xf(t)\ dt}$

For $\displaystyle\sf {x=1,}$

$\displaystyle\longrightarrow\sf {F(1)=\int\limits_{-1}^1f(t)\ dt}$

Using (1),

$\displaystyle\longrightarrow\sf {F(1)=0}$

But given,

$\displaystyle\sf {\longrightarrow\lim_{x\to1}\dfrac{F(x)}{G(x)}=\dfrac{1}{14}}$

Here $\displaystyle\sf {F(x)\to0}$ when $\displaystyle\sf {x\to1}$ but still the limit $\displaystyle\sf {\lim_{x\to1}\dfrac{F(x)}{G(x)}}$ exists as non-zero.

That means $\displaystyle\sf {G(x)\to0}$ and so L'hospital's Rule is applied in this limit, i.e.,

$\displaystyle\sf{\longrightarrow\lim_{x\to1}\dfrac{F'(x)}{G'(x)}=\dfrac{1}{14}\quad\quad\dots(2)}$

Consider,

$\displaystyle\longrightarrow\sf {F(x)=\int\limits_{-1}^xf(t)\ dt}$

By Leibniz Rule,

$\displaystyle\longrightarrow\sf {F'(x)=f(x)-f(-1)}$

$\displaystyle\longrightarrow\sf {F'(x)=f(x)+\dfrac{1}{2}}$

At $\displaystyle\sf {x=1,}$

$\displaystyle\longrightarrow\sf {F'(1)=f(1)+\dfrac{1}{2}}$

$\displaystyle\longrightarrow\sf {F'(1)=\dfrac{1}{2}+\dfrac{1}{2}}$

$\displaystyle\longrightarrow\sf {F'(1)=1}$

So (2) becomes,

$\displaystyle\sf{\longrightarrow\dfrac{1}{G'(1)}=\dfrac{1}{14}}$

$\displaystyle\sf{\longrightarrow G'(1)=14}$

Consider,

$\displaystyle\sf {\longrightarrow G(x)=\int\limits_{-1}^xt|f(f(t))|\ dt}$

By Leibniz Rule,

$\displaystyle\sf {\longrightarrow G'(x)=x|f(f(x))|+|f(f(-1))|}$

At $\displaystyle\sf {x=1,}$

$\displaystyle\sf {\longrightarrow G'(1)=|f(f(1))|+|f(f(-1))|}$

$\displaystyle\sf {\longrightarrow 14=\left|f\left(\dfrac{1}{2}\right)\right|+\left|f\left(-\dfrac{1}{2}\right)\right|}$

Since $\displaystyle\sf {f}$ is odd,

$\displaystyle\sf {\longrightarrow 14=\left|f\left(\dfrac{1}{2}\right)\right|+\left|-f\left(\dfrac{1}{2}\right)\right|}$

Given that $\displaystyle\sf {f}$ vanishes exactly at one point, and we see $\displaystyle\sf {f(0)=0,}$ i.e., it vanishes at $\displaystyle\sf {x=0}$ and there is no other $\displaystyle\sf {x}$ such that $\displaystyle\sf {f(x)=0.}$

Since $\displaystyle\sf {f(1)>0,}$ the function $\displaystyle\sf {f}$ should be strictly increasing and is positive in the interval $\displaystyle\sf {[0,\ 1].}$ Else there may be another $\displaystyle\sf {x\in(0,\ 1)}$ such that $\displaystyle\sf {f(x)=0.}$

Thus $\displaystyle\sf {f\left(\dfrac {1}{2}\right)>0.}$ So,

$\displaystyle\sf {\longrightarrow 14=f\left(\dfrac{1}{2}\right)+f\left(\dfrac{1}{2}\right)}$

$\displaystyle\sf {\longrightarrow 14=2f\left(\dfrac{1}{2}\right)}$

$\displaystyle\sf {\longrightarrow\underline{\underline{f\left(\dfrac{1}{2}\right)=7}}}$

Hence 7 is the answer.