If 3x-2y-1 acts as a tangent to y-fx at x-12 and if p-limx-0xx-1fe2x2-fe-2x2- then -r-1-pr-

Slope of 3x+2y=1 is 32 ⇒f'(12)= 32 Using L Hospital's rule we will get p=lim_x→02x 1f'(e^2x2)e^2x+f'(e^ 2x2)e^ 2x= 1 32 32 = 1 3=13⇒∑_r=1^∞p^r=131 (13)=12 ...

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Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

If 3x+2y=1 ac...

Question

If $3 x + 2 y = 1$ acts as a tangent to $y = f (x)$ at $x = \frac{1}{2}$ and if $p = lim x \to 0 \frac{x (x - 1)}{f (\frac{e^{2 x}}{2}) - f (\frac{e^{- 2 x}}{2})},$ then $\infty \sum r = 1 p^{r} =$ ______

\frac{1}{2}

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\frac{1}{3}

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\frac{1}{6}

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\frac{1}{7}

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Solution

The correct option is A $\frac{1}{2}$
Slope of $3 x + 2 y = 1$ is $\frac{- 3}{2}$
$\Rightarrow f^{'} (\frac{1}{2}) = \frac{- 3}{2}$
Using L-Hospital's rule we will get
$p = lim x \to 0 \frac{2 x - 1}{f^{'} (\frac{e^{2 x}}{2}) e^{2 x} + f^{'} (\frac{e^{- 2 x}}{2}) e^{- 2 x}} = \frac{- 1}{\frac{- 3}{2} - \frac{3}{2}}$
$= \frac{- 1}{- 3} = \frac{1}{3} \Rightarrow \infty \sum r = 1 p^{r} = \frac{\frac{1}{3}}{1 - (\frac{1}{3})} = \frac{1}{2}$

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If 3x+2y=1 acts as a tangent to y=f(x) at x=12 and if p=limx→0x(x−1)f(e2x2)−f(e−2x2), then ∞∑r=1pr=______

The correct option is A 12Slope of 3x+2y=1 is −32 ⇒f′(12)=−32 Using L-Hospital's rule we will get p=limx→02x−1f′(e2x2)e2x+f′(e−2x2)e−2x=−1−32−32 =−1−3=13⇒∞∑r=1pr=131−(13)=12

If $3 x + 2 y = 1$ acts as a tangent to $y = f (x)$ at $x = \frac{1}{2}$ and if $p = lim x \to 0 \frac{x (x - 1)}{f (\frac{e^{2 x}}{2}) - f (\frac{e^{- 2 x}}{2})},$ then $\infty \sum r = 1 p^{r} =$ ______