If fx is the anti-derivative of x-1x2ex and fx-xaekx-C- where a-k - and C is an integration constant- Then which of the following isare correct

As, f(x) is anti derivative of x+1x^2e^x ⇒ f(x)=∫x+1x^2e^x dx ⇒ f(x)=∫e^x(x+1)x^2e^2xdx=∫e^x(x+1)(xe^x)^2dx Substituting, xe^x=z ⇒ e^x(x+1)dx=dz ⇒ I=∫dzz^2= 1z ...

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Standard XII

Physics

Introduction

If fx is the ...

Question

If $f (x)$ is the anti-derivative of $\frac{x + 1}{x^{2} e^{x}}$ and $f (x) = - x^{a} e^{k x} + C,$ where $a, k \in R$ and $C$ is an integration constant. Then which of the following is(are) correct

k + a = 0

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k - a = 0

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| 3 k + a | = 4

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k^{2} + a^{2} = 2

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Solution

The correct option is D $k^{2} + a^{2} = 2$
As, $f (x)$ is anti-derivative of $\frac{x + 1}{x^{2} e^{x}}$
$\Rightarrow f (x) = \int \frac{x + 1}{x^{2} e^{x}} d x \Rightarrow f (x) = \int \frac{e^{x} (x + 1)}{x^{2} e^{2 x}} d x = \int \frac{e^{x} (x + 1)}{(x e^{x})^{2}} d x$
Substituting, $x e^{x} = z$
$\Rightarrow e^{x} (x + 1) d x = d z$
$\Rightarrow I = \int \frac{d z}{z^{2}} = - \frac{1}{z} + C$
$\Rightarrow f (x) = - \frac{1}{x e^{x}} + C = - x^{- 1} e^{- x} + C$
$∴ a = - 1, k = - 1$

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Introduction

Standard XII Physics

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If f(x) is the anti-derivative of x+1x2ex and f(x)=−xaekx+C, where a,k∈R and C is an integration constant. Then which of the following is(are) correct

The correct option is D k2+a2=2As, f(x) is anti-derivative of x+1x2ex ⇒f(x)=∫x+1x2exdx⇒f(x)=∫ex(x+1)x2e2xdx=∫ex(x+1)(xex)2dx Substituting, xex=z ⇒ex(x+1)dx=dz ⇒I=∫dzz2=−1z+C ⇒f(x)=−1xex+C=−x−1e−x+C ∴a=−1,k=−1

If $f (x)$ is the anti-derivative of $\frac{x + 1}{x^{2} e^{x}}$ and $f (x) = - x^{a} e^{k x} + C,$ where $a, k \in R$ and $C$ is an integration constant. Then which of the following is(are) correct