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  • Value of the parameter a such that the area bounded by co-ordin

Value of the parameter a such that the area bounded by y=a^{2}x^{2}+ax+1, co-ordinate axes and the line x=1, attains it's least value, is equal to

Option: 1 -0.25

Option: 2 -0.5

Option: 3 -0.75

Option: 4 -1

Answers (1)

best_answer

 

Indefinite integrals for Algebraic functions -

 \frac{\mathrm{d}}{\mathrm{d} x} \frac{\left ( x^{n+1} \right )}{n+1}=x^{n} so \int x^{n}dx=\frac{x^{n+1}}{n+1}

- wherein

Where  n\neq-1

 

 

Introduction of area under the curve -

The area between the curve y= f(x),x axis and two ordinates at the point  x=a\, and \,x= b\left ( b>a \right ) is given by

A= \int_{a}^{b}f(x)dx=\int_{a}^{b}ydx

- wherein

 

 

y=a^{2}x^{2}+ax+1

\therefore area\; A=\int_{0}^{1}(a^{2}x^{2}+ax+1)dx

=\left ( a^{2}\frac{x^{3}}{3}+\frac{ax^{2}}{2}+x \right )\mid _{0}^{1}=\frac{a^{2}}{3}+\frac{a}{2}+1

=\frac{2a^{2}+3a+6}{6}=\frac{1}{3}\left ( a^{2}+\frac{3}{2}a+\frac{9}{16} \right )+1-\frac{9}{48}

=\frac{1}{3}\left ( a+\frac{3}{4} \right )^{2}+\frac{39}{48} attains its least value, when a=-\frac{3}{4}

Posted by

Kuldeep Maurya

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