The line y=m x bisects the area enclosed by lines x=0, y=0 and x=3 / 2 and the curve y=1+4 x- ^ 2 Then the value of m is

Correct answer: A. (13)/(6) y=1+4 x-x^ 2 =5-(x-2)^ 2 We have ∫_ 0 ^ 3 / 2 (1+4 x-x^ 2 ) dx=2 ∫_ 0 ^ 3 / 2 mx dx=(3)/(2)+2((9)/(4))-(1)/(3)((27)/(8))=m · (9)/(4) On solving we get m=(13)/(6)

BITSAT 2020

Exam: BITSAT
Year: 2020
Subject: Mathematics
Chapter: Area Under Curves

BITSAT 2020 · Mathematics · Area Under Curves

The line y=m x bisects the area enclosed by lines x=0, y=0 and x=3...

The line

y=m x

bisects the area enclosed by lines

x=0, \mathrm{y}=0

and

\mathrm{x}=3 / 2

and the curve

\mathrm{y}=1+4 \mathrm{x}-\mathrmx^{2}

Then the value of

\mathrm{m}

Options

A $\frac{13}{6}$
B $\frac{13}{2}$
C $\frac{13}{5}$
D $\frac{13}{7}$

Correct answer

A. $\frac{13}{6}$

Step-by-step solution

y=1+4 x-x^{2}=5-(x-2)^{2}

We have

\int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) \mathrm{dx}=2 \int_{0}^{3 / 2} \operatorname{mx} \mathrm{dx}=\frac{3}{2}+2\left(\frac{9}{4}\right)-\frac{1}{3}\left(\frac{27}{8}\right)=\mathrm{m} \cdot \frac{9}{4}

On solving we get

\mathrm{m}=\frac{13}{6}

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