MHT CET · Maths · Area Under Curves
If a curve \(y=\mathrm{a} \sqrt{x}+\mathrm{b} x\) passes through the point \((1,2)\) and the area bounded by the curve, line \(x=4\) and \(\mathrm{X}\)-axis is 8 sq. units, then
- A \(\mathrm{a}=3, \mathrm{~b}=-1\)
- B \(\mathrm{a}=3, \mathrm{~b}=1\)
- C \(\mathrm{a}=-3, \mathrm{~b}=1\)
- D \(\mathrm{a}=-3, \mathrm{~b}=-1\)
Answer & Solution
Correct Answer
(A) \(\mathrm{a}=3, \mathrm{~b}=-1\)
Step-by-step Solution
Detailed explanation
The given curve passes through \((1,2)\).
\(\therefore \quad 2=a+b\)
According to the given condition,
\(\int_0^4(a \sqrt{x}+b x) d x=8\) ...(i)
\(\Rightarrow \frac{2 \mathrm{a}}{3}\left[x^{3 / 2}\right]_0^4+\frac{\mathrm{b}}{2}\left[x^2\right]_0^4=8 \Rightarrow \frac{2 \mathrm{a}}{3} \cdot 8+8 \mathrm{~b}=8\)
\(\Rightarrow 2 a+3 b=3\) ...(ii)
From (i) and (ii), we get
\(\mathrm{a}=3, \mathrm{~b}=-1\)
\(\therefore \quad 2=a+b\)
According to the given condition,
\(\int_0^4(a \sqrt{x}+b x) d x=8\) ...(i)
\(\Rightarrow \frac{2 \mathrm{a}}{3}\left[x^{3 / 2}\right]_0^4+\frac{\mathrm{b}}{2}\left[x^2\right]_0^4=8 \Rightarrow \frac{2 \mathrm{a}}{3} \cdot 8+8 \mathrm{~b}=8\)
\(\Rightarrow 2 a+3 b=3\) ...(ii)
From (i) and (ii), we get
\(\mathrm{a}=3, \mathrm{~b}=-1\)
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