Answer: (a) \(\displaystyle y=\frac{10^{4x^2-3}-1}{2}\); (b) \(y=0\); (c) \(x=\pm\sqrt{\frac{3+\lg5}{4}}\approx\pm0.962\).

Answer: (a) \(\displaystyle y=\frac{10^{4x^2-3}-1}{2}\); (b) \(y=0\); (c) \(x\approx\pm0.962\).

Let \(Y=\lg(2y+1)\) and \(X=x^2\). The straight line passes through \((1,1)\) and \((2,5)\), so its gradient is

\(\frac{5-1}{2-1}=4.\)

Using \(Y=4X+c\) and the point \((1,1)\),

\(1=4(1)+c,\)

so \(c=-3\). Therefore

\(\lg(2y+1)=4x^2-3.\)

Hence

\(2y+1=10^{4x^2-3},\)

and

\(y=\frac{10^{4x^2-3}-1}{2}.\)

When \(x=\frac{\sqrt3}{2}\), we have \(x^2=\frac34\), so

\(4x^2-3=4\cdot\frac34-3=0.\)

Thus

\(y=\frac{10^0-1}{2}=0.\)

When \(y=2\),

\(\lg(2y+1)=\lg5.\)

So

\(4x^2-3=\lg5,\)

giving

\(x=\pm\sqrt{\frac{3+\lg5}{4}}\approx\pm0.962.\)