Additional Mathematics 0606 / Straight-line graphs / 3.193.19: Normal to ln\lnln curve; gradient of BCBCBC0606/12/M/J/22 — Question 9 · 9 marksMark as done · Save for later · Show all solutionsThe normal to the curve y=ln (x2+3)+12x2y=\ln\!\left(x^{2}+3\right)+\dfrac{1}{2}x^{2}y=ln(x2+3)+21x2 at the point AAA where x=0x=0x=0 meets the xxx-axis at BBB. Point CCC has coordinates (0,3ln2)(0, 3\ln 2)(0,3ln2). Find the gradient of the line BCBCBC in terms of ln2\ln 2ln2.[9]▸ Answer▸ Official mark scheme← 3.18: Trigonometric relation yyy in terms of xxx3.90: Straight-line log graph for lg(2y+1) →