Q.1 (A) (iv) The equation of normal to the curve y=3x²-x+1 at (1, 3) is _____. A. x-5y-16=0 B. x+5y-16=0 C. x-5y+16=0 D. -5y-x-16=0

The equation of normal to the curve \(y=3x^2-x+1\) at (1, 3) is _____.

Q.1 (A) (iv) The equation of normal to the curve \(y=3x^2-x+1\) at (1, 3) is _____.

A. \(x-5y-16=0\)

B. \(x+5y-16=0\)

C. \(x-5y+16=0\)

D. \(-5y-x-16=0\)

Doubtly Asked 28 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

Best answer

Answer: B. \(x+5y-16=0\)

Answer: (b) \(x+5y-16=0\)

\(\frac{dy}{dx}=6x-1\), so the slope of the tangent at (1, 3) is \(6(1)-1=5\). Slope of the normal \(=-\frac{1}{5}\).

Equation of normal: \(y-3=-\frac{1}{5}(x-1)\Rightarrow 5y-15=-x+1\Rightarrow x+5y-16=0\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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