Question

If the tangent drawn at the point \( (x_1,y_1) \), \(x_1,y_1 \in N \) on the curve \( y = x^4 - 2x^3 + x^2 + 5x \) passes through origin, then \( x_1+y_1 = \)

• A. 5
• B. 4
• C. 7
• D. 6

Hint:

1. Find the derivative \( \frac{dy}{dx} \) of the curve to get the slope of the tangent. 2. The equation of the tangent at \( (x_1,y_1) \) is \( Y-y_1 = (\frac{dy}{dx}|_{(x_1,y_1)})(X-x_1) \). 3. If this tangent passes through the origin (0,0), substitute \(X=0, Y=0\) into the tangent equation. This gives a relation \( -y_1 = m(-x_1) \implies y_1 = mx_1 \). 4. Use the fact that \( (x_1,y_1) \) also lies on the original curve \( y_1 = f(x_1) \). 5. Solve the system of equations for \(x_1\), then find \(y_1\). Use any constraints on \(x_1, y_1\) (like being natural numbers).

Answer 1

The Correct Option is D

The curve is \( y = x^4 - 2x^3 + x^2 + 5x \).
First, find the derivative \( \frac{dy}{dx} \) to get the slope of the tangent.
\[ \frac{dy}{dx} = 4x^3 - 6x^2 + 2x + 5 \] The slope of the tangent at \( (x_1, y_1) \) is \( m = 4x_1^3 - 6x_1^2 + 2x_1 + 5 \).
The equation of the tangent at \( (x_1, y_1) \) is \( Y - y_1 = m(X - x_1) \).
\[ Y - y_1 = (4x_1^3 - 6x_1^2 + 2x_1 + 5)(X - x_1) \] This tangent passes through the origin (0,0).
Substitute \( X=0, Y=0 \): \[ 0 - y_1 = (4x_1^3 - 6x_1^2 + 2x_1 + 5)(0 - x_1) \] \[ -y_1 = -(4x_1^3 - 6x_1^2 + 2x_1 + 5)x_1 \] \[ y_1 = (4x_1^3 - 6x_1^2 + 2x_1 + 5)x_1 \] \[ y_1 = 4x_1^4 - 6x_1^3 + 2x_1^2 + 5x_1 \] Also, the point \( (x_1, y_1) \) lies on the curve, so \( y_1 = x_1^4 - 2x_1^3 + x_1^2 + 5x_1 \).
Equating the two expressions for \( y_1 \): \[ 4x_1^4 - 6x_1^3 + 2x_1^2 + 5x_1 = x_1^4 - 2x_1^3 + x_1^2 + 5x_1 \] \[ 3x_1^4 - 4x_1^3 + x_1^2 = 0 \] Factor out \( x_1^2 \): \[ x_1^2 (3x_1^2 - 4x_1 + 1) = 0 \] Factor the quadratic: \( 3x_1^2 - 3x_1 - x_1 + 1 = 0 \implies 3x_1(x_1-1) - 1(x_1-1) = 0 \).
So, \( (3x_1-1)(x_1-1) = 0 \).
The equation becomes \( x_1^2 (3x_1-1)(x_1-1) = 0 \).
Possible values for \( x_1 \) are \( 0, 1/3, 1 \).
Given that \( x_1 \in N \) (natural numbers), so \( x_1 = 1 \).
Now find \( y_1 \) using the curve equation with \( x_1 = 1 \): \[ y_1 = (1)^4 - 2(1)^3 + (1)^2 + 5(1) = 1 - 2 + 1 + 5 = 5 \] So the point \( (x_1, y_1) \) is \( (1,5) \).
Given \( y_1 \in N \), which \( 5 \in N \) satisfies.
Then \( x_1+y_1 = 1+5 = 6 \).
This matches option (4).