12th Maths Public Exam (August 2021)

Detailed Solution for Question 37

Question 37: Application of Derivatives

Show that the point on the curve $ y = x^2 - 5x + 4 $ at which the tangent is parallel to the line $ 3x + y = 7 $ is $ (1, 0) $.

Step 1: Find the Slope of the Given Line

The equation of the given line is:

$$ 3x + y = 7 $$

We rewrite this in the slope-intercept form ($ y = mx + c $):

$$ y = -3x + 7 $$

Comparing this with $ y = mx + c $, the slope of the line ($ m $) is:

$$ m = -3 $$

Since the tangent to the curve is parallel to this line, the slope of the tangent must also be -3.

Step 2: Find the Slope of the Tangent to the Curve

The equation of the curve is:

$$ y = x^2 - 5x + 4 $$

To find the slope of the tangent at any point $(x, y)$, we differentiate $y$ with respect to $x$:

$$ \frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(5x) + \frac{d}{dx}(4) $$ $$ \frac{dy}{dx} = 2x - 5 $$

Step 3: Equate the Slopes

Since the slope of the tangent ($\frac{dy}{dx}$) is equal to the slope of the line ($m = -3$), we set them equal:

$$ 2x - 5 = -3 $$

Step 4: Solve for x

Now, we solve for $x$:

$$ 2x = -3 + 5 $$ $$ 2x = 2 $$ $$ x = \frac{2}{2} $$ $$ x = 1 $$

Step 5: Find the y-coordinate

Substitute $ x = 1 $ back into the original curve equation to find the corresponding $ y $ value:

$$ y = (1)^2 - 5(1) + 4 $$ $$ y = 1 - 5 + 4 $$ $$ y = 0 $$

Conclusion

The coordinates of the point where the tangent is parallel to the given line are:

$$ (x, y) = (1, 0) $$

Result: The point is proved to be $ (1, 0) $.