Chapter 6 Application of Derivatives Ex 6.3 Solutions
Question - 1 : - Find the slope of the tangent to thecurve y = 3x4 − 4x at x =4.
Answer - 1 : -
The given curve is y = 3x4 −4x.
Then, the slope of the tangent to the givencurve at x = 4 is given by,
Question - 2 : - Find the slope of thetangent to the curve
, x ≠ 2 at x =10.
Answer - 2 : - The given curve is
Thus, the slope of the tangent at x =10 is given by,
Hence, the slope of thetangent at x = 10 is 
Find the slope of the tangent to curve y = x3 − x +1 at the point whose x-coordinate is 2.
Answer - 3 : - The given curve is
The slope of the tangent toa curve at (x0, y0)is
It is given that x0 =2.
Hence, the slope of the tangent at the pointwhere the x-coordinate is 2 is given by,
Find the slope of the tangent to thecurve y = x3 − 3x +2 at the point whose x-coordinate is 3.
Answer - 4 : - The given curve is
The slope of the tangent toa curve at (x0, y0)is Hence, the slope of the tangent at the pointwhere the x-coordinate is 3 is given by,
Answer - 5 : -
It is given that x = acos3θ and y = asin3θ.
Therefore, the slope of thetangent at is given by,
Hence, the slope of thenormal at
Question - 6 : - Find the slope of thenormal to the curve x = 1 − a sin θ, y = b cos2θ at
Answer - 6 : -
It is given that x = 1− a sin θ and y = b cos2θ.
Therefore, the slope of thetangent at is given by,
Hence, the slope of thenormal at
Question - 7 : - Find points at which the tangent to thecurve y = x3 − 3x2 −9x + 7 is parallel to the x-axis.
Answer - 7 : - The equation of the givencurve is
Now, the tangent is parallel to the x-axisif the slope of the tangent is zero.
When x = 3, y =(3)3 − 3 (3)2 − 9 (3) + 7 = 27− 27 − 27 + 7 = −20.
When x = −1, y =(−1)3 − 3 (−1)2 − 9 (−1) + 7 = −1− 3 + 9 + 7 = 12.
Hence, the points at which the tangent isparallel to the x-axis are (3, −20) and
(−1, 12).
Find a point on the curve y =(x − 2)2 at which the tangent is parallel to the chordjoining the points (2, 0) and (4, 4).
Answer - 8 : -
If a tangent is parallel to the chord joining the points(2, 0) and (4, 4), then the slope of the tangent = the slope of the chord.
The slope of the chord is
Now, the slope of the tangent to the givencurve at a point (x, y) is given by,
Since the slope of the tangent = slope of the chord, wehave:
Hence, the required point is (3, 1).
Question - 9 : - Find the point on the curve y = x3 −11x + 5 at which the tangent is y = x −11.
Answer - 9 : -
The equation of the given curve is y = x3 −11x + 5.
The equation of the tangent to the givencurve is given as y = x − 11 (which is of theform y = mx + c).
∴Slope of the tangent = 1
Now, the slope of the tangent to the given curve atthe point (x, y) is given by, 
Then, we have:
When x = 2, y =(2)3 − 11 (2) + 5 = 8 − 22 + 5 = −9.
When x = −2, y =(−2)3 − 11 (−2) + 5 = −8 + 22 + 5 = 19.
Hence, the required points are (2, −9) and (−2, 19). But,both these points should satisfy the equation of the tangent as there would bepoint of contact between tangent and the curve. ∴ (2, −9) is the required point as (−2, 19) is notsatisfying the given equation of tangent.
Answer - 10 : - The equation of the givencurve is
The slope of the tangents to the given curveat any point (x, y) is given by,
If the slope of the tangent is −1, then we have:
When x = 0, y =−1 and when x = 2, y = 1.
Thus, there are two tangents to the given curve havingslope −1. These are passing through the points (0, −1) and (2, 1).
∴The equation of the tangent through (0, −1) is given by,
∴The equation of the tangent through (2, 1) is given by,
y − 1 =−1 (x − 2)
⇒ y − 1 = − x + 2
⇒ y + x − 3 = 0
Hence, the equations of the required linesare y + x + 1 = 0 and y + x −3 = 0.