2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3

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Karnataka 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.3

Question 1.
Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4.
Answer:

Question 2.
Find the slope of the tangent to the curve
\(y=\frac{x-1}{x-2}, x \neq 2 \text { at } x=10 \)
Answer:

Question 3.
Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x- coordinate is 2.
Answer:

Question 4.
Find the slope of the tangent to the curve y = x³ – 3x + 2 at the point whose x-coordinate is 3.
Answer:
\(\frac{d y}{d x}\) = 3x² – 3 dx
slope at x = 3 is 3 (9) – 3 = 24.

Question 5.
Find the slope of the normal to the curve
\(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta \text { at } \theta=\frac{\pi}{4}\)
Answer:

Question 6.
Find the slope of the normal to the curve
\(x=1-a \sin \theta, y=b \cos ^{2} \theta \text { at } \theta=\frac{\pi}{2}\)
Answer:

Question 7.
Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to
the x-axis.
Answer:
\(\frac{d y}{d x}\) = 3x² – 6x – 9, slope of the tangent
since the tangent is parallel to x-axis \(\frac{d y}{d x}\) = 0
3x² – 6x – 9 = 0
3 (x + 1) (x – 3) = 0, x = 3, x = -1
when x = 3, y = 27 – 27 – 27 4- 7 = -20
when x = -1, y = -1 -3 + 9 + 7 = 12
The points at which the tangent parallel to x – axis are (3, -20) and (-1, 12).

Question 8.
Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Answer:

Question 9.
Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.
Answer:
Slope of the line y = x – 11 is 1
slope of the curve \(\frac{d y}{d x}\) = 3x² – 11
∴ 3x² – 11 = 1                   ‘
3x² = 12 ⇒ x² = 4, x = +2
when x = 2, y = (2)³ – 11 (2) + 5 = -9
when x = -2, y = -8 + 22 + 5 = 19
points are (2, -9) and (-2, 19)
equation of tangent at (2, -9) and slope is 1
y + 9 = 1 (x – 2)
y = x- 11 equation of tangent at (-2, 19)
y – 19 = 1 (x + 2) ⇒ y = x + 211
∴ (2, -9) is the only point at which the tangent is y = x – 11.10.

Question 10.
Find the equation of all lines having slope – 1 that are tangents to the curve
\(y=\frac{1}{x-1}, x \neq 3\)
Answer:

x = 2, x = 0
when x = 2, y = 1
when x = 0, y = -1
The points are (2, 1) and (0,-1)
equation of line though (2,1) having slope = -1
is y – 1 = -1 x (x – 2) ⇒ y + x = 3
or x + y – 3 = 0 and equation of line through (0,- 1)y + 1 = -1 (x – 0)
∴  y + x + 1 = 0.

Question 11.
Find the equation of all lines having slope 2 which are tangents to the curve
\(y=\frac{1}{x-3}, x \neq 3\)
Answer:
slope of the line = 2
slope of the curve = \(\frac{-1}{(x-3)^{2}}\)
\(\frac{-1}{(x-3)^{2}}\) =2
⇒ 2 (x -3)² = -1
⇒ 2 (x² – 6x + 9) = -1
⇒ 2x² – 12x + 19 = 0
which has no real roots b² – 4ac < 0
hence there is no point on the curve.

Question 12.
Find the equations of all lines having slope 0 which are tangent to the curve
\(y=\frac{1}{x^{2}-2 x+3}\)
Answer:

Question 13.
Find the points of curve \(\frac{x^{2}}{9}+\frac{y^{2}}{16}=1\) which the tangents are
(i) parallel to x-axis
(ii) parallel to y-axis.
Answer:
16x² + 9y² = 144
16 x 2 x + 9 x 2y x \(\frac{d y}{d x}\) = 0

(0,4) and (0, -4) are the points on the curve at which tangent is parallel to x – axis.

(ii) For tangents parallel to y – axis \(\frac{d y}{d x}\) = \(\frac{1}{0}\)
\(\frac{-16 x}{9 y}=\frac{1}{0} \Rightarrow y=0\)
when y = 0, x² = 9x = ± 3 point is (3,0) and (-3,0)
(3, 0) and (-3, 0) are the points at which the curve is parallel to y – axis.

Question 14.
Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)
Answer:
y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5) dy
\(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10 dx
slope at (0,5) = -10
∴ Equation of tangent at (0, 5) is
y – 5 = -10 (x – 0)
y – 5 = -10 x 10 x + y – 5 = 0
slope of the normal at (0,5)
\((0,5)=\frac{-1}{-10}=\frac{1}{10}\)
∴ equation of normal is
y – 5 =\(\frac{1}{10}\) (x – 0) = 10y – 50 = x
x – 10y + 50 = 0

(ii) y = x⁴ – 6x³ + 13x² – 10x + 5 at (1, 3)
Answer:
\(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10 dx
slope of the tangent at x = 1

(iii) y = x³ at (1,1)
Answer:
\(\frac{d y}{d x}=3 x^{2}\) 
∴ slope of the tangent at x = 1 = 3
∴ Equation of tangent is y – 1 = 3(x – 1)
3x – y – 2 = 0
∴ slope of normal = \(\frac{-1}{3} \)
∴ equation of normal is y – 1 = \(\frac{-1}{3} \)(x – 1)
3y – 3 = – (x – 1)
x + 3y – 4 = 0.

(iv) y = x² at (0, 0)
Answer:
\(\frac{d y}{d x}=2 x\)
∴ slope at x = 0 =0
∴ Equation of tangent is y – 0 = 0 (x – 0) ⇒ y = 0
slope of the normal = -1/(0)
∴ equation of normal \(y – 0=\frac{-1}{0}(x-0) \Rightarrow x=0\)

(v) x = cos t, y = sin t at \(\frac{\pi}{4}\)
Answer:

Question 15.
Find the equation of the tangent line to the curve y = x² – 2x +7 which is
(a) parallel to the line 2x – y + 9 = 0
Answer:
\(\frac{d y}{d x}\) = 2x – 2 also slope = 2
2 x – 2 = 2 ⇒2x = 4 ⇒ x = 2
when x = 2, y = (2)² – 2 (2) + 7 = 7
∴ equation of tangent is y – 7 = 2 (x – 2)
2x – y + 3 = 0 ⇒ 2x – y + 3 = 0.