Class 12 Maths Ncert Solutions Chapter 6 Pdf
NCERT Solutions for Class 12 Maths Exercise 6.3 Chapter 6 Application of Derivatives – FREE PDF Download
Free PDF download of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 (Ex 6.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.3) Exercise 6.3
1. Find the slope of tangent to the curve
at
Ans. Given: Equation of the curve ……….(i)
Slope of the tangent to the curve = Value of
at the point
Slope of the tangent at point
to the curve (i)
=
= 768 – 4 = 764
2. Find the slope of tangent to the curve
at
Ans. Given: Equation of the curve
……….(i)
=
=
……….(ii)
Slope of the tangent at point
to the curve (i)
=
=
3. Find the slope of tangent to the curve
at the given point whose
coordinate is 2.
Ans. Given: Equation of the curve ……….(i)
Slope of the tangent at point
to the curve (i)
=
= 12 – 1 = 11
4. Find the slope of tangent to the curve
at the given point whosecoordinate is 3.
Ans. Given: Equation of the curve……….(i)
Slope of the tangent at point
to the curve (i)
=
= 27 – 3 = 24
5. Find the slope of the normal to the curve
at
Ans. Given: Equations of the curves are
and
=
and
and
=
Slope of the tangent at
=
And Slope of the normal at
=
= 1
6. Find the slope of the normal to the curve
at
Ans. Given: Equations of the curves are
and
and
and
=
Slope of the tangent at
=
And Slope of the normal at
=
=
7. Find the point at which the tangent to the curve
is parallel to theaxis.
Ans. Given: Equation of the curve ……….(i)
Since, the tangent is parallel to theaxis, i.e.,
From eq. (i), when
when
Therefore, the required points are
and
8. Find the point on the curve
at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
Ans. Let the given points are A (2, 0) and B (4, 4).
Slope of the chord AB =
Equation of the curve is
Slope of the tangent at
=
If the tangent is parallel to the chord AB, then Slope of tangent = Slope of chord
Therefore, the required point is (3, 1).
9. Find the point on the curve
at which the tangent is
Ans. Given: Equation of the curve ……….(i)
Equation of the tangent
……….(ii)
From eq. (i),
= Slope of the tangent at
But from eq. (ii), the slope of tangent =
From eq. (i), when
And when
Since
does not satisfy eq. (ii), therefore the required point is
10. Find the equation of all lines having slope
that are tangents to the curve
Ans. Given: Equation of the curve
……….(i)
=
= Slope of the tangent at
But according to question, slope =
=
or
From eq. (i), when 
And when
Points of contact are (2, 1) and
And Equation of two tangents are
=
and
=
11. Find the equations of all lines having slope 2 which are tangents to the curve
Ans. Given: Equation of the curve
=
= Slope of the tangent at
But according to question, slope =
= 2
which is not possible.
Hence, there is no tangent to the given curve having slope 2.
12. Find the equations of all lines having slope 0 which are tangents to the curve
Ans. Given: Equation of the curve
……….(i)
=
=
But according to question, slope = 0
= 0
From eq. (i),
Therefore, the point on the curve which tangent has slope 0 is
Equation of the tangent is
13. Find the points on the curve
at which the tangents are:
(i) parallel toaxis
(ii) parallel to
axis
Ans. Given: Equation of the curve ……….(i)
……….(ii)
(i) If tangent is parallel toaxis, then Slope of tangent = 0
= 0
From eq. (i),
Therefore, the points on curve (i) where tangents are parallel toaxis are
.
(ii) If the tangent parallel toaxis.
Slope of the tangent =
d x d y = 0 dxdy=0
From eq. (ii),
From eq. (i),
Therefore, the points on curve (i) where tangents are parallel toaxis are
.
14. Find the equations of the tangent and normal to the given curves at the indicated points:
(i)
at (0, 5)
(ii) at (1, 3)
(iii)
at (1, 1)
(iv)
at (0, 0)
(v)
at
Ans. (i) Equation of the curve
slope of tangent =at (0, 5)
=
(say)
Slope of the normal at (0, 5) is
Equation of the tangent at (0, 5) is y − 5 = − 10 ( x − 0 ) y−5=−10(x−0)
y − 5 = − 10 x y−5=−10x
And Equation of the normal at (0, 5) is
(ii) Equation of the curve
slope of tangent = at (1, 3)
=
(say)
Slope of the normal at (1, 3) is
Equation of the tangent at (1, 3) is
And Equation of the normal at (1, 3) is
(iii) Equation of the curve ……….(i)
slope of tangent = at (1, 1)
=
=
(say)
Slope of the normal at (1, 1) is
Equation of the tangent at (1, 1) is
And Equation of the normal at (1, 1) is
(iv) Equation of the curve ……….(i)
slope of tangent = at (0, 0)
=
= (say)
Equation of the tangent at (0, 0) is
which is equaion of x-axis
thus normal at (0, 0) isaxis.
equation of normal is
x=0
(v) Equation of the curves are
and
Slope of the tangent at =
(say)
Slope of the normal at is
Point =
=
=
Equation of the tangent is
And Equation of the normal is
15. Find the equation of the tangent line to curve
which is:
(a) parallel to the line
(b) perpendicular to the line
Ans. Given: Equation of the curve……….(i)
Slope of tangent =
…….(ii)
(a) Slope of the line is
Slope of tangent parallel to this line is also = 2
From eq. (ii),
From eq. (i),
Therefore, point of contact is (2, 7).
Equation of the tangent at (2, 7) is
(b) Slope of the line
is
=
Slope of the required tangent perpendicular to this line =
From eq. (ii),
From eq. (i),
=
=
Therefore, point of contact is
Equation of the required tangent is
16. Show that the tangents to the curve
at the points where and
are parallel.
Ans. Given: Equation of the curve
Slope of tangent at =
At the point Slope of the tangent =
At the point Slope of the tangent =
Since, the slopes of the two tangents are equal.
Therefore, tangents at and are parallel.
17. Find the points on the curve at which the slope of the tangent is equal to thecoordinate of the point.
Ans. Given: Equation of the curve………(i)
Slope of tangent at
=……….(ii)
According to question, Slope of the tangent =coordinate of the point
or
or
From eq. (i), at The point is (0, 0).
And From eq. (i), at
The point is (3, 27).
Therefore, the required points are (0, 0) and (3, 27).
18. For the curve
find all points at which the tangent passes through the origin.
Ans. Given: Equation of the curve
……….(i)
Slope of the tangent at passing through origin (0, 0)
=
=
Substituting this value of
in eq. (i), we get,
or
or
From eq. (i), at
From eq. (i), at
From eq. (i), at 
Therefore, the required points are (0, 0), (1, 2) and
19. Find the points on the curve
at which the tangents are parallel toaxis.
Ans. Equation of the curve ……….(i)
[tangent is parallel toaxis]
From eq. (i),
Therefore, the required points are (1, 2) and
20. Find the equation of the normal at the point
for the curve
Ans. Given: Equation of the curve
……….(i)
Slope of the tangent at the point
=
=
Slope of the normal at the point =
Equation of the normal at
=
21. Find the equations of the normal to the curve
which are parallel to the line
Ans. Given: Equation of the curve ….(i)
Slope of the tangent at
=
Slope of the normal to the curve at
=
……….(ii)
But Slope of the normal (given) =
=
From eq. (i), at 
at 
Therefore, the points of contact are (2, 18) and
Equation of the normal at (2, 18) is
And Equation of the normal at
is
22. Find the equation of the tangent and normal to the parabola
at the point
Ans. Given: Equation of the parabola ……….(i)
Differentiating eq. (i) w.r.t
Slope of the tangent at the point
=
Slope of the normal =
Equation of the tangent at the point
=
And Equation of the normal at the point
=
23. Prove that the curves
and
cut at right angles if
Ans. Given: Equations of the curves are …..(i) and……….(ii)
Substituting the value of in eq. (ii), we get
Putting the value of in eq. (i), we get
Therefore, the point of intersection is =
……….(iii)
Differentiating eq. (i) w.r.t
……….(iv)
Differentiating eq. (ii) w.r.t
……….(v)
According to the question,
[From eq. (iii)]
[Cubing both sides]
24. Find the equations of the tangent and normal to the hyperbola x 2 a 2 − y 2 b 2 = 1 x2a2−y2b2=1 at the point
Ans. Given: Equation of the hyperbola
……….(i)
……….(ii)
Slope of tangent at
is
Equation of the tangent at is
……….(iii)
Since lies on the hyperbola (i), therefore,
From eq. (iii),
Now, Slope of normal at =
Equation of the normal at is y − y 0 = − a 2 y 0 b 2 x 0 ( x − x 0 ) y−y0=−a2y0b2x0(x−x0)
Dividing both sides by
25. Find the equation of the tangent to the curve
which is parallel to the line
Ans. Given: Equation of the curve ……….(i)
Slope of the tangent at point is
d y d x = 1 2 ( 3 x − 2 ) − 1 / 2 d ( 3 x − 2 ) d x dydx=12(3x−2)−1/2d(3x−2)dx
=
……….(ii)
Again slope of the line
is
……….(iii)
According to the question,
[Parallel lines have same slope]
From eq. (i),
=
=
Therefore, point of contact is
.
Equation of the required tangent is
Choose the correct answer in Exercises 26 and 27.
26. The slope of the normal to the curve
at is:
(A) 3
(B)
(C)
(D)
Ans. Given: Equation of the curve……….(i)
Slope of the tangent at point is
Slope of the tangent at 
(say)
Slope of the normal =
Therefore, option (D) if correct.
27. The line
is a tangent to the curve
at the point:
(A) (1, 2)
(B) (2, 1)
(C)
(D)
Ans. Given: Equation of the curve
……….(i)
Slope of the tangent at point is 2 y d y d x = 4 2ydydx=4
……….(ii)
Slope of the line
is……….(iii)
From eq. (ii) and (iii),
From eq. (i),
Therefore, required point is (1, 2).
Therefore, option (A) is correct.
Class 12 Maths Ncert Solutions Chapter 6 Pdf
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