Prasanna

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1

Question 1.
Find all the values of x such that
(i) -10π ≤ x ≤ 10π and sin x = 0
(ii) -8π ≤ x ≤ 8π and sin x = -1
Solution:
(i) sin x = 0
⇒ x = nπ
where n = 0, ±1, ±2, ±3, ……., ±10
(ii) sin x = -1
⇒ x = (4n – 1) \(\frac{\pi}{2}\), n = 0, ±1, ±2, ±3, 4

Question 2.
Find the period and amplitude of
(i) y = sin 7x
(ii) y = -sin(\(\frac{1}{3}\)x)
(iii) y = 4 sin(-2x)
Solution:
(i) y = sin 7x
Period of the function sin x is 2π
Period of the function sin 7x is \(\frac{2 \pi}{7}\)
The amplitude of sin 7x is 1.

(ii) y = -sin\(\frac{1}{3}\)x
Period of sin x is 2π
So, period of sin\(\frac{1}{3}\)x is 6π and the amplitude is 1.

(iii) y = 4 sin(-2x) = -4 sin 2x
Period of sin x is 2π
π Period of sin 2x is π and the amplitude is 4.

Question 3.
Sketch the graph of y = sin(\(\frac{1}{3}\)x) for 0 ≤ x < 6π.
Solution:
The period of sin(\(\frac{1}{3}\)x) is 6π and the amplitude is 1.

The graph is

Question 4.
Find the value of
(i) \(\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)\)
(ii) \(\sin ^{-1}\left(\sin \left(\frac{5 \pi}{4}\right)\right)\)
Solution:

Question 5.
For that value of x does sin x = sin^-1 x?
Solution:
sin x = sin^-1 x is possible only when x = 0 (∵ x ∈ R)

Question 6.
Find the domain of the following
(i) \(f(x)=\sin ^{-1}\left(\frac{x^{2}+1}{2 x}\right)\)
(ii) \(g(x)=2 \sin ^{-1}(2 x-1)-\frac{\pi}{4}\)
Solution:
(i) \(f(x)=\sin ^{-1}\left(\frac{x^{2}+1}{2 x}\right)\)
The range of sin-1 x is -1 to 1
\(-1 \leq \frac{x^{2}+1}{2 x} \leq 1\)
⇒ \(\frac{x^{2}+1}{2 x} \geq-1\) or \(\frac{x^{2}+1}{2 x} \leq 1\)
⇒ x² + 1 ≥ -2x or x² + 1 ≤ 2x
⇒ x² + 1 + 2x ≥ 0 or x² + 1 – 2x ≤ 0
⇒ (x+ 1)² ≥ 0 or (x – 1)² ≤ 0 which is not possible
⇒ -1 ≤ x ≤ 1 or
(ii) \(g(x)=2 \sin ^{-1}(2 x-1)-\frac{\pi}{4}\)
-1 ≤ (2x – 1) ≤ 1
0 ≤ 2x ≤ 2
0 ≤ x ≤ 1
x ∈ [0, 1]

Question 7.
Find the value of \(\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} \sin \frac{\pi}{9}\right)\)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.1 Additional Problems

Question 1.

Solution:

Question 2.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.6

Solve the following differential equations:

Question 1.

Solution:

This is a Homogeneous differential equation

Seperating the variables

Question 2.
(x³ + y³) dy – x²y dx = 0
Solution:
(x³ + y³) dy – x²y dx = 0
(x³ + y³) dy = x³y dx

This is a Homogeneous differential equation.

Separating the variables

Question 3.

Solution:

This is a Homogeneous differential equation

Seperating the variables

Question 4.
2xydx + (x² + 2y²) dy = 0
Solution:
2xy dx + (x² + 2y²) dy = 0
(x² + 2y²) dy = – 2xy dx

This is a Homogeneous differential equation


Separating the variables

Question 5.
(y² – 2xy)dx = (x² – 2xy)dy
Solution:
(y² – 2xy) dx = (x² – 2xy) dy

This is a Homogeneous differential equation


Seperating the variables

Question 6.

Solution:

Seperating the variables

Question 7.

Solution:

This is a Homogeneous differential equation

Separating the variables

Question 8.
(x² + y²) dy = xy dx. It is given that y(1) = 1 and y(x₀) = e. Find the value of x₀.
Solution:

This is a Homogeneous differential equation

Seperating the variables

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.6 Additional problems

Question 1.
Solve: (2\(\sqrt{x y}\))dy + y dx = 0
Solution:

Question 2.
Solve: (x³ + 3xy²)dx + (y³ + 3x²y)dy = 0
Solution:


Integrating, we have

Question 3.

Solution:

Which is in terms of v alone.
⇒ the given problem comes under homogeneous type.

Question 4.
Solve: (x² + y²)dy = xy dx
Solution:

Which is a function in v alone.
⇒ the given problem comes under homogeneous type.

Question 5.
Find the equation of the curve passing through (1, 0) and which has slope \(1+\frac{y}{x}\) at (x, y).
Solution:


Given the curve passes through (1, 0) ⇒ at x = 1, y = 0
0 = log 1 +c ⇒ c = 0

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10

Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
If \(\vec{a}\) and \(\vec{b}\) are parallel vector, then \([\vec{a}, \vec{c}, \vec{b}]\) is equal to ………………..
(a) 2
(b) -1
(c) 1
(d) 0
Solution:
(d) 0
Hint:
\(\vec{a}\) and \(\vec{b}\) are parallel vectors, so \(\vec{a} \times \vec{b}\) = 0
then \([\vec{a} \vec{c} \vec{b}]=-[\vec{a} \vec{b} \vec{c}]=-(\vec{a} \times \vec{b}) \cdot \vec{c}\) = 0

Question 2.
If a vector \(\vec{\alpha}\) lies in the plane of \(\vec{\beta}\) and \(\vec{\gamma}\), then …………

Solution:
(c) \([\vec{\alpha}, \vec{\beta}, \vec{\gamma}]\) = 0
Hint:
Since \([\vec{\alpha}, \vec{\beta}, \vec{\gamma}]\) are lie in the same plane
so \([\vec{\alpha}, \vec{\beta}, \vec{\gamma}]\) = 0

Question 3.
If \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}\)= 0, then the value of \(|[\vec{a}, \vec{b}, \vec{c}]|\) is ……………….
(a) \(|\vec{a}||\vec{b}||\vec{c}|\)
(b) \(\frac{1}{3}|\vec{a}||\vec{b}||\vec{c}|\)
(c) 1
(d) -1
Solution:
(a) \(|\vec{a}||\vec{b}||\vec{c}|\)
Hint:

Question 4.
If \(\vec{a}, \vec{b}, \vec{c}\) are three unit vectors such that \(\vec{a}\) is perpendicular to \(\vec{b}\), and is parallel to \(\vec{c}\) then \(\vec{a} \times(\vec{b} \times \vec{c})\)is equal to ………………….
(a) \(\vec{a}\)
(b) \(\vec{b}\)
(c) \(\vec{c}\)
(d) \(\vec{0}\)
Solution:
(b) \(\vec{b}\)
Hint:

Question 5.

(a) 1
(b) -1
(c) 2
(d) 3
Solution:
(a) 1
Hint:

Question 6.
The volume of the parallelepiped with its edges represented by the vectors \(\hat{i}+\hat{j}, i+2 \hat{j}\), \(\hat{i}+\hat{j}+\pi \hat{k}\) is ……………

Solution:
(c) π
Hint:
Volume = \([\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}{1} & {1} & {0} \\ {1} & {2} & {0} \\ {1} & {1} & {\pi}\end{array}\right|\)
= π(2 – 1) = π cubic units

Question 7.
If \(\vec{a}\) and \(\vec{b}\) are unit vectors such that \([\vec{a}, \vec{b}, \vec{a} \times \vec{b}]=\frac{\pi}{4}\) then the angle between \(\vec{a}\) and \(\vec{b}\) is …………

Solution:
(a) \(\frac{\pi}{6}\)
Hint:

Question 8.

(a) 0
(b) 1
(c) 6
(d) 3
Solution:
(a) 0
Hint:

Equate corresponding coefficients on both sides
λ + µ = 0 and λ = -1 this gives µ = 1
∴ Then the value of λ + µ = 0.

Question 9.
If \(\vec{a}, \vec{b}, \vec{c}\) are three non-coplanar vectors such that \(\vec{a} \times(\vec{b} \times \vec{c})=\frac{\vec{b}+\vec{c}}{\sqrt{2}}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is …………
(a) 81
(b) 9
(c) 27
(d) 18
Solution:
(a) 81
Hint:

Question 10.

Solution:
(b) \(\frac{3 \pi}{4}\)
Hint:

Question 11.
If the volume of the parallelepiped with \(\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}\) as coterminous edges is 8 cubic units, then the volume of the parallelepiped with \((\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c}),(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})\) and \((\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})\) as coterminous edges is ………………
(a) 8 cubic units
(b) 512 cubic units
(c) 64 cubic units
(d) 24 cubic units
Solution:
(c) 64 cubic units
Hint:

Question 12.
Consider the vectors \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) such that \((\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\overrightarrow{0}\). Let P₁ and P₂ be the planes determined by the pairs of vectors \(\vec{a}, \vec{b}\) and \(\vec{c}, \vec{d}\) respectively. Then the angle between P₁ and P₂ is ……………..
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Solution:
(a) 0°
Hint:

Question 13.

(a) perpendicular
(b) parallel
(c) inclined at an angle \(\frac{\pi}{3}\)
(d) inclined at an angle \(\frac{\pi}{6}\)
Solution:
(b) parallel
Hint:

Question 14.
If , then a vector perpendicular to \(\vec{a}\) and lies in the plane containing \(\vec{b}\) and \(\vec{c}\) is

Solution:
(d) \(-17 \hat{i}-21 \hat{j}-97 \hat{k}\)
Hint:

Question 15.
The angle between the lines is ………..

Solution:
(d) \(\frac{\pi}{2}\)
Hint:

Question 16.
If the line lies in the plane x + 3 + αz + β = 0, then (α, β) is …………..
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Solution:
(d) (-6, 7)
Hint:
d.c.s of the first line = (3, -5, 2)
d.c.s of the line perpendicular to plane = (1, 3, -α)
a₁a₂ + b₁b₂ + c₁c₂ = 0
3 – 15 – 2α = 0 => -12 – 2α = 0
-2α =12 => α = -6
Plane passes through the point (2, 1, -2) so
2 + 3 + 6(-2) + β = 0 => β = 7
(α, β) = (-6, 7)

Question 17.
The angle between the line \(\vec{r}=(\hat{i}+2 \hat{j}-3 \hat{k})+t(2 \hat{i}+\hat{j}-2 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}+\hat{j})+4\) = 0 is ……………
(a) 0°
(b) 30°
(c) 45°
(d) 90°
Solution:
(c) 45°
Hint:

Question 18.
The coordinates of the point where the line \(\vec{r}=(6 \hat{i}-\hat{j}-3 \hat{k})+t(-\hat{i}+4 \hat{k})\) meets the plane \(\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})\) are
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(c) (5, -1, 1)
Solution:
(d) (5, -1, 1)
Hint:
Cartesian equation of the line
\(\frac{x-6}{-1}=\frac{y+1}{0}+\frac{z+3}{4}\) = λ
(-λ + 6, -1, 4λ – 3)
This meets the plane x + y – z = 3
-λ + 6 – 1 – 41 + 3 = 3 ⇒ -5λ = -5
λ = 1
The required point (5, -1, 1).

Question 19.
Distance from the origin to the plane 3x – 6y + 2z + 7 = 0 is ………………
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:
Distance from the origin (0, 0, 0) to the plane

Question 20.
The distance between the planes x + 2y + 3z + 7 = 0 and 2x + 4y + 6z + 7 = 0 is ……………

Solution:
(a) \(\frac{\sqrt{7}}{2 \sqrt{2}}\)
Hint:
x + 2y + 3z + 1 = 0; 2x + 4y + 6z + 7 = 0
Multiplying 2 on both sides
2x + 4y + 6z + 14 = 0 .
a = 2, b = 4, c = 6, d₁ = 14, d₂ = ?

Question 21.
If the direction cosines of a line are \(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\), then ……………….
(a) c = ±3
(b) c = ± \(\sqrt{3}\)
(c) c > 0
(d) 0 < c < 1
Solution:
(b) c = ± \(\sqrt{3}\)
Hint:
We know that sum of the squares of direction cosines = 1

Question 22.
The vector equation \(\vec{r}=(\hat{i}-2 \hat{j}-\hat{k})+t(6 \vec{j}-\hat{k})\) represents a straight line passing through the points ……………. (a) (0, 6, -1) and (1, -2, -1) (b) (0, 6, -1) and (-1, -4, -2) (c) (1, -2, -1) and (1, 4, -2) (d) (1, -2, -1) and (0, -6, 1)
Solution:
(c) (1, -2, -1) and (1, 4, -2)
Hint:
The required vector equation is \(\vec{r}=\vec{a}+t(\vec{b}-\vec{a})\)

From (1) and (2) The points are (1, -2, -1) and (1, 4, -2)

Question 23.
If the distance of the point (1, 1, 1) from the origin is half of its distance from the plane x + y + z + k= 0, then the values of k are ………….. (a) ±3 (b) ±6 (c) -3, 9 (d) 3, -9
Solution:
(d) 3, -9
Hint:

Question 24.
If the planes \(\vec{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3 \text { and } \vec{r} \cdot(4 \hat{i}+\hat{j}-\mu \hat{k})\) = 5 are parallel, then the value of λ and µ are ……………
Solution:
(c) \(-\frac{1}{2}\), -2
Hint:

Question 25.
If the length of the perpendicular from the origin to the plane 2x + 3y + λz = 1, λ > 0 is \(-\frac{1}{5}\), then the value of λ is …………..
(a) \(2 \sqrt{3}\)
(b) \(3 \sqrt{2}\)
(c) 0
(d) 1
Solution:
(a) \(2 \sqrt{3}\)
Hint:
Given length of perpendicular from origin to the plane = \(-\frac{1}{5}\)

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.10 Additional Problems

Question 1.

(a) 6
(b) 10
(c) 12
(d) 24
Solution:
(c) 12
Hint:

Question 2.

(a) only x
(b) only y
(c) Neither x nor y
(d) Both x and y
Solution:
(c) Neither x nor y
Hint:

Question 3.

(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 3
Hint:

Question 4.
The value of = ………………
(a) 1
(b) 3
(c) -3
(d) 0
Solution:
(b) 3
Hint:

Question 5.
Let a, b, c be distinct non-negative numbers. If the vectors lie in a plane, then c is ……………..
(a) the A.M. of a and b
(b) the G.M. of a and b
(c) the H.M. of a and b
(d) equal to zero.
Solution:
(b) the G.M. of a and b
Hint:

Question 6.
The value of \(\hat{i} \cdot(\hat{j} \times \hat{k})+(\hat{i} \times \hat{k}) \cdot \hat{j}\) ………….
(a) 1
(b) -1
(c) 0
(d) \(\hat{j}\)
Solution:
(c) 0
Hint:

Question 7.
The value of \((\hat{i}-\hat{j}, \hat{j}-\hat{k}, \hat{k}-\hat{i})\) is …………..
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:

Question 8.

Solution:
(c) \(\vec{u}=\overrightarrow{0}\)
Hint:

Question 9.
The area of the parallelogram having a diagonal \(3 \vec{i}+\vec{j}-\vec{k}\) and a side \(\vec{i}-3 \vec{j}+4 \vec{k}\) is ………………

Solution:
(d) \(3 \sqrt{30}\)
Solution:

Question 10.
If , then ……………….

Solution:
(d) \(\vec{x}=\overrightarrow{0} \text { or } \vec{y}=\overrightarrow{0} \text { or } \vec{x} \text { and } \vec{y}\) are parallel
Hint:

Question 11.
If \(\overrightarrow{\mathrm{PR}}=2 \vec{i}+\vec{j}+\vec{k}, \overrightarrow{\mathrm{Question}}=-\vec{i}+3 \vec{j}+2 \vec{k}\), then the area of the quadrilateral PQRS is ………………..

Solution:
(c) \(\frac{5 \sqrt{3}}{2}\)
Hint:

Question 12.

Solution:
(c) \(\vec{c}\) parallel to \(\vec{a}\)
Hint:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7

Question 1.
Find intervals of concavity and points of inflexion for the following functions:
(i) f(x) = x (x – 4)³
(ii) f(x) = sin x + cos x, 0 < x < 2π

Solution:
(i) f(x) = x(x – 4)³
f(x) = x[x³ – 12x² + 48x – 64]
(i.e.,) f(x) = x⁴ – 12x³ + 48x² – 64x
f'(x) = 4x³ – 36x² + 96x – 64
f”(x) = 12x² – 72x + 96
f”(x) = 0 ⇒ 12(x² – 6x + 8) = 0
⇒ 12 (x – 2) (x – 4) = 0 ⇒ x = 2 or 4
Marking the points in the number line

in the interval (-∞, 2) the curve concave upwards
when x ∈ (2, 4), f”(x) = (3 – 2) (3 – 4)
[say x = 3] = -ve
⇒ The curve concave upwards in (2,4)
when x ∈ (4, ∞), f”(x) = (5 – 2) (5 – 4)
[say x = 5] = (+) (+) = +ve ⇒ The curve concave downwards is (4, ∞) in (-∞, 2) concave upwards and in (2, 4) the concave downwards
⇒ x = 2 is a point of inflection at x = 2, f(2) = -16
So (2, -16) is a point of inflection
Again in (2, 4) the curve concave downwards and in (4, ∞) the curve concave upwards ⇒ x = 4 is a point of inflection f(4) = 0
∴ (4, 0) is a point of inflection
So points of inflection are (2, -16) and (4, 0)

(ii) f(x) = sin x + cos x
f'(x) = cos x – sin x
f”(x) = – sin x – cos x
f”(x) = 0 ⇒ – sin x = cos x



So the intervals are (-∞, 0), (0, ∞) when x ∈ (-∞, 0) f”(x) is negative
⇒ the curve concave downwards and when x ∈ (0, ∞) f”(x) is positive
⇒ the curve concave upwards
⇒x = 0 is a point of inflection f(0) = \(\frac{1}{2}\)(1 – 1) = 0
So (0, 0) is the point of inflection and the curve concave upwards in (0, ∞) and curve concave downwards in (-∞, 0) and (0, 0) is the point of inflection.

Question 2.
Find the local extrema for the following functions using second derivative test:
(i) f(x) = – 3x⁵ + 5x³
(ii) f(x) = x log x
(iii) f(x) = x² e^-2x
Solution:
(i) f(x) = – 3x⁵ + 5x³
f'(x) = 0, f”(x) = -ve at x = a
⇒ x = a is a maximum point
f'(x) = 0, f”(x) = +ve at x = 6
⇒ x = b is a minimum point
f(x) = – 3x⁵ + 5x³
f’ (x) = -15x⁴ + 15x²
f”(x) = -60x³ + 30x
f'(x) = 0 ⇒ – 15x² (x² – 1) = 0
⇒ x = 0, +1, -1
at x = 0, f”(x) = 0
at x = 1, f”(x) = -60 + 30 = – ve
at x = -1, f”(x) = 60 – 30 = + ve
So at x = 1, f'(x) = 0 and f”(x) = -ve
⇒ x = 1 is a local maximum point.
and f(1) = 2
So the local maximum is (1, 2)
at x = -1, f'(x) = 0 and f”(x) = +ve
⇒ x = -1 is a local maximum point and f(-1) = -2.
So the local minimum point is (-1, -2)
∴ local minimum is -2 and local maximum is 2.

(ii) f(x) = x log x

(iii) f(x) = x² e^-2x
f'(x) = x²[-2e^-2x] + e^-2x (2x)
= 2e^-2x (x – x²)
f”(x) = 2e^-2x(1 – 2x) + (x – ²) (-4e^-2x)
= 2e^-2x [(1 – 2x) + (x – x²) (- 2)]
= 2e^-2x [2x² – 4x + 1]
f'(x) = 0 ⇒ 2e^-2x(x – x²) = 0
⇒ x (1 – x) = 0
⇒ x = 0 or x = 1
at x = 0, f”(x) = 2 × 1 [0 – 0 + 1] = +ve
⇒ x = 0 is a local minimum point and the minimum value is f(0) = 0 at x = 1,
f”(x) = 2e^-2 [2 – 4 + 1] = -ve
⇒ x = 1 is a local maximum point and the maximum value is f(1) = \(\frac{1}{e^{2}}\)
Local maxima \(\frac{1}{e^{2}}\) and local minima = 0

Question 3.
For the function f(x) = 4x³ + 3x² – 6x + 1 find the intervals of monotonicity, local extrema, intervals of concavity and points of inflection.
Solution:
4x³ + 3x² – 6x + 1
f'(x) = 12x² + 6x – 6
f”(x) = 24x + 6
f'(x) = 0 ⇒ 6(2x² + x – 1) = 0
6(x + 1)(2x – 1) = 0

From (1) and (2)
x = -1 is a maximum point,
and f(-1)= -4 + 3 + 6 + 1 = 6
So local maximum is 6

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.7 Additional Problems

Question 1.
Discuss the curves y = x⁴ – 4x³ with respect to concavity and points of inflection.
Solution:
f(x) = x⁴ – 4x³ ⇒ f'(x) = 4x³ – 12x²
f”(x) = 12x² – 24x = 12x (x – 2)
Since f”(x) = 0 when x = 0 or 2, we divide the real line into three intervals.
(-∞, 0) , (0, 2), (2, ∞) and complete the following chart.

The point (0, f(0) i.e., (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also (2, f(2)) i.e., (2, -16) is an inflection point since the curve changes from concave downward to concave upward there.

Question 2.
Find the intervals of concavity and the points of inflection of the following functions.
f(x) = 2x³ + 5x² – 4x
Solution:
y = f (x) = 2x³ + 5x² – 4x
f'(x) = 6x² + 10x – 4
f”(x) = 12x + 10; f'” (x) = 12
f”(x) = 0 ⇒ 12x + 10 = 0

Consider x in (-∞, -5/6) say x = -1
f”(x)= -12 + 10 = -2 < 0 ⇒ the curve convex upward in the interval (-∞, -5/6) Consider x in (-5/6, ∞) say x = 0 f”(x) = 0 + 10 = 10 > 0
⇒ the curve is concave upward in (-5/6, -∞)
Thus, the curve is concave upward in (-5/6, ∞) and convex upward in (-∞, – 5/6)

Question 3.
f(x) = x⁴ – 6x²
Solution:

So, the points of inflection are at x = ± 1.
f”(x) = 0 ⇒ x² = 1 ⇒ x = ± 1
So, the intervals are (-∞, -1), (-1, 1), (1, ∞)
When x ∈ (-∞, -1), say x = -2
f”(x) = 12(-2)² – 12 = 48 – 12 = 36 > 0
⇒ f(x) is concave upward in (-∞, -1) when x ∈ (-1, 1) say x = 0.
f”(x) = 0 – 12 = -12 < 0
⇒ f(x) convex upward.
⇒ x = -1 is a point of inflection. Again when x ∈ (-1, 1), the curve is convex upward and when x ∈ (1, ∞) say x = 2.
f”(x) = (1, ∞) say x = 2.
f”(x) = 12 (4) – 12 > 0
⇒ the curve is concave upward.
⇒ x = 1 is a point of inflection.
Thus x = ± 1 are the points of inflection.
At x = 1, f(x) = 1 – 6 = -5
At x = -1, f(x) = (-1)⁴ – 6(-1)² = 1 – 6 = -5
The curve concave upward in (-∞, -1) u (1, ∞) and convex upward in (-1, 1) and the points of inflection are (1, -5) and (-1, -5).

Question 4.
f(θ) = sin 2θ in (θ, π)
Solution:
f'(θ) = sin 2θ
f’ (θ) = (cos 2θ) (2) = 2 cos 2θ
f”(θ)= 2[-sin 2θ] (2) = – 4 sin 2θ
f”(θ) = 0 ⇒ – 4 sin 2θ = 0

Question 5.
y = 12x² – 2x³ – x⁴
Solution:

So the intervals are (-∞, -2), (-2, 1)(1, ∞)
When x ∈ (-∞, -2) say x = -3

The curve is concave upward.
When x ∈ (-2, 1) say x = 0.

⇒ The curve is concave upward.

⇒ The curve is convex upward. So, x = 1 is a point of inflection.
At x = – 2, y = 12(4) – 2(-8) – (16) = 48 + 16 – 16 = 48
At x = 1, y = 12(1) – 2(1) – 1 = 12 – 2 – 1 = 9
So, the points of inflection are (1, 9) and (-2, 48)

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8

Question 1.
Find two positive numbers whose sum is 12 and their product is maximum.
Solution:
Let the two numbers be x, 12 – x.
Their product p = x (12 – x) = 12x – x²
To find the maximum product.
p'(x) = 12 – 2x
p”(x) = -2
p'(x) = 0 ⇒ 12 – 2x = 0 ⇒ 2x = 12
⇒ x = 6
at x = 6, p”(x) = -2 = -ve
⇒ p is maximum at x = 6
when x = 6, 12 – x = 12 – 6 = 6
So the two numbers are 6, 6

Question 2.
Find two positive numbers whose product is 20 and their sum is minimum.
Solution:

but x is positive (given)

Question 3.
Find the smallest possible value of x² + y² given that x + y = 10.
Solution:
Given x + y = 10 ⇒ 7 = 10 – x
To find the smallest value of x² + y²

at x = 5, f”(x) = 4 at x = 5, y = 10 – 5 = 5 = +ve
x = 5 is a minimum point.
So the minimum value of x² + y² = 5² + 5² = 50

Question 4.
A garden is to be laid out in a rectangular area and protected by wire fence. What is the largest possible area of the fenced garden with 40 metres of wire.
Solution:
Perimeter = 40 m
2(l + b) = 40 ⇒ l + b = 20
Let l = x m
b = (20 – x)m
Area = l × b = x(20 – x) = 20x – x²
To find the maximum area
A(x) = 20x – x²
A'(x) = 20 – 2x
A”(x) = -2
A'(x) = 0 ⇒ 20 – 2x = 0
⇒ x = 10
x = 10 is a maximum point
:. Maximum Area = x (20 – x)
= 10(20 – 10)
= 10 × 10 = 100 sq.m.

Question 5.
A rectangular page is to contain 24 cm² of print. The margins at the top and bottom of the page are 1.5 cm and the margins at other sides of the page is 1 cm. What should be the dimensions of the page so that the area of the paper used is minimum.
Solution:
Let the length of the printed page be = x cm
and breadth = y cm
Now xy = 24
⇒ y = \(\frac{24}{x}\)
The length of the paper = y + 3
Area A = (x + 2)(y + 3)
= xy + 3x + 2y + 6
= 24 + 3x + 2y + 6
= 3x + 2y + 30 ……. (2)
Substituting (1) in (2) we get

∴ Dimensions of the paper are
x + 2 = 4 + 2 = 6 cm
and y + 3 = 6 + 3 = 9 cm

Question 6.
A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?
Solution:
Given Area = 180000 sq. mtrs
Let length be = x
and breadth be = y

Question 7.
Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius 10 cm.
Solution:
P is a point on the circumference of a circle of radius 10 cm P = (10 cos α, 10 sin α)
∴ PQ = 20 sinα and
PS = 20 cos α
A = area of PQRS = (20 sin α) (20 cos α)
= 400 sin α cosc α
= (200) (2 sin α cos α)

Question 8.
Prove that among all the rectangles of the given perimeter, the square has the maximum area.
Solution:
Let the length and breadth of the rectangle be x and y respectively.
P = 2(x + y) [given]

Substitute (3) in (2) we get

so x = 6 cm and y = 3 cm

Question 9.
Find the dimensions of the largest rectangle that can be inscribed in a semi circle of radius r cm.
Solution:
Let θ be the angle made by OP with the positive direction of x -axis.
Then the area of the rectangle A is

A(θ) = (2 r cos θ)(r sin θ)
= r² 2 sin θ cos θ = r² sin 2θ
Now A(θ) is maximum when sin 2θ is maximum. The maximum value of

Question 10.
A manufacturer wants to design an open box having a square base and a surface area of 108 sq.cm. Determine the dimensions of the box for the maximum volume.
Solution:
Let the side of the square base be = x cm and the height be = y cm
Surface area = 108 sq cm

Substituting (3) in (2) we get

so x = 6 cm and y = 3 cm

Question 11.
The volume of a cylinder is given by the formula V = πr²h. Find the greatest and least values of V if r + h = 6.
Solution:
V = πr²h
Given r + h = 6 ⇒ h = 6 – r
V = πr²(6 – r) = 6πr² – πr³

3πr(4 – r) = 0 ⇒ r = 0 or 4
when r = 4, h = 2
So v = π(16)(2) = 32π
when r = 0, V = 0
So the maximum volume = 32π and the minimum volume = 0

Question 12.
A hollow cone with base radius a cm and height b cm is placed on a table. Show that the
volume of the largest cylinder that can be hidden underneath is — times volume of the cone.
Solution:
The height of cone = h = b
The base radius = r = a
The base radius of cylinder = r
The height of cylinder = h

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.8 Additional Problems

Question 1.
The top and bottom margins of a poster are each 6 cms and the side margins are each 4 cms. If the area of the printed material on the poster is fixed at 384 cms², find the dimension of the poster with the smallest area.
Solution:
Let x and y be the length and breadth of printed area, then the area xy = 384
Dimensions of the poster area are (x + 8) and (y + 12) respectively.
Poster area A = (x + 8) (y + 12)
= xy + 12x + 8y + 96
= 12x + 8y + 480


But x > 0
∴ x = 16
when x = 16, A” > 0
when x = 16, the area is minimum
y = 24
∴ x + 8 = 24,
y + 12 = 36
Hence the dimensions are 24 cm and 36 cm

Question 2.
Show that the volume of the largest right circular cone that can be inscribed in a sphere of radius a is \(\frac{8}{27}\) (volume of the sphere).
Solution:
Given that a is the radius of the sphere and let x be the base radius of the cone. If h is the height of the cone, then its volume is


where OC = y so that height h = a + y
From the diagram x² + y² = a²
Using (2) in (1) we have

For the volume to be maximum:

⇒ 3y = + a or y = -a

Question 3.
A closed (cuboid) box with a square base is to have a volume of 2000 c.c. The material for the top and bottom of the box is to cost Rs. 3 per square cm and the material for the sides is to cost Rs. 1.50 per square cm. If the cost of the materials is to be the least, find the dimensions of the box.
Solution:
Let x, y respectively denote the length of the side of the square base and depth of the box. Let C be the cost of the material
Area of the bottom = x²
Area of the top = x²
Combined area of the top and bottom = 2x²
Area of the four sides = 4xy
Cost of the material for the top and bottom = 3(2x)²
Cost of the material for the sides = (1.5)(4xy) = 6xy

Question 4.
Find two numbers whose sum is 100 and whose product is a maximum.
Solution:
Let the two numbers be x and y.
x + y = 100
⇒ y = 100 – x
Product = xy = x(100 – x)
f = x(100 – x) = 100x – x²
We have to find x at which f is maximum.

∴ f is maximum at x = 50
So, y = 100 – x = 100 – 50 = 50
So, the two numbers are 50, 50.

Question 5.
Find two positive numbers whose product is 100 and whose sum is minimum.
Solution:
Let the two numbers be x and y.

To find x at which f is maximum

So the two numbers are 10, 10.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.9

Question 1.
Find the asymptotes of the following curves:

Solution:

y = 1 is a horizontal asymptote
So the asymptotes are x = -1, x = +1, y = 1

(ii) Since the numerator is of higher degree than the denominator we have a slant asymptote to find that asymptote we have to divide the numerator by the denominator So the slant asymptote is y = x – 1



∴ y = 3 and y = -3 are the horizontal asymptotes and there is no slant asymptote
(iv) Since the numerator is of highest degree than the denominator. We have a slant asymptote to find it we have to divide numerator by the denominator.

So the equation of asymptotes is y = x – 9 and x = -3
(v) Since the numerator is of highest degree than the denominator. We have a slant asymptote to find it we have to divide the numerator by the denominator.

Question 2.
Sketch the graphs of the following functions:

Solution:

Factorizing we get

• The domain and the range of the given function f(x) are the entire real line.
• Putting y = 0 we get x = 1, 1, -2. Hence the x intercepts are (1, 0) and (-2, 0) and by putting x = 0 we get y = \(-\frac{2}{3}\)

∴ The function is concave upward in the negative real line.
• Since f”(x) = 0 at x = 0 and f”(x) changes its sign when passing through x = 0, x = 0 is a point of inflection is \(\left(0,-\frac{2}{3}\right)\)
• The curve has no asymptotes.

(ii)


The curve concave downward in the negative real line
• No point of inflection exists.
• as x ➝ ∞, y = ± ∞ and so the curve does not have any asymptotes

(iii)



• Putting y = 0. x is unreal hence there is no ‘x’ intercept. By putting x = 0 we get

• No points of reflection
• When x = ± 2, y = ∞, Vertical asymptotes are x = 2 and x = -2 and horizontal asymptote is y = 1

f'(x) = 0 ⇒ e^-x = 0 which is not possible hence there is no extremum.
• No vertical asymptote for the curve and the horizontal asymptotes are y = 1 and y = 0

(v)

• The curve exists only for positive values of (x > 0)



• No point of inflection.
• No horizontal asymptote is possible.
But the vertical asymptote is x = 0(y axis).

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2

Question 1.
Find the slope of the tangent to the curves at the respective given points.
(i) y = x⁴ + 2x² – x at x = 1
(ii) x = a cos³ t, sin ³ t at t = \(\frac{\pi}{2}\)
Solution:
(i) y = x⁴ + 2x² – x at x = 1

Question 2.
Find the point on the curve y = x² – 5x + 4 at which the tangent is parallel to the line 3x + y = 7.
Solution:
Let (x₁, y₁) be the required point.

Given tangent is similarly to the line ⇒ m₁ = m₂
⇒ 2x₁ – 5 = -3
⇒ 2x₁ = -3 + 5 = 2
⇒ x₁ = 1
Substituting x₁ = 1 in the curve.
y₁ = 1 – 5 + 4 = 0
So the required point is (1, 0)

Question 3.
Find the points on the curves = x³ – 6x² + x + 3 where the normal is parallel to the line x + y = 1729.
Solution:
Let (x₁,y₁) be the required point
y = x³ – 6x² + x + 3

Substituting x₁ values in the curve
when x₁ = 4, y₁ = 3
when x₁ = 4, y₁ = 4³ – 6(4)² + 4 + 3 = 64 – 96 + 4 + 3 = -25
So the required points are (0, 3) and (4, -25)

Question 4.
Find the points on the curve y² – 4xy = x² + 5 for which the tangent is horizontal.
Solution:

Given that tangent is horizontal (i.e) tangent is parallel to x – axis
⇒ Equation of tangent will be of the form y = c

from (1) and (2) we get

Substituting x = -2y in the equation of the curve we get
y² – 4(-2y)(y) = 4y² + 5
⇒ y² + 8y² – 4y² = 5
⇒ 5y² = 5 ⇒ y² = 1
⇒ y = ± 1
when y = 1, x = -2 and when y = -1, x = 2 So the points are (2, -1), (-2, 1)

Question 5.
Find the tangent and normal to the following curves at the given points on the curve.

Solution:

Now m = -2, (x₁, y₁) = (1, 0)
So equation of the tangent is
y – y₁ = m(x – x₁)
(i. e) y – 0 = -2(x – 1)
y = -2x + 2
2x+ y = 2
Slope of tangent = m = -2

Question 6.
Find the equations of the tangents to the curve y = 1 + x³ for which the tangent is orthogonal with the line x + 12y = 12.
Solution:


Substituting x₁ values in the curve
when x₁ = 2, y₁ = 9; when x₁ = -2, y₁ = -1
So the points are (2, 9) and (-2, -7)
To find the equations of tangents:
Tangents are orthogonal to x + 12y = 12
So equations of tangents will be of the form 12x – y = k
The tangent passes through (2, 9) ⇒ 24 – 9 = k ⇒ k = 15 .
∴ Equation of tangent is 12x – y = 15
The tangent passes through (-2, -7) ⇒ 12 (-2) + 7 = k ⇒ -17
So equation of tangent is 12x – y = -17

Question 7.
Find the equations of the tangents to the curve y = \(\frac{x+1}{x-1}\)which are parallel to the line x + 2y = 6.
Solution:

Tangent is parallel the line x + 2y = 6
⇒ Slope of tangent = Slope of line


when x₁ = -1, y₁ = 0; when x₁ = 3, y₁ = 2
So the points are (-1, 0) and (3, 2). The tangents are parallel to x + 2y = 6. So equation of tangents will be of the form x + 2y = k.
∴ Equation of tangent is x + 2y = -1
The tangent passes through (-1, 0) ⇒ -1 = k
The tangent passes through (3, 2) ⇒ 3 + 4 = k ⇒ k = 7
∴ Equation of tangent is x + 2y = 7

Question 8.
Find the equation of tangent and normal to the curve given by x = 7 cos t and y = 2sin t, t ∈ R at any point on the curve.
Solution:

(2 cos t)y – 4 sint cos t = (7 sin t)x – 49 sin t cos t
x(7 sin t) – y(2 cos t) = 45 sin t cos t

Question 9.
Find the angle between the rectangular hyperbola xy = 2 and the parabola x² + 4y = 0.
Solution:
Solving the given two equations. To find the point of intersection:

Question 10.
Show that the two curves x² – y² = r² and xy = c² where c, r are constants, cut orthogonally.
Solution:
Let (x₁, y₁) be the point of intersection of the two curves
I Curve: x² – y² = r²
Differentiating w.r.to x

II Curve: xy = c²
Differentiating w.r.to x

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2 Additional Questions

Question 1.
Prove that the sum of the intercepts on the co-ordinate axes of any tangent to the curve x = a cos⁴ θ, y = a sin⁴ θ, 0 < θ < \(\frac{\pi}{2}\) is equal to a.
Solution:

Question 2.
Find the equations of a normal to y = x³ – 3x that is parallel to 2x + 18y – 9 = 0.
Solution:
Let (x₁, y,) be a point on the curve.

It is given that the normal is parallel to the line ⇒ m₁ = m₂

9y + 18 = -x – 2
x + 9y + 20 = 0

Question 3.
Prove that the curves 2x² + 4y² = 1 and 6x² – 12y² = 1 cut each other at right angles.
Solution:
Solving the given two equations,


Similarly it can be proved at the other points also.

Question 4.
Show that the equation of the normal to the curve x = a cos³ θ, y = a sin³ θ at ‘θ’ is x cos θ – y sin θ = a cos 2θ
Solution:

So, equation of the normal is

i.e. y sin θ – a sin⁴ θ = x cos θ – a cos⁴ θ
x cos θ – y sin θ = a cos⁴ θ – a sin⁴ θ
i.e.x cos θ – y sin θ = a[cos² θ + sin²θ][cos²θ – sin² θ]
= a[cos 2 θ]
So, the equation of the normai is x cos θ – y sin θ = a cos 2θ

Question 5.
If the curve y² = x and xy = k are orthogonal, then prove that 8k² = 1.
Solution:
y² = x, xy = k
Solving the two equations, we get, (y²) (y) = k
y³ = k

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9

Question 1.
Find the equation of the plane passing through the line of intersection of the planes \(\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})\) = 3 and 3x – 5y + 4z + 11 = 0, and the point (-2, 1, 3).
Solution:
Given planes are
\(\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})\) = 3
2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0
Equation of a plane which passes through the line of intersection of the planes
(2x – 7y + 4z – 3) + λ(3x – 5y + 4z + 11) = 0 …………… (1)
This passes through the point (-2, 1, 3).
(1) ⇒ (-4 – 7 + 12 – 3) + λ(-6 – 5 + 12 + 11) = 0
-2 + λ(12) = 0 ⇒ 12λ = 2
λ = \(\frac{2}{12}\) ⇒ λ = \(\frac{1}{6}\)
The required equation is
(1) ⇒ (2x – 7y + 4z – 3) + \(\frac{1}{6}\) (3x – 5y + 4z + 11) = 0
12x – 42y + 24z – 18 + 3x – 5y + 4z + 11 = 0
15x – 47y + 28z – 7 = 0

Question 2.
Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z + 11 = 3, and at a distance \(\frac{2}{\sqrt{3}}\) from the point (3, 1, -1).
Solution:
Equation of a plane which passes through the line of intersection of the plane
(x + 2y + 3z – 2) + λ(x – y + z – 3) = 0
(1 + λ)x + (2 – λ)y + (3 + λ)z + (-2 – 3λ) = 0 ………….. (1)

Question 3.
Find the angle between the line \(\vec{r}=(2 \hat{i}-\hat{j}+\hat{k})+t(\hat{i}+2 \hat{j}-2 \hat{k})\) and the plane \(\vec{r} \cdot(6 \hat{i}+3 \hat{j}+2 \hat{k})\) = 8.
Solution:
Angle between the line and a plane

Question 4.
Find the angle between the planes \(\vec{r} \cdot(\hat{i}+\hat{j}-2 \hat{k})\) = 3 and 2x – 2y + z = 2.
Solution:
Angle between given two planes

Question 5.
Find the equation of the plane which passes through the point (3, 4, -1)and is parallel to the plane 2x – 3y + 5z + 7 = 0. Also, find the distance between the two planes.
Solution:
The required equation parallel to the plane
2x – 3y + 5z + 7 = 0 ………….. (1)
2x – 3y + 5z + λ = 0 ………….. (2)
This passes through (3, 4, -1)
(2) ⇒ 2(3) – 3(4) + 5(-1) + λ = 0
6 – 12 – 5 + 1 = 0
λ = 11
(2) ⇒ The required equation is 2x – 3y + 5z + 11 =0 …………… (3)
∴ Now, distance between the above parallel lines (1) and (3)
a = 2, b = -3, c = 5, d₁ = 7, d₂ = 11

Question 6.
Find the length of the perpendicular from the point (1, -2, 3) to the plane x – y + z = 5.
Solution:
Perpendicular length from the point (x₁, y₁, z₁) to the plane ax + by + cz + d = 0 is

Question 7.
Find the point of intersection of the line x – 1 = \(\frac{y}{2}\) = z + 1 with the plane 2x – y + 2z = 2. Also, find the angle between the line and the plane.
Solution:
Any point on the line x – 1 = \(\frac{y}{2}\) = z + 1 is
x – 1 = \(\frac{y}{2}\) = z + 1 = λ,(say)
(λ + 1, 2λ, λ – 1)
This passes through the plane 2x – y + 2z = 2
2(λ + 1) – 2λ + 2(λ – 1) = 2
2λ + 2 – 2λ + 2λ – 2 = 2
λ = 1
∴ The required point of intersection is (2, 2, 0)
Angle between the line and the plane is

Question 8.
Find the coordinates of the foot of the perpendicular and length of the perpendicular from the point (4, 3, 2) to the plane x + 2y + 3z = 2
Solution:

Direction of the normal plane (1, 2, 3)
d.c.s of the PQ is \(\frac{x_{1}-4}{1}=\frac{y_{1}-3}{2}=\frac{z_{1}-2}{3}\) = k
x₁ = k + 4, y₁ = 2k + 3, z₁ = 3k + 2
This passes through the plane x + 2y + 3z = 2
k + 4 + 2(2k + 3) + 3(3k + 2) = 2
k + 4 + 4k + 6 + 9k + 6 = 2
14k = 2 – 16 ⇒ 14k = -14
k = -1
∴ The coordinate of the foot of the perpendicular is (3, 1, -1)
∴ Length of the perpendicular to the plane is

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 Additional Problems

Question 1.
Find the point of intersection of the line passing through the two points (1, 1, – 1); (-1, 0,1) and the xy-plane.
Solution:
The equation of the line passing through (1, 1,-1) and (-1, 0, 1) is

Question 2.
Find the co-ordinates of the point where the line \(\vec{r}=(\vec{i}+2 \vec{j}-5 \vec{k})+t(2 \vec{i}-3 \vec{j}+4 \vec{k})\) meets the plane \(\vec{r} \cdot(2 \vec{i}+4 \vec{j}-\vec{k})\) = 3.
Solution:
The equation of the straight line in the cartesian form is
\(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}\) = λ (say)
∴ Any point on this line is of the form (2λ + 1, -3λ, + 2, 4λ – 5)
The cartesian equation of the plane is 2x + 4y – z – 3 = 0
But the required point lies on this plane.
∴ 2(2λ + 1) + 4(-3λ + 2) – (4λ – 5 ) – 3 = 0 ⇒ λ = 1
∴ The required point is (3, -1, -1).

Question 3.
Find the point of intersection of the line \(\vec{r}=(\vec{j}-\vec{k})+s(2 \vec{i}-\vec{j}+\vec{k})\) and xz-plane.
Solution:
The given point is (0, 1, -1); parallel vector is \(2 \vec{i}-\vec{j}+\vec{k}\) .
The equation of the line passing through the point (0, 1, -1) and parallel to the vector \(2 \vec{i}-\vec{j}+\vec{k}\) is

∴ The required point is (2, 0, 0)

Question 4.
Find the meeting point of the line \(\vec{r}=(2 \vec{i}+\vec{j}-3 \vec{k})+t(2 \vec{i}-\vec{j}-\vec{k})\) and the plane x – 2y + 3z + 7 = 0.
Solution:
The equation of the straight line in cartesian form is
\(\frac{x-2}{2}=\frac{y-1}{-1}=\frac{z+3}{-1}\) = λ (say)
Any point on the line is (2λ, + 2, -λ + 1, λ – 3)
The point lies on the plane x – 2y + 3z + 7 = 0
⇒ 2λ, + 2 – 2(-λ + 1) + 3(-λ, – 3) + 7 = 0 .
2λ + 2 + 2λ – 2 – 3λ – 9 + 7 = 0
λ – 2 = 0 ⇒ λ = 2
So 2λ + 2 = 6; -λ + 1 = -1; -λ – 3 = – 5
∴ The required point is (6, -1, -5)

Question 5.
Show that the following planes are at right angles:

Solution:
The normal vectors are
\(\bar{n}_{1}=2 \bar{i}-\bar{j}+\bar{k}\) and \(\vec{n}_{2}=+\vec{i}-\vec{j}-3 \vec{k}\)
\(\vec{n}_{1} \cdot \vec{n}_{2}\) = (2)(1) + (-1)(-1) + (1)(-3) = 2 + 1 – 3 = 0
⇒ \(\vec{n}_{1}\) ⊥r to \(\vec{n}_{2}\), i.e., the normals to the planes are at right angles. So, the planes are at right angles.

Question 6.
The planes \(\vec{r} \cdot(2 \vec{i}+\lambda \vec{j}-3 \vec{k})\) = 10 and \(\vec{r} \cdot(\lambda \vec{i}+3 \vec{j}+\vec{k})\) = 5 are perpendicular. Find λ.
Solution:
Since the planes are perpendicular, the angle between the normals = 90°.
The normals are \(\vec{n}_{1}=2 \vec{i}+\lambda \vec{j}-3 \vec{k}\) and \(\vec{n}_{2}=\lambda \vec{i}+3 \vec{j}+\vec{k}\)
\(\vec{n}_{1} \cdot \vec{n}_{2}\) = 0 [∵ θ = π/2]
⇒ (2)(λ) + (λ)(3) + (-3)(1) = 0 ⇒ 2λ + 3λ – 3 = 0
5λ – 3 = 0 ⇒ 5λ = 3 ⇒ λ = 3/5

Question 7.
Find the angle between the line \(\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{-2}\) and the plane 3x + 4y + z = 0.
Solution:
The angle between the line \(\vec{r}=\vec{a}+t \vec{b}\) and the plane \(\vec{r} \cdot \vec{n}\) = p is given by the formula

Question 8.
Find the angle between the line \(\vec{r}=\vec{i}+\vec{j}+3 \vec{k}+\lambda(2 \vec{i}+\vec{j}-\vec{k})\) and the plane \(\vec{r} \cdot(\vec{i}+\vec{j})\) = 1,
Solution:

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.8

Question 1.
Show that the straight lines and are coplanar. Find the vector equation of the plane in which they lie.
Solution:

Question 2.
Show that lines are coplanar. Also, find the plane containing these lines.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (2, 3, 4) and (x₂, y₂, z₂) = (1, 4, 5)
(b₁, b₂, b₃) = (1, 1, 3) and (d₁, d₂, d₃) = (-3, 2, 1)
Condition for coplanarity

(x – 2)[1 – 6] – (y – 3)[1 + 9] + (z – 4)[2 + 3] = 0
-5(x – 2) – 10(y – 3) + 5(z – 4) = 0
-5x + 10 – 10y + 30 + 5z – 20 = 0
-5x – 10y + 5z + 20 = 0
(÷ by -5) ⇒ x + 2y – 2z – 4 = 0

Question 3.
If the straight lines are coplanar, find the district real values of m.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (1, 2, 3) and (x₂, y₂, z₂) = (3, 2, 1)
(b₁, b₂, b₃) = (1, 2, m²) and (d₁, d₂, d₃) = (1, m², 2)
Condition for coplanarity
\(\left|\begin{array}{ccc}{x_{2}-x_{1}} & {y_{2}-y_{1}} & {z_{2}-z_{1}} \\ {b_{1}} & {b_{2}} & {b_{3}} \\ {d_{1}} & {d_{2}} & {d_{3}}\end{array}\right|=0\)

Question 4.
If the straight lines are coplanar, find λ and equation of the planes containing these two lines.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (1, -1, 0) and (x₂, y₂, z₂) = (-1, -1, 0)
(b₁, b₂, b₃) = (2, λ, 2) and (d₁, d₂, d₃) = (5, 2, λ)
Condition for coplanarity

\(\left|\begin{array}{ccc}{x-1} & {y+1} & {z} \\ {2} & {-2} & {2} \\ {5} & {2} & {-2}\end{array}\right|\) = 0
(x – 1)[0] – (y + 1)[-4 – 10] + z[4 + 10] = 0
14(y + 1) + 14z = 0 ⇒ 14y + 14 + 14z = 0
(÷ by 14) ⇒ y + z + 1 = 0

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.8 Additional Problems

Question 1.
Show that the straight lines.

are coplanar. Find the vector equation of the plane in which they lie.
Solution:

Question 2.
If the straight lines are coplanar. Find λ.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (1, 1, 1) and (b₁, b₂, b₃) = (1, λ, 1)
(x₂, y₂, z₂) = (0, 4, 2) and (d₁, d₂, d₃) = (2, λ, 3)
Condition for coplanarity
\(\left|\begin{array}{ccc}{x_{2}-x_{1}} & {y_{2}-y_{1}} & {z_{2}-z_{1}} \\ {b_{1}} & {b_{2}} & {b_{3}} \\ {d_{1}} & {d_{2}} & {d_{3}}\end{array}\right|\) = 0
\(\left|\begin{array}{ccc}{-1} & {3} & {1} \\ {1} & {\lambda} & {1} \\ {2} & {\lambda} & {3}\end{array}\right|\) = 0
-1(3λ – λ) – 3(3 – 2) + 1(λ – 2λ) = 0 ⇒ -2λ – 3 – λ = 0
-3λ = 3 ⇒ λ = -1

Question 3.
If the lines are coplanar, then find the value of k.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (2, 3, 4) and (b₁, b₂, b₃) = (1, 1, -1)
(x₂, y₂, z₂) = (1, 4, 5) and (d₁, d₂, d₃) = (k, 2, 1)
Condition for coplanarity

-1(1 + 2k) – 1(1 + k²) + 1(2 – k) = 0
-1 – 2k – 1 – k² + 2 – k = 0
-k² – 3k = 0
k² + 3k = 0
k(k + 3) = 0
k = 0 or k = -3

Question 4.
Show that the lines are coplanar, Also find the equation of the plane containing these two lines.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (-3, 1, 5) and (b₁, b₂, b₃) = (-3, 1, 5)
(x₂, y₂, z₂) = (-1, 2, 5) and (d₁, d₂, d₃) = (-1, 2, 5)
Condition for coplanarity

Given two lines are coplanar

(x + 3)[5 – 10] – (y – 1)[-15 + 5] + (z – 5)[-6 + 1] = 0
5(x + 3) + 10(y – 1) – 5(z – 5) = 0
(÷ by 5) ⇒ (x + 3) -2(y – 1) + (z – 5) = 0
x + 3 – 2y + 2 + z – 5 = 0
x – 2y + z = 0
or

(x + 1)(5 – 10) – (y – 2)(-15 + 5) + (z – 5)(-6 + 1) = 0
-5(x + 1) + 10(y – 2) – 5(z – 5) = 0
(x + 1) – 2(y – 2) + (z – 5) = 0
x + 1 – 2y + 4 + z – 5 = 0
x – 2y + z = 0

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.7

Question 1.
Find the non-parametric form of vector equation, and Cartesian equation of the plane passing through the point (2, 3, 6) and parallel to the straight lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-3}{1}\) and \(\frac{x+3}{2}=\frac{y-3}{-5}=\frac{z+1}{-3}\)
Solution:

Question 2.
Find the parametric form of vector equation, and Cartesian equations of the plane passing through the points (2, 2, 1), (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 9.
Solution:
The required plane passes through the points

Question 3.
Find the parametric form of vector equation and Cartesian equations of the plane passing through the points (2, 2, 1), (1, -2, 3) and parallel to the straight line passing through the points (2, 1, -3) and (-1, 5, -8).
Solution:
Equation of the straight line passing through the points (2, 1, -3) and (-1, 5, -8) is

(x – 2) [20 – 8] – (y – 2) [5 + 6] + (z – 1) [-4 – 12] = 0
12(x – 2) – 11(y – 2) – 16(z – 1) = 0
12x – 24 – 11y + 22 – 16z + 16 = 0
12x – 11y – 16z + 14 = 0

Question 4.
Find the non-parametric form of vector equation of the plane passing through the point (1, -2, 4) and perpendicular to the plane x + 2y – 3z = 11 and parallel to the line \(\frac{x+7}{3}=\frac{y+3}{-1}=\frac{z}{1}\)
Solution:
The required plane passing through the point \(\vec{a}=\vec{i}-2 \vec{j}+4 \vec{k}\) and parallel to the plane \(\vec{b}=\vec{i}+2 \vec{j}-3 \vec{k}\) and parallel to the line \(\vec{c}=3 \vec{i}-\vec{j}+\vec{k}\)

Question 5.
Find the parametric form of vector equation, and Cartesian equations of the plane containing the line \(\vec{r}=(\hat{i}-\hat{j}+3 \hat{k})+t(2 \hat{i}-\hat{j}+4 \hat{k})\) and perpendicular to the plane \(\vec{r} \cdot(\hat{i}+2 \hat{j}+\hat{k})\) = 8.
Solution:
The required plane passing through the point \(\vec{a}=\vec{i}-\vec{j}+3 \vec{k}\) and parallel to \(\vec{b}=2 \vec{i}-\vec{j}+4 \vec{k}\) and \(\vec{c}=\vec{i}+2 \vec{j}+\vec{k}\)
Parametric form of vector equation

(x – 1) [-1 – 8] – (y + 1)[z – 4] + (z – 3) [4 + 1] = 0
-9(x- 1) + 2(y + 1) + 5(z- 3) = 0
-9x + 9 + 2y + 2 + 5z – 15 = 0
-9x + 2y + 5z – 4 = 0
9x – 2y – 5z + 4 = 0

Question 6.
Find the parametric vector, non-parametric vector and Cartesian form of the equation of the plane passing through the point (3, 6, -2), (-1, -2, 6) and (6, 4, -2).
Solution:
The required plane passing through the points

Question 7.
Find the non-parametric form of vector equation, and Cartesian equations of the plane

Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.7 Additional Problem

Question 1.
Find the vector and cartesian equations of the plane containing the line \(\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{3}\) and parallel to the line \(\frac{x+1}{3}=\frac{y-1}{2}=\frac{z+1}{1}\)
Solution:
The required plane passes through the point A(2, 2, 1) and parallel to

i.e., (x – 2)[3 – 6] – (y – 2)[2 – 9] + (z – 1)[4 – 9] = 0
i.e., (x – 2)(-3) + (y – 2)(7) – 5(z – 1) = 0
-3x + 6 + 7y – 14 – 5z + 5 = 0
-3x + 7y – 5z – 3 = 0 i.e. 3x – 7y + 5z + 3 = 0

Question 2.
Find the vector and cartesian equation of the plane passing through the point (1, 3, 2) and parallel to the lines
Solution:
The required plane passes through the point A = (1, 3, 2) and parallel to the vectors

i.e. (x – 1)(-2 – 6) – (y – 3)(4 – 3) + (z – 2)(4 + 1) = 0
—8(x – 1) – 1(y – 3) + 5(z – 2) = 0
-8x + 8 – y + 3 + 5z – 10 = 0
-8x – y + 5z + 1 = 0 i.e. 8x + y – 5z – 1 = 0

Question 3.
Find the vector and cartesian equations of the plane passing through the point (-1, 3, 2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + y + 2z = 8
Solution:
The normal vector to the given planes x + 2y + 2z = 5 and 3x + y + 2z = 8 are respectively \(\vec{i}+2 \vec{j}+2 \vec{k}\) and \(3 \vec{i}+\vec{j}+2 \vec{k}\). These vectors are parallel to the required plane
The required plane passes through the point A(-1, 3, 2) and parallel to the vectors

i.e., (x + 1)(4 – 2) – (y – 3)(2 – 6) + (z – 2)(1 – 6) = 0
2(x + 1) + 4(y – 3) – 5(z – 2) = 0
2x + 2 + 4y – 12 – 5z + 10 = 0
2x + 4y – 5z = 0

Question 4.
Find the vector and cartesian equations of the plane passing through the points A(1, -2, 3) and B(-1, 2, -1) and is parallel to the line \(\frac{x-2}{2}=\frac{y+1}{3}=\frac{z-1}{4}\)
Solution:
The vector equation of the plane is

i.e, (x – 1)[16 + 12] – (y + 2)(-8 + 8) + (z – 3)(-6 – 8) = 0
28(x – 1) – 14(z – 3) = 0
28x – 28 – 14z + 42 = 0
28x – 14z + 14 = 0
(÷ by 14) ⇒ 2x – z + 1 = 0

Question 5.
Find the vector and cartesian equations of the plane through the points (1, 2, 3) and (2,3,1) and perpendicular to the plane 3x – 2y + 4z – 5 = 0.
Solution:
The vector normal to the plane 3x – 2y + 4z – 5 = 0 is \(3 \vec{i}-2 \vec{j}+4 \vec{k}\)
The required plane is parallel to the vector \(3 \vec{i}-2 \vec{j}+4 \vec{k}\)
∴ Vector equation of the plane passing through two points and parallel to one vector is

i.e. (x – 1)[4 – 4] – (y – 2)[4 + 6] + (z – 3)[-2 – 3] = 0
-10(y – 2) – 5(z – 3) = 0
-10y + 20 – 5z + 15 = 0
-10y – 5z + 35 = 0
10y + 5z – 35 = 0
(÷ by 5) ⇒ 2y + z – 1 = 0

Question 6.
Find the vector and cartesian equations of the plane containing the line \(\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{-2}\) and passing through the point (-1, 1, -1).
Solution:
The required plane passes through the points A(-1, 1,-1) and B(2, 2, 1) and parallel to the vectors \(\vec{c}=2 \vec{i}+3 \vec{j}-2 \vec{k}\) .
The required equation is

i.e. (x + 1)[-2 – 6] – (y – 1)[-6 – 4] + (z + 1)[9 – 2] = 0
-8(x + 1) + 10(y – 1) + 7(z + 1) = 0
-8x – 8 + 10y – 10 + 7z + 7 = 0
-8x + 10y + 7z – 11 = 0
i.e., 8x – 10y – 7z + 11 = 0

Question 7.
Find the vector and cartesian equations of the plane passing through the points with position vectors .
Solution:
Vector equation of the plane passing through three given non-collinear points is

i.e. (x – 3)[6 – 12] – (y – 4)[1 + 12] + (z – 2)[4 + 24] = 0
-6(x – 3) – 13(y – 4) + 28(z – 2) = 0
-6x + 18 – 13y + 52 + 28z – 56 = 0
-6x – 13y + 28z + 14 = 0 .
i.e. 6x + 13y – 28z – 14 = 0

Question 8.
Derive the equation of the plane in the intercept form.
Solution:
Let the required plane makes intercepts on X, Y, Z-axes respectively as a, b and c.
i.e. A = (a, 0, 0)
B = (0, b, 0)
C = (0, 0, c)
The equation of the plane passing through three non-collinear points A, B and C is

Question 9.
Find the cartesian form of the following plane:

Solution:

x(0 + 2) – (y – 3) (1 + 4) + z(-1) = 0
2x – 5(y – 3) – z = 0
2x – 5y + 15 – z = 0
2x – 5y – z + 15 = 0
which is the cartesian equation of the plane.