Prasanna

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.5

Evaluate the following:

Question 1.

Solution:
u = x³ ; dv = e^-2
Applying Bernoulli’s formula, we get,

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.4 Additional Problems

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.3

Question 1.
Evaluate the following definite integrals :
(i)
Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

Question 2.
Evaluate the following integrals using properties of integration :
(i)
Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

(vii)
Solution:

(viii)
Solution:

(ix)
Solution:

(x)
Solution:

(xi)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.3 Additional Problems

Question 1.

Solution:
Let f (x) = x³ sin² x = x³ (sin x)²
∴f(- x) = (- x)³ (sin (- x))² = (- x)³ (- sin x)²
= – x³ sin² x = -f(x)
f(-x) = -f(x) ∴ f(x) is an odd function.

Question 2.

Solution:

Question 3.

Solution:
Let f(x) = x sin x
f(-x) = (-x)sin(-x)
= x sin x (∵ sin(-x) = – sin x)
∴ f(x) is an even function

Question 4.

Solution:

Question 5.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.6

Choose the correct or the most suitable answer from the given four alternatives.
Question 1.
The equation of the circle passing through (1,5) and (4, 1) and touching y-axis is x² + y² – 5x – 6y + 9 + λ (4x + 3y – 19) = 0 where λ is equal to …………
(a) 0, \(-\frac{40}{9}\)
(b) 0
(c) \(\frac{40}{9}\)
(d) \(-\frac{40}{9}\)
Solution:
(a) 0, \(-\frac{40}{9}\)
Hint:
x² + y² – 5x – 6y + 9 + λ (4x + 3y -19) = 0
ie., x² + y² + x(4λ – 5) + y(3λ – 6) + 9 — 19λ = 0
Since it touches y axis x = 0
⇒ y² + y(3λ – 6) + 9 – 19λ = 0
It is a quadratic in y ,
Since the circle touches y axis the roots must be equal ⇒ b² – 4ac = 0
ie.,(3λ – 6)² – 4(1) (9 – 19λ) = 0
Solving we get λ = o or \(-\frac{40}{9}\)

Question 2.
The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci is ………..

Solution:
(c) \(\frac{2}{\sqrt{3}}\)
Hint:
Given \(\frac{2 b^{2}}{a}\) = 8 and 2b = ae

Question 3.
The circle x² + y² = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if …………..
(a) 15 < m < 65
(b) 35 < m < 85
(c) -85 < m < -35
(d) -35 < m < 15
Solution:
(d) -35 < m < 15
Hint:
x² + y² – 4x – 8y – 5 =0
3x – 4y = m Solving (1) and (2) we get
from (2) ⇒ 3x = 4y + m
x = \(\frac{4 y+m}{3}\)
Substituting x value in (1) we get
\(\left(\frac{4 y+m}{3}\right)^{2}\) + y² – \(4\left(\frac{4 y+m}{3}\right)\) – 8y – 5 = 0
It is a quadratic in y and given that the roots are distinct
⇒ b² – 4ac > 0
On simplifying we get
⇒ – 9m² – 18w + 4725 >0 ⇒ m² + 20m – 525 < 0
⇒ (m + 35) (m – 15) ≤ 0 ⇒ m lies between -35 and 15 ie„ – 35 < m < 15

Question 4.
The length of the diameter of the circle which touches the x -axis at the point (1,0) and passes through the point (2, 3) …………

Solution:
(c) \(\frac{10}{3}\)
Hint:
Since radius is ⊥ r to tangent it passes through the centre (1, y)
AC = CB = radius
AC² = CB² ⇒ y² = 1 + (y – 3)²
6y = 10 ⇒ y = \(\frac{5}{3}\)
Diameter 2y = \(\frac{10}{3}\)

Question 5.
The radius of the circle 3x² + by² + 4bx – 6by + b² = 0 is ……………
(a) 1
(b) 3
(c) \(\sqrt{10}\)
(d) \(\sqrt{11}\)
Solution:
(c) \(\sqrt{10}\)
Hint:
Here b = 3
3x² + 3y² + 12x- 18y + 9 = 0
(÷ 3) zz> x² + y² + 4x – 6y + 3 = 0
Comparing with general form
g = 2,f = -3, c = 3
∴ radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+9-3}=\sqrt{10}\)

Question 6.
The centre of the circle inscribed in a square formed by the lines x² – 8x – 12 = 0 and y² – 14y + 45 = 0 is ………..
(a) (4, 7)
(b) (7, 4)
(c) (9, 4)
(d) (4, 9)

Solution:
(a) (4, 7)
Hint:
(x – 6) (x – 2) = 0 ⇒ x = 2 or 6
(y – 9) (y – 5) = 0 ⇒ y = 5 or 9

Question 7.
The equation of the normal to the circle x² + y² – 2x – 2y + 1 = 0 which is parallel to the line 2x + 4y = 3 is …………..
(a) x + 2y = 3
(b) x + 2y + 3 = 3
(c) 2x + 4y + 3 = 0
(d) x – 2y + 3 = 0
Solution:
(a) x + 2y = 3
Hint:
Centre of the circle = (1, 1) and radius = r = 1
Normal is parallel to 2x + 4y = 3
So equation of normal will be the form 2x + 4y = k
It passes throug the centre (1, 1) ⇒ 2 + 4 = 6 = k
So equation of the normal is x + 4y = 6
(÷ by 2) x + 2y = 3

Question 8.
If P(x, y) be any point on 16x² + 25y² = 400 with foci F₁(3, 0) and F₂(- 3, 0) then PF₁ + PF₂ is
(a) 8
(b) 6
(c) 10
(d) 12
Solution:
(c) 10
Hint:
F₁P + F₂P = 2a
Here the equation is 16x² + 25y² = 400
(÷ by 400) ⇒ \(\frac{16 x^{2}}{400}+\frac{25 y^{2}}{400}\) = 1 ⇒ \(\frac{x^{2}}{25}+\frac{y^{2}}{16}\) = 1
a² = 25 => a = 5
∴ 2a = 10

Question 9.
The radius of the circle passing through the point (6, 2) two of whose diameter are x + y = 6 and x + 2y = 4 is …………..
(a) 10
(b) \(2 \sqrt{5}\)
(c) 6
(d) 4
Solution:
(b) \(2 \sqrt{5}\)
Hint:
The point of intersection of the diameters is the centre.
Now solving x + y = 6 …………. (1)
x + 2y = 4 ………….. (2)
(1) – (2) ⇒ -y = 2 ⇒ y = -2
Substituting y = -6 in (1)
x – 2 = 6 ⇒ x = 8
Centre = (8, -2)
The circle passes through (6, 2)
∴ radius = \(\sqrt{(8-6)^{2}+(-2-2)^{2}}=\sqrt{4+16}\)
= \(\sqrt{20}=\sqrt{4 \times 5}=2 \sqrt{5}\)

Question 10.
The area of quadrilateral formed with foci of the hyperbolas \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = -1 is ……………
(a) 4(a² + b²)
(b) 2(a² + b²)
(c) a² + b²
(d) \(\frac{1}{2}\)(a² + b²)
Solution:
(b) 2(a² + b²)
Hint:
The foci are (± ae, 0) and (0, ± ae)

The above diagram is a rhombus.
Its area = \(\frac{1}{2}\)d₁d₂
= \(\frac{1}{2}\) (2ae) (2ae) = 2a² e²
But we know b² = a² (e² – 1)
i.e., b² = a² e² – a²
⇒ b² + a² = a² e²
So area = 2(a² + b²)

Question 11.
If the normals of the parabola y² = 4x drawn at the end points of its latus rectum are tangents to the circle (x – 3)² + (y + 2)² = r² , then the value of r² is ………………..
(a) 2
(b) 3
(c) 1
(d) 4
Solution:
(a) 2
Hint:
The normals are x ± y = 3
Distance from (3, -2) on both normal is r
(i.e.,) r = \(\left|\frac{3-2-3}{\sqrt{2}}\right|=\sqrt{2}\) ⇒ r² = 2

Question 12.
If x + y = k is a normal to the parabola y² = 12x, then the value of k is ……………
(a) 3
(b) -1
(c) 1
(d) 9
Solution:
(d) 9
Hint:
Slope of normal = -1 and Slope of tangent = -1
The condition for y = mx + c to be a tangent to the parabola
y² = 4ax is c = \(\frac{a}{m}\)
Here the parabola is y² = 12x ⇒ a = 3 and m = 1
∴ c = \(\frac{3}{1}\) = 3
∴ Equation of tangent is y = x + 3
Solving the parabola and tangent we get the points (3, 6) and (3,-6)
∴ k = -3 or k = 9

Question 13.
The ellipse E₁ : \(\frac{x^{2}}{9}+\frac{y^{2}}{4}\) = 1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E₂ passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse is …………….

Solution:
(c) \(\frac{1}{2}\)
Hint:

Question 14.
Tangents are drawn to the hyperbola \(\frac{x^{2}}{9}-\frac{y^{2}}{4}\) = 1 parallel to the straight line 2x – y = 1. One of the points of contact of tangents on the hyperbola is ……………

Solution:
(c) \(\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
Hint:
The tangents are parallel to 2x – y = 1
So equation of tangents will be of the form
2x – y = k ⇒ y = 2x – k
Comparing this equation with y = mx + c we get m = 2 and c = -k
The given hyperbola is \(\frac{x^{2}}{9}-\frac{y^{2}}{4}\) = 1
⇒ a² = 9 and b² = 4
Now the condition for the line y = mx + c to be a tangent to

Question 15.
The equation of the circle passing through the foci of the ellipse \(\frac{x^{2}}{16}-\frac{y^{2}}{9}\) = 1 having centre at (0, 3) is ……………….
(a) x² + y² – 6y – 7 = 0
(b) x² + y² – 6y + 7 = 0
(c) x² + y² – 6y – 5 = 0
(d) x² + y² – 6y + 5 = 0
Solution:
(a) x² + y² – 6y – 7 = 0

The distance between centre and foci = radius of the circle
ie., r = \(\sqrt{(0 \pm \sqrt{7})^{2}+(3-0)^{2}}=\sqrt{7+9}=\sqrt{16}\) = 4
Now Centre = (0, 3); r = 4
Equation of the circle is (x – 0)2 + (y -3)2 = 42
ie., x² + y² – 6y + 9 – 16 = 0
x² + y² – 6y – 7 = 0

Question 16.
Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centered at (0, y) passing through the origin and touching the circle C externally, then the radius of T is equal ……………..

Solution:
(d) \(\frac{1}{4}\)
Hint:
Equation of circle is x² + y² + 2gx + 2fy + c = 0
i.e., (1 + y²)² = (1 – y)² = 1
Solving we get y = \(\frac{1}{4}\)

Question 17.
Consider an ellipse whose centre is of the origin and its major axis is along x-axis. If its eccentricity is \(\frac{3}{5}\) and the distance between its foci is 6, then the area of the quadrilateral inscribed in the ellipse with diagonals as major and minor axis of the ellipse is ……………
(a) 8
(b) 32
(c) 80
(d) 40
Solution:
(d) 40
Hint:

Question 18.
Area of the greatest rectangle inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 is …………….
(a) 2ab
(b) ab
(c) \(\sqrt{a b}\)
(d) \(\frac{a}{b}\)
Solution:
(a) 2ab
Hint:
Area of the greatest rectangle is (\(\sqrt{2}\) a) (\(\sqrt{2}\) b) = 2ab

Question 19.
An ellipse has OB as semi minor axes, F and F’ its foci and the angle FBF’ is a right angle. Then the eccentricity of the ellipse is …………..

Solution:
(a) \(\frac{1}{\sqrt{2}}\)
Hint:

Question 20.
The eccentricity of the ellipse (x – 3)² + (y – 4)² = \(\frac{y^{2}}{9}\) is ……………

Solution:
(b) \(\frac{1}{3}\)
Hint:
The given equation is of the form FP² = e²PM²
i.e., (x – 3)² + (y – 4)² = \(\frac{1}{9}\)(y²)
⇒ e² = \(\frac{1}{9}\) ⇒ e = \(\frac{1}{3}\)

Question 21.
If the two tangents drawn from a point P to the parabola y² = 4x are at right angles then the locus of P is ………………….
(a) 2x + 1 = 0
(b) x = -1
(c) 2x – 1 = 0
(d) x = 1
Solution:
(b) x = -1
Hint:
When the tangents at drawns from a point on the directrix are at right angles.
So equation of directrix to y² = 4x is x = -1

Question 22.
The circle passing through (1, -2) and touching the axis of x at (3, 0) passing through the point …………..
(a) (-5, 2)
(b) (2, -5)
(c) (5, -2)
(d) (-2, 5)
Solution:
(c) (5, -2)
Hint:

Let the centre be (3, h)
r = k
Equation of the circle is
(x – 3)² + (y – h)² = r² = (k)²
It passes through (1, -2) ⇒ k² = 8
Substituting (5, -2) the equation satisfies.

Question 23.
The locus of a point whose distance from (-2, 0) is \(\frac{2}{3}\) times its distance from the line x = \(\frac{-9}{2}\) is ……….
(a) a parabola
(b) a hyperbola
(c) an ellipse
(d) a circle
Solution:
(c) an ellipse
Hint:
Here e = \(\frac{2}{3}\) < 1
So the conic is an ellipse

Question 24.
The values of m for which the line y = mx + 2\(\sqrt{5}\) touches the hyperbola 16x² – 9y² = 144 are the roots of x² – (a + b)x -4 = 0, then the value of (a + b) is
(a) 2
(b) 4
(c) 0
(d) -2
Solution:
(c) 0
Hint:

c² = a² m² – b²
Here the given line is y = mx + 2 \(\sqrt{5}\)
⇒ m = m and c = 2\(\sqrt{5}\)
The condition is c² = a²m² – b²
ie., (2\(\sqrt{5}\))²= 9(m²)- 16
⇒ 9m² = 20 + 16 = 36 ⇒ m² = 4
i.e., m = ± 2
a = 2, b = -2 (say)
So a + b = 2 – 2 = 0

Question 25.
If the coordinates at one end of a diameter of the circle x² + y² – 8x – 4y + c = 0 are (11, 2), the coordinates of the other end are ……………
(a) (-5, 2)
(b) (-3, 2)
(c) (5, -2)
(d) (-2, 5)
Solution:
(b) (-3, 2)
Hint:

Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.6 Additional Problems

Choose the most appropriate answer:
Question 1.
The parabola y² = 4ax passes through the point (2, -6), the the length of its latus rectum is …………..
(a) 9
(b) 16
(c) 18
(d) 6
Solution:
(c) 18
Hint:
The given parabola is y² = 4ax
Since it passes through (2, -6)
∴ (-6)² = 4a(2)
⇒ 36 = 8a a = \(\frac{36}{8}=\frac{9}{2}\)
Length of the latus rectum = 4a = \(4\left(\frac{9}{2}\right)\) = 18

Question 2.
The vertex of the parabola x² + 12x – 9y = 0 is …………
(a) (6, -1)
(b) (-6, 4)
(c) (6, 4)
(d) (-6, -4)
Solution:
(d) (-6, -4)
Hint:
Given parabola is x² + 12x – 9y = 0
⇒ (x² + 12x + 36) – 36 – 9y = 0
⇒ (x + 6)² = 9y + 36 ⇒ (x + 6)² = 9 (y + 4)
∴ Vertex = (-6, -4)

Question 3.
The length of the major axis of an ellipse is three times the length of minor axis, its eccentricity is ……………

Solution:
(d) \(\frac{2 \sqrt{2}}{3}\)
Hint:
Length of major axis = 2a and Length of minor axis = 2b
2a = 3(2b) ⇒ a = 3b

Question 4.
The eccentricity of the ellipse 9x² + 4y² = 36 is …………..

Solution:
(d) \(\frac{\sqrt{5}}{3}\)
Hint:
Given equation of the ellipse 9x² + 4y² = 36
On dividing both the sides by 36, we get

Question 5.
S and T are the foci of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 and B is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is …………..

Solution:
(c) \(\frac{1}{2}\)
Hint:

Question 6.
The sum of the focal distances from any point of the ellipse 9x² + 16y² = 144 is …………….
(a) 32
(b) 18
(c) 16
(d) 8
Solution:
(d) 8
Hint:
We have, 9x² + 16y² = 144
⇒ \(\frac{x^{2}}{16}+\frac{y^{2}}{9}\) = 1 (Dividing both sides by 144)
Here a² = 16, b² = 9 ⇒ a = 4, b = 3
Length of major axis = 2a = 2(4) = 8
Since the sum of the focal distances from any point on the ellipse is equal to its major axis
∴ Required sum = 8

Question 7.
If the eccentricities of two ellipse \(\frac{x^{2}}{169}+\frac{y^{2}}{25}\) = 1 and \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 are equal then \(\frac{a}{b}\) = ………………

Solution:
(c) \(\frac{13}{5}\)
Hint:

Question 8.
Equation of the hyperbola, whose eccentricity \(\frac{3}{2}\) and foci at (±2, 0) is ……………

Solution:
(a) \(\frac{x^{2}}{4}-\frac{y^{2}}{5}=\frac{4}{9}\)
Hint:
Eccentricity e = \(\frac{c}{a}=\frac{3}{2}\)
Foci are (± c, 0) = (± 2, 0) (Given c = 2)

Question 9.
If e₁ is the eccentricity of the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1 and if e₂ is the eccentricity of the hyperbola 9x² – 16y² = 144, then e₁e₂ is …………….
(a) \(\frac{16}{25}\)
(b) 1
(c) Greater than 1
(d) Less than \(\frac{1}{2}\)
Solution:
(b) 1
Hint:

Question 10.
The point of intersection of the tangents at t₁ = t and t₂ = 3 t to the parabola y² = 8x is ………….
(a) (6t², 8t)
(b) (8t, 6t²)
(c) (t², 4t)
(d) (4t, t²)
Solution:
(a) (6t², 8t)
Hint:
Point of intersection of the tangents at t₁ and t₂ to y² = 4ax is [a t₁ t₂, a(t₁ + t₂)]
Here, t₁ = t, t₂ = 3t, a = latex]\frac{8}{4}[/latex] = 2
So a t₁ t₂ = 2(t) (3t) = 6t²
a(t₁ + t₂) = 2(t + 3t) = 8t
∴ Point = (6t², 8t²)

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 9 Applications of Integration Ex 9.2

Question 1.
Evaluate the following integrals as the limits of sums:

Solution:
We use the formula

(ii) \(\int_{1}^{2}\left(4 x^{2}-1\right) d x\)
Solution:
We use the formula

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.2 Additional Problems

Question 1.
Evaluate as the limit of sums:
Solution:
Let f(x) = 2x² +5; a = 1; b = 3 and nh = 3 – 1 = 2

Here, f(a) = f(1) = 2(1)² + 5
f(a + h) = f(1 + h) = 2(1 + h)² + 5
f(a + 2h) = f(1 + 2h) = 2 (1 + 2h)² + 5
f[a+(n – 1)h = f[1 + (n – 1)h] = 2[1 + (n – 1)h]² + 5

Question 2.
Evaluates as the limit of sums: \(\int_{1}^{2}\left(x^{2}-1\right) d x\)
Solution:
Let f(x) = x² – 1 for 1 ≤ x ≤ 2
We divide the interval [1, 2] into n equal sub-intervals each of length h.
We have a = 1, b = 2

Here, f(a) = f(1) = (1)² – 1 = 0
f(a + h) = f(1 + h) = (1 + h)² – 1 = 1 + h² + 2h – 1 = h² + 2h
f(a + 2h) = f(1 + 2h) = (1 + 2h)² – 1 = 1 + 4h² + 4h – 1 = 4h² + 4h
f[a + (n -1)h] = f[1 + (n – 1)h] = [1 + (n – 1)h]² – 1
= 1 + (n – 1)² h² + 2(n – 1 )h – 1 = (n – 1)² h² + 2 (n – 1 )h

Question 3.
Evalute:
Solution:
Let f(x) = x² + x + 2, for 0 ≤ x ≤ 2 Here, a = 0,b = 2
We divide the closed interval [0, 2] into n subintervals each of length h.

By definition,

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 9 Applications of Integration Ex 9.1

Question 1.
Find an approximate value of \(\int_{1}^{1.5} x d x\) by applying the left-end rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}
Solution:
Here a = 1, b = 1.5, n = 5, f(x) = x
So, the width of each subinterval is

The left hand rule for Riemann sum,
S = [f(x₀) + f(x₁)) + f(x₂) + f(x₃) + f(x₄)] ∆x
= [f(1) + f(1.1) + f(1.2) + f(1.3) + f(1.4)] (0.1)
= [1 + 1.1 + 1.2 + 1.3 + 1.4] (0.1)
= [6] (0.1)
= 0.6.

Question 2.
Find an approximate value of \(\int_{1}^{1.5} x^{2} d x\) by applying the right-end rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}.
Solution:
Here a = 1;
b = 1.5;
n = 5;
f(x) = x²
So, the width of each subinterval is

The Right hand rule for Riemann sum,
S = [f(x₁) +f(x₂) +f(x₃) + f(x₄) + f(x₅)] ∆x
= [f(1.1) + f(1.2) + f(1.3) + f(1.4) + f(1.5)] (0.1)
= [1.21 + 1.44 + 1.69 + 1.96 + 2.25] (0.1)
= [8.55] (0.1)
= 0.855.

Question 3.
Find an approximate value of \(\int_{1}^{1.5}(2-x) d x\) by applying the mid-point rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}.
Solution:
Here a = 1;
b = 1.5;
n = 5;
f(x) = 2 – x
So, the width of each subinterval is

The mid-point rule for Riemann sum,

= [f(1.05) + f(1 .15) + f(1.25) +f(1.35) + f(1 .45)] (0.1)
= [0.95 + 0.85 + 0.75 + 0.65 + 0.55] (0.1)
= [3.75] (0.1)
= 0.375.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.8

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
A circular template has a radius of 10 cm. The measurement of radius has an approximate error of 0.02 cm. Then the percentage error in calculating area of this template is …….
(a) 0.2%
(b) 0.4%
(c) 0.04%
(d) 0.08%
Solution:
(b) 0.4%
Hint:
r = 10 cm
dr = 0.02

Question 2.
The percentage error of fifth root of 31 is approximately how many times the percentage error in 31?

Solution:
\(\frac{1}{5}\)
Hint:
We know that the percentage error in the «th root of a number is approximately \(\frac{1}{n}\) times the percentage error in the number.

Question 3.


Solution:
(b) 2xu
Hint:

Question 4.


Solution:
(d) 1
Hint:

Question 5.
If w (x, y) = x^y, x > 0, then \(\frac{\partial w}{\partial x}\) is equal to ……

Solution:
(c) yx^{y – 1}
Hint:

Question 6.

(a) xye^xy
(b) (1 + xy)e^xy
(c) (1 + y)e^xy
(d) (1 + x)e^xy
Solution:
(b) (1 + xy)e^xy
Hint:

Question 7.
If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in our calculation of the volume is …….
(a) 0.4 cu.cm
(b) 0.45 cu.cm
(c) 2 cu.cm
(d) 4.8 cu.cm
Solution:
(d) 4.8 cu.cm
Hint.
a = 4 cm
da = 0.1 cm
v = a³
dv = 3a²da
= 3(4)² (0.1)
= 4.8 cu. cm

Question 8.
The change in the surface area S = 6x² of a cube when the edge length varies from x₀ to x₀ + dx is …….
(a) 12x₀ + dx
(b) 12x₀dx
(c) 6x₀dx
(d) 6x₀ + dx
Solution:
(b) 12x₀dx

Question 9.
the approximate change in the volume: V of a cube of side x mentres caused by increasing the side by 1% is ……..
(a) 0.3 xdx m³
(b) 0.03 x m³
(c) 0.03 x² m³
(d) 0.03 x³m³
Solution:
(d) 0.03 x³m³
Hint:
Let the side of the cube be x units
v = x³
when dx = 0.01x
dv = 3x²dx
= 3x²(0.01 x)
= 0.03 x³m³

Question 10.
If g(x, y) = 3x² – 5y + 2y², x(t) = e^t and y(t) = cos t, then \(\frac{d g}{d t}\) is equal to ……..
(a) 6e^2t + 5 sin t – 4 cos t sin t
(b) 6e^2t – 5 sin t + 4 cos t sin t
(c) 3e^2t + 5 sin t + 4 cos t sin t
(d) 3e^2t – 5 sin t + 4 cos t sin t
Solution:
(a) 6e^2t + 5 sin t – 4 cos t sin t

Question 11.


Solution:
(b) \(\frac{1}{(x+1)^{2}} d x\)
Hint:

Question 12.

(a) -4
(b) -3
(c) -7
(d) 13
Solution:
(c) -7
Hint:

Question 13.


Solution:
(b) \(-x+\frac{\pi}{2}\)
Hint:

Question 14.
If w(x, y, z) = x² (y – z) + y² (z – x) + z² (x – y), then
(a) xy + yz + zx
(a) xy + yz + zx
(b) x(y + z)
(c) y(z + x)
(d) 0
Solution:
(d) 0
Hint:

Question 15.
If f(x, y, z) = xy + yz + zx, then f_x – f_z is equal to ……….
(a) z – x
(b) y – z
(c) x – z
(d) y – x
Solution:
(a) z – x
Hint:
f_x = y + z
f_z = y + x
f_x – f_z = y + z – y – x = z – x

Additional Problems

Question 1.
If u = x^y then is equal to ……..
(a) yx^{y – 1}
(b) u log x
(c) u log y
(d) xy^{x – 1}
Solution:
(a) yx^{y – 1}
Hint:

Question 2.

(a) 0
(b) 1
(c) 2
(d) 4
Solution:
(c) 2
Hint:

Question 3.


Solution:
(d) -u
Hint:

Question 4.
The curve y² (x- 2) = x² (1 +x) has …….
(a) an asymptote parallel to x-axis
(b) an asymptote parallel to y-axis
(c) asymptotes parallel to both axis
(d) no asymptotes
Solution:
(b) an asymptote parallel to y-axis
Hint:

Question 5.
If x = r cos θ, y = r sin θ, then \(\frac{\partial r}{\partial x}\) is equal to …….
(a) sec θ
(b) sin θ
(c) cos θ
(d) cosec θ
Solution:
(c) cos θ
Hint:

Question 6.

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

Question 7.
The percentage error in the 11^th root of the number 28 is approximately …. times the precentage error is 28.

Solution:
\(\frac{1}{11}\)
Hint:

Take log on both sides

The percentage error is approximately \(\frac{1}{11}\) times the percentage error is 28.

Question 8.
The curve a²y² = x² (a² – x²) has ……
(a) only one loop between x = 0 and x = a
(b) two loops between x = 0 and x = a
(c) two loops between x = – a and x = a
(d) no loop
Solution:
(c) two loops between x = – a and x = a
Hint.

Question 9.
An asymptote to the curve y² (a + 2x) = x² (3a – x) is ………
(a) x = 3a
(b) x = – a/2
(c) x = a/2
(d) x = 0
Solution:
(b) x = – a/2
Hint.

Question 10.
In which region the curve y² (a + x) = x² (3a – x) does not lie?
(a) x > 0
(b) 0 < x < 3a (c) x ≤ – a and x > 3a
(d) – a < x < 3a Solution: (c) x ≤ – a and x > 3a
Hint.
y² (a + x) = x² (3a – x)
Taking y = 0 (1) ⇒ 0 = x² (3a – x)
= x = 0, x = 3a
∴ The points are (0, 0) (3a, 0)
There is a loop between x = 0 and x = 3a

when x > 3 a, y ➝ imaginary
the curves does not exist beyond x = 3a i.e., x > 3a
the curve has asymptote at x = -a
the curve does not exist when x < -a
the curve exists in the region – a < x < 3a

Question 11.
If M = y sin x, then \(\frac{\partial^{2} u}{\partial x \partial y}\) is equal to …….
(a) cos x
(b) cos y
(c) sin x
(d) 0
Solution:
(a) cos x
Hint:

Question 12.

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

Question 13.
The curve 9y² = x² (4 – x²) is symmetrical about …….
(a) y – axis
(b) x – axis
(c) y = x
(d) both the axes
Solution:
(d) both the axes
Hint.
Replace x by – x and y by -y
9 (-y²) = (-x)²(4-(-x)²)
The equation is unaltered
∴ the curve is symmetrical about both the axes.

Question 14.
The curve ay² = x² (3a – x) cuts the y – axis at …….
(a) x = – 3a, x = 0
(b) x = 0, x = 3a
(c) x = 0, x = a
(d) x = 0
Solution:
(d) x = 0
Hint:
Given ay² = x² (3a – x)
The point of intersection with y-axis by putting x = O
In (1) = ay² = 0 (3a – 0)
ay = 0 ; y = 0
∴ The curve intersects y-axis at the origin is x = 0

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.5

Question 1.
Compute P(X = k) for the binomial distribution, B(n, p) where

Solution:

Question 2.
The probability that Mr. Q hits a target at any trial is \(\frac{1}{4}\). Suppose he tries at the target 10 times. Find the probability that he hits the target
(i) exactly 4 times
(ii) at least one time.
Solution:
Let ‘p’ be the probability of hitting the trial

and number of trials ‘n’ = 10
Probability of ‘x’ success in ‘n’ trials is

(i) Probability that Mr.Q hits the target exactly 4 times is

(ii) Probability that Mr.Q hits the target atleast one time is
P(X ≥ 1) = 1 – P(X < 1)
= 1 – P(X = 0)

Question 3.
Using binomial distribution find the mean and variance of X for the following experiments
(i) A fair coin is tossed 100 times, and X denote the number of heads.
(ii) A fair die is tossed 240 times, and X denote the number of times that four appeared.
Solution:
(i) n = 100, ‘X’ denotes the number of heads.

(ii) n = 240, ‘X’ denotes the number of times four appeared.

Question 4.
The probability that a certain kind of component will survive a electrical test is \(\frac{3}{4}\). Find the probability that exactly 3 of the 5 components tested survive.
Solution:
Given n = 5

Question 5.
A retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 5%. The inspector of the retailer randomly picks 10 items from a shipment. What is the probability that there will be
(i) at least one defective item
(ii) exactly two defective items.
Solution:
Given n = 10

Let ‘X’ be the random variable denotes the number of defective items.
∴ Probability of ‘x’ successes in ‘n’ trials is

Question 6.
If the probability that a fluorescent light has a useful life of at least 600 hours is 0.9, find the probabilities that among 12 such lights.
(i) exactly 10 will have a useful life of at least 600 hours;
(ii) at least 11 will have a useful life of at least 600 hours;
(iii) at least 2 will not have a useful life of at least 600 hours.
Solution:
Given n = 12
Probability that a fluorescent light has a life of atleast of 600 hours is p = 0.9
∴ q = 1 – p = 1 – 0.9 = 0.1
Let ‘X’ be the number of bulbs.
∴ Probability of ‘x’ successes in ‘n’ trials is

(i) Probability that exactly 10 bulbs will have a useful life of atleast 600 hours

(ii) Probability that atleast 11 will have a useful life of atleast 600 hours is

(iii) Probability that atleast 2 will not have a useful life of 600 hours is

Question 7.
The mean and standard deviation of a binomial variate X are respectively 6 and 2. Find
(i) the probability mass function
(ii) P(X = 3)
(iii) P(X ≥ 2) .
Solution:

Question 8.
If X ~ B(n, p) such that 4P(X = 4) = P(x = 2) and n = 6. Find the distribution, mean and standard deviation.
Solution:

Question 9.
In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the mean and variance of the distribution.
solution:
Number of trials n = 5
Probability of ‘x’ successes in ‘n’ trials is

Given P(X = 1) = 0.4096 and P (X = 2) = 0.2048

Dividing Eq.(1) by Eq.(2)

Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.5 Additional Problems

Question 1.
In a Binomial distribution if n = 5 and P(X = 3) = 2P(X = 2) find p.
Solution:

Question 2.
If the sum of mean and variance of a Binomial Distribution is 4.8 for 5 trials find the distribution.
Solution:
np + npq = 4.8 ⇒ np (1 + q) = 4.8
5p [1 + (1 – p)] = 4.8
p² – 2p + 0.96 = 0 ⇒ p = 1.2, 0.8
∴ p = 0.8 ; q = 0.2 [∵ p cannot be greater than 1]
∴ The Binomial distribution is P[X = x] = 5C_x (0.8)^x (0.2)^{5 – x}, x = 0 to 5

Question 3.
If on an average 1 ship out of 10 do not arrive safely to ports. Find the mean and the standard deviation of the ships returning safely out of a total of 500 ships.
Solution:
Probability of a ship arriving safely

Question 4.
The overall percentage of passes in a certain examination is 80. If six candidates appear in the examination what is the probability that at least five pass the examination.
Solution:
Pass percentage = 80%
∴ Probability of a candidate passing in the examination

Question 5.
In a hurdle race a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6. What is the probability that he will knock down less than 2 hurdles?
Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.3

Identify the type of conic section for each of the equations.
Question 1.
2x² – y² = 7
Solution:
Comparing this equation with the general equation of the conic
Ax² + Bxy + Cy² + Dx + Ey + F = 0
We get A = 2, C = – 1
Elere A ≠ C also A and C are of opposite signs.
So the conic is a hyperbola.

Question 2.
3x² + 3y² – 4x + 3y + 10 = 0
Sol. Comparing this equation with the general equation of the conic
Ax² + Bxy + Cy² + Dx + Ey + F = 0
We get A = C also B = 0
So the given conic is a circle.

Question 3.
3x² + 2y² = 14
Solution:
Comparing this equation with the general equation of the conic
Ax² + Bxy + Cy² + Dx + Ey + F = 0
We get A ≠ C also C are of the same sign.
So the given conic is an ellipse.

Question 4.
x² + y² + x – y = 0
Solution:
Comparing this equation with the general equation of the conic
Ax² + Bxy + Cy² + Dx + Ey + F = 0
We get A = C and B = 0
So the given conic is a circle.

Question 5.
11x² – 25y² – 44x + 50y – 256 = 0
Solution:
Comparing this equation with the general equation of the conic
Ax² + Bxy + Cy² + Dx + Ey + F = 0
We get A ≠ C. Also A and C are of opposite sign.
So the conic is a hyperbola.

Question 6.
y² + 4x + 3y + 4 = 0
Solution:
Comparing this equation with the general equation of the conic
Ax² + Bxy + Cy² + Dx + Ey + F = 0
We get A = 0 and B = 0
So the conic is a parabola.

Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.3 Additional Problems

Identify the type of conic section for each of the following equations
(i) x² – 4y² + 6x + 16y – 11 = 0
(ii) y² – 8y + 4x – 3 = 0
(iii) 4x² – 9y² = 36
(iv) 16x² + 25y² = 400
(v) 16x² + 9y² + 32x – 36y – 92 = 0
(vi) x² + 4y² – 8x – 16y – 68 = 0
(vii) x² + y² – 4x + 6y – 17 = 0
Solution:
(i) Hyperbola
(ii) Parabola
(iii) Hyperbola
(iv) Ellipse
(v) Ellipse
(vi) Ellipse
(vii) Circle

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.5

Question 1.
If w(x, y) = x³ – 3xy + 2y², x, y ∈ R, find the linear approximation for w at (1, -1) .
Solution:
w(x, y) = x³ – 3xy + 2y² ; at (1, -1)
Linear approximation is given by

(1) ⇒ Let(x, y) = 6 + 6(x – 1) – 7(y + 1)
= 6x – 6 – 7y – 7
= = 6x – 7y – 7

Question 2.
Let z(x, y) = x²y + 3xy⁴, x, y ∈ R, Find the linear approximation for z at (2, -1).
Solution:

Linear approximation is given by

Question 3.
If v(x, y) = x² – xy + \(\frac{1}{4}\)y² + 7, x, y ∈ R, find the differential dv.
Solution:

The differential is dv = (2x -y) dx + (- x + \(\frac{y}{2}\))dy

Question 4.
Let W(x, y, z) = x² – xy + 3sinz,x,y, z ∈ R. Find the linear approximation at (2, -1, 0).
Solution:
w (x, y, z) = x² – xy + 3 sin z
Here(x₀, y₀, y₀) = (2, -1, 0)

Linear approximation is given by

Question 5.
Let V (x, y, z) = xy + yz + zx, x, y, z ∈ R. Find the differential dV.
Solution:
V(x, y, z) = xy + yz + zx
V_x = y + z
V_y = x + z
V_z = y + x
The differential is dV = (y + z) dx + (x + z) dy + (y + x) dz

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.4

Question 1.
Find the partial derivatives of the following functions at the indicated points.
(i) f(x, y) = 3x² – 2xy + y² + 5x + 2, (2, -5)
(ii) g(x, y) = 3x² + y² + 5x + 2, (1, -2)

(iv) G (x, y) = e*^{x + 3y} log (x² + y²), (-1, 1)
Solution:

Question 2.
For each of the following functions find the f_x, f_y and show that f_xy = f_yx

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
For each of the following functions find the g_xy, g_xx, g_yy and g_yx.
(i) g(x, y) = xe_y + 3x₂y
(ii) g(x, y) = log(5x + 3y)
(iii) g(x, y) = x₂ + 3xy – 7y + cos(5x)
Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.
A firm produces two types of calculators each week, x number of type A and j number of type B . The weekly revenue and cost functions (in rupees) are
R(x, y) = 80x + 90y + 0.04xy – 0.05.x² – 0.05y² and C(x, y) = 8x + 6y + 2000 respectively.
(i) Find the profit function P(x,y),

Solution:
(i) Profit = Revenue – Cost
= (80x + 90y + 0.04 xy – 0.05 x² – 0.05y²) – (8x + 6y + 2000)
= 80x + 90y + 0.04 xy – 0.05 x² – 0.05y² – 8x – 6y – 2000
P(x, y) = 72x + 84y + 0.04 xy – 0.05 x² – 0.05y² – 2000

Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.4 Additional Problems

Question 1.

Solution:

Question 2.
If U = (x – y) (y – z) (z – x) then show that U_x + U_y + U_z = 0
Solution:
U_x = (y – z) {(x – y)(-1) + (z – x). 1}
= (y – z) [(z – x) – (x – y)]
Similarly U_y = (z – x) [(x – y) – (y – z)]
_z = (x – y)[(y – z) – (z – x]
U_x + U_y + U_z = (y – z) [(z – x) – (z – x)] + (x – y) [- (y – z) + (y – z)] + (z – x) [(x – y) – (x – y)]
∴ U_x + U_y + U_z = 0.
Hence proved.

Question 3.

Solution:

Question 4.

Solution: