Class 12

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Ex 12(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Exercise 12(b)

Question 1.
Each question given below has four possible answers, out of which only one is correct. Choose the correct one.
(i) (2î – 4ĵ) . (î + ĵ + k̂) = _______.
(a) -3
(b) +2
(c) -1
(d) -2
Solution:
(d) -2

(ii) If a = î + 2ĵ – k̂, b = î + ĵ + 2k̂, c = 2î – ĵ; then
(a) \(\vec{a} \perp \vec{b}\)
(b) \(\vec{b} \perp \vec{c}\)
(c) \(\vec{a} \perp \vec{c}\)
(d) no pair of vectors are perpendicular.
Solution:
(c) \(\vec{a} \perp \vec{c}\)

(iii) (-3, λ, 1) ⊥ (1, 0, -3) ⇒ λ = _______.
(a) 0
(b) 1
(c) impossible to find
(d) any real number
Solution:
(c) impossible to find

(iv) If \(\vec{a} \cdot \vec{b}=\vec{c} \cdot \vec{a}\) for all vectors \(\vec{a}\), then
(a) \(\vec{a} \perp(\vec{b}-\vec{c})\)
(b) \(\vec{b}-\vec{c}\) = 0
(c) \(\vec{b} \neq \vec{c}\)
(d) \(\vec{b}+\vec{c}=\overrightarrow{0}\)
Solution:
(b) \(\vec{b}-\vec{c}\) = 0

Question 2.
Find the scalar product of the following pairs of vectors and the angle between them.
(i) 3î – 4ĵ and -2î + ĵ
Solution:

(ii) 2î – 3ĵ + 6k̂ and 2î – 3ĵ – 5k̂
Solution:

(iii) î – ĵ and ĵ + k̂
Solution:

(iv) \(\vec{a}\) = (2, -2, 1) and \(\vec{b}\) = (0, 2, 4)
Solution:

Question 3.
If A, B, C are the points (1, 0, 2), (0, 3, 1) and (5, 2, 0) respectively, find m∠ABC.
Solution:

Question 4.
Find the value of λ so that the vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other.
(i) \(\vec{a}\) = 3î + 4ĵ, \(\vec{b}\) = -5î + λĵ
Solution:
If \(\vec{a}\) and \(\vec{b}\) are perpendicular \(\vec{a} \cdot \vec{b}\) = 0
⇒ (3î + 4ĵ) . (-5î + λĵ) = 0
⇒ -15 + 4λ = 0
⇒ λ = \(\frac{15}{4}\)

(ii) \(\vec{a}\) = î + ĵ + λk̂, \(\vec{b}\) = 4î – 3k̂
Solution:
If \(\vec{a}\) and \(\vec{b}\) are perpendicular \(\vec{a} \cdot \vec{b}\) = 0
⇒ ( î + ĵ + λk̂) . (4î – 3k̂) = 0
⇒ 4 + 0 – 3λ = 0
⇒ λ = \(\frac{4}{3}\)

(iii) \(\vec{a}\) = 2î – ĵ – k̂, \(\vec{b}\) = λî + ĵ + 5k̂
Solution:
If \(\vec{a}\) and \(\vec{b}\) are perpendicular \(\vec{a} \cdot \vec{b}\) = 0
⇒ (2î – ĵ – k̂) . (λî + ĵ + 5k̂) = 0
⇒ 2λ – 1 – 5 = 0
⇒ 2λ = 6
⇒ λ = 3

(iv) \(\vec{a}\) = (6, 2, -3), \(\vec{b}\) = (1, -4, λ)
Solution:
If \(\vec{a}\) and \(\vec{b}\) are perpendicular \(\vec{a} \cdot \vec{b}\) = 0
⇒ (6, 2, -3) . (1, -4, λ) = 0
⇒ 6 – 8 – 3λ = 0
⇒ -2 – 3λ = 0
⇒ λ = –\(\frac{2}{3}\)

Question 5.
Find the scalar and vector projections of \(\vec{a}\) on \(\vec{b}\).
(i) \(\vec{a}\) = î, \(\vec{b}\) = ĵ
Solution:

(ii) \(\vec{a}\) = î + ĵ, \(\vec{b}\) = ĵ + k̂
Solution:

(iii) \(\vec{a}\) = î – ĵ – k̂, \(\vec{b}\) = 3î + ĵ + 3k̂
Solution:

Question 6.
In each of the problems given below, find the work done by a force \(\overrightarrow{F}\) acting on a particle, such that the particle is displaced from a point A to a point B.
(i) \(\overrightarrow{F}\) = 4î + 2ĵ + 3k̂
A (1, 2, 0), B (2, -1, 3)
Solution:
Displacement of the particle \(\overrightarrow{S}=\overrightarrow{AB}\)
= (2 – 1)î + (-1 – 2)ĵ + (3 – 0)k̂
=î – 3ĵ + 3k̂
Work done = \(\overrightarrow{F} \cdot \overrightarrow{S}\)
= (4î + 2ĵ + 3k̂) . (î – 3ĵ + 3k̂)
= 4 – 6 + 9
= 7 units.

(ii) \(\overrightarrow{F}\) = 2î + ĵ – k̂
A (0, 1, 2), B (-2, 3, 0)
Solution:
Displacement
\(\vec{S}\) = (-2 – 0)î + (3 – 1)ĵ + (0 – 2)k̂
= -2î + 2ĵ – 2k̂
Work done = \(\overrightarrow{F} \cdot \overrightarrow{S}\)
= (2î + ĵ – k̂) . (-2î + 2ĵ – 2k̂)
= -4 + 2 + 2
= 0 units.

(iii) \(\overrightarrow{F}\) = 4î – 3k̂
A (1, 2, 0), B (0, 2, 3)
Solution:
Displacement \(\vec{S}\) = -î + 3k̂
Work done = \(\overrightarrow{F} \cdot \overrightarrow{S}\)
= (4î – 3k̂) . (-î + 3k̂)
= -4 – 9
= -13 units.

(iv) \(\overrightarrow{F}\) = 3î – ĵ – 2k̂
A (-3, -4, 1), B (-1, -1, -2)
Solution:
Displacement \(\vec{S}\) = 2î + 3ĵ – 3k̂
Work done \(\overrightarrow{F} \cdot \overrightarrow{S}\)
= (3î – ĵ – 2k̂) . (2î + 3ĵ – 3k̂)
= 6 – 3 + 6
= 9 units.

Question 7.
If \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 0 show that \(|\vec{a}|=|\vec{b}|\).
Solution:

Question 8.
(i) If a and b are perpendicular vectors show that

Solution:

(ii) Prove that two vectors are perpendicular iff \(|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2\)
Solution:

Question 9.
If \(\vec{a}, \vec{b}, \vec{c}\) are mutually perpendicular vectors of equal magnitude, show that \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to \(\vec{a} \cdot \vec{b} \cdot \vec{c}\)
Solution:

Question 10.
Prove the following by vector method.
(i) Altitudes of a triangle are concurrent;
Solution:
Let ABC be a triangle.

⇒ CF is perpendicular to AB.
Hence the altitudes of a triangle are concurrent.

(ii) Median to the base of an isosceles triangle is perpendicular to the base;
Solution:

⇒ OD is perpendicular to the base AB.
Hence the median to the base of an isosceles triangle is perpendicular to the base. (Proved)

(iii) The parallelogram whose diagonals are equal is a rectangle;
Solution:

⇒ m∠COA = 90°
Hence OABC is a rectangle. (Proved)

(iv) The diagonals ofa rhombus are at right angles;
Solution:

Hence the diagonals of a rhombus are at right angles. (Proved)

(v) An angle inscribed in a semi-circle is a right angle;
Solution:

∴ m∠ABC = 90°
Hence the angle inscribed in a semi-circle is a right-angle. (Proved)

(vi) In any triangle ABC; a = b cos C + c cos B;
Solution:

(vii) In a triangle AOB, m∠AOB = 90°. If P and Q are the points of trisection of AB, prove that OP2 + OQ2 = \(\frac{5}{9}\) AB2;
Solution:

(viii) Measure of the angle between two diagonals of a cube is cos-1\(\frac{1}{3}\).
Solution:

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Ex 12(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Exercise 12(c)

Question 1.
Each question given below has four possible answers out of which only one is correct. Choose the correct one.
(i) (î + k̂) × (î + ĵ + k̂) = ______.
(a) î – k̂
(b) k̂ – î
(c) k̂ – 2î – ĵ
(d) 2
Solution:
(î + k̂) × (î + ĵ + k̂) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= î (0 – 1) – ĵ (1 – 1) + k̂ (1 – 0)
= -î + k̂ = k̂ – î

(ii) A vector perpendicular to the vectors î + ĵ and î + k̂ is ______.
(a) î – ĵ – k̂
(b) ĵ – k̂ + î
(c) k̂ – ĵ – î
(d) ĵ + k̂ + î
Solution:
A vector perpendicular to the vectors î + ĵ and î + k̂ is
(î + ĵ) × (î + k̂) = \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 0 \\
1 & 0 & 1
\end{array}\right|\)
= î (1 – 0) – ĵ (1 – 0) + k̂ (0 – 1)
= î – ĵ – k̂

(iii) The area of the triangle with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1) is ______.
(a) \(\frac{1}{2}\)
(b) 1
(c) \(\frac{\sqrt{3}}{2}\)
(d) 2
Solution:

(iv) If â and b̂ are unit vectors such that â × b̂ is a unit vector, then the angle between â and b̂ is ______.
(a) of any measure
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{2}\)
(d) π
Solution:
|a × b| = ab sin θ = sin θ
⇒ sin θ = 1
⇒ θ = \(\frac{\pi}{2}\)

(v) If \(\vec{a}, \vec{b} \text { and } \vec{c}\) are non-zero vectors, then \(\vec{a} \times \vec{b}=\vec{a} \times \vec{c}\) ______.
(a) \(\vec{b}=\vec{c}\)
(b) \(\vec{a} \|(\vec{b}-\vec{c})\)
(c) \(\vec{b} \| \vec{c}\)
(d) \(\vec{b} \perp \vec{c}\)
Solution:

Question 2.
Let \(\vec{a}\) = 2î + ĵ, \(\vec{b}\) = -î + 3ĵ + k̂ and \(\vec{c}\) = î + 2ĵ + 5k̂ be three vectors. Find
(i) \(\vec{c} \times \vec{a}\)
Solution:

(ii) \(\vec{a} \times(-\vec{b})\)
Solution:

(iii) \((\vec{a}-2 \vec{b}) \times \vec{c}\)
Solution:

(iv) \((\vec{a}-\vec{c}) \times \vec{c}\)
Solution:

(v) \((\vec{a}-\vec{b}) \times(\vec{c}-\vec{a})\)
Solution:

Question 3.
Find the unit vectors perpendicular to the vectors
(i) î, k̂
Solution:

(ii) î + ĵ, î – k̂
Solution:

(iii) 2î + 3k̂, î – 2ĵ
Solution:

(iv) 2î – 3ĵ + k̂, -î + 2ĵ – k̂
Solution:

Question 4.
Determine the area of parallelogram whose adjacent sides are the vectors
(i) 2î, ĵ
Solution:

(ii) î + ĵ, -î + 2ĵ
Solution:

(iii) 2î + ĵ + 3k̂, î – ĵ
Solution:

(iv) (1, – 3, 1), (1, 1, 1).
Solution:

Question 5.
Calculate the area of the traingle ABC (by vector method) where
(i) A (1, 2, 4), B (3, 1, -2), C (4, 3, 1)
Solution:

(ii) A (1, 1, 2), B (2, 2, 3), C (3, -1, -1).
Solution:

Question 6.
Determine the sine of the angle between the vectors
(i) 5î – 3ĵ, 3î – 2k̂
Solution:

(ii) î – 3ĵ + k̂, î + ĵ + k̂
Solution:

Question 7.
Show that \((\vec{a} \times \vec{b})^2\) = a²b² – \((\vec{a}, \vec{b})^2\)
Solution:

Question 8.
If \(\vec{a} \times \vec{b}=\vec{b} \times \vec{c} \neq \overrightarrow{0}\), prove that \(\vec{a}+\vec{c}=m \vec{b}\), where m is a scalar.
Solution:

Question 9.
If \(\vec{a}\) = 2î + ĵ – k̂, \(\vec{b}\) = -î + 2ĵ – 4k̂, \(\vec{c}\) = î + ĵ + k̂, find \((\vec{a} \times \vec{b}) \cdot(\vec{a} \times \vec{c})\).
Solution:

Question 10.
If \(\vec{a}\) = 3î + ĵ – 2k̂, \(\vec{b}\) = 2î – 3ĵ + 4k̂ then verify that \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).
Solution:

Question 11.
Find the area of the parallelogram whose diagonals are vectors 3î + ĵ – 2k̂ and î – 3ĵ + 4k̂.
Solution:

Question 12.
Show that \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\). Interpret this result geometrically.
Solution:

= Vector area of the parallelogram ABCD.
Hence twice the vector area of a parallelogram ABCD is equal to the vector area of the parallelogram whose adjacent sides are the diagonals of the parallelogram ABCD.

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Additional Exercise Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Additional Exercise

(A) Multiple Choice Questions (Mcqs) With Answers

Question 1.
If f is an odd function, then write the value of \(\int_{-a}^a \frac{f(\sin x)}{f(\cos x)+f\left(\sin ^2 x\right)}\) dx
(a) 1
(b) 0
(c) -1
(d) 2
Solution:
(b) 0

Question 2.
If p and q are respectively degree and order of the differential equation y = e^dy/dx then write the relation between p and q.
(a) p ≠ q
(c) p ≡ q
(b) p = q
(d) None of these
Solution:
(b) p = q

Question 3.
Write the value of \(\int_0^1\){x} dx where {x} stands for fractional part of x.
(a) \(\frac{1}{2}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{2}{3}\)
Solution:
(a) \(\frac{1}{2}\)

Question 4.
Write the value of:
\(\int_0^{\pi / 2} \frac{\sin x}{\sin x+\cos x}\) dx – \(\int_0^{\pi / 2} \frac{\cos x}{\sin x+\cos x}\) dx
(a) 1
(b) 2
(c) 0
(d) π
Solution:
(c) 0

Question 5.
Write the value of \(\int_{\frac{\pi}{4}}^{\frac{\pi}{4}}\)sin⁵ x cos x dx
(a) 0
(b) 1
(c) cos x
(d) sin x
Solution:
(a) 0

Question 6.
Write the particular solution of the equation \(\frac{d y}{d x}\) = sin x given that y(π) = 2
(a) y = cos x + 1
(b) y = -cos x + 1
(c) y = -cos x – 1
(d) y = -sin x + 1
Solution:
(b) y = -cos x + 1

Question 7.
Write the degree of the following differential equation:
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = \(\frac{2 y^3+\left(\frac{d y}{d x}\right)^4}{\sqrt{\frac{d^2 y}{d x^2}}}\)
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3

Question 8.
Write the order ofthe following differential equation:
\(\frac{d^2 y}{d x^2}\) = \(\frac{2 y^3+\left(\frac{d y}{d x}\right)^4}{\sqrt{\frac{d^2 y}{d x^2}}}\)
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(c) 2

Question 9.
What is F(x) if F(x) = \(\int_0^x\)e^2t sin 3t dt?
(a) e^2x sin 3x
(b) e^2x cos 3x
(c) e^x sin 3x
(d) e^2x sin x
Solution:
(a) e^2x sin 3x

Question 10.
\(\int \frac{d x}{\cos ^2 x \sin ^2 x}\) = ?
(a) -2 cos 2x + C
(b) -2 cot 2x + C
(c) -2 sin 2x + C
(d) 2 cot 2x + C
Solution:
(b) -2 cot 2x + C

Question 11.
If \(\int_1^2\)f(x) dx= λ, then what is the value of \(\)f(3 – x) dx?
(a) λ
(b) λ²
(c) 1λ
(d) 2λ
Solution:
(a) λ

Question 12.
What is the value of \(\int_{-1}^1 \frac{d x}{1+x^2}\)?
(a) \(\frac{2 \pi}{2}\)
(b) 2π
(c) π
(d) \(\frac{\pi}{2}\)
Solution:
(d) \(\frac{\pi}{2}\)

Question 13.
Write the order of the following differential equation:
\(\frac{d^3 y}{d x^3}\) = \(\left(\frac{d^2 y}{d x^2}\right)^2\) + \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^4\) + y
(a) 1
(b) 3
(c) 2
(d) 0
Solution:
(b) 3

Question 14.
Write the degree of the following differential equation:
\(\frac{d^3 y}{d x^3}\) = \(\left(\frac{d^2 y}{d x^2}\right)^2\) + \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^4\) + y
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(a) 1

Question 15.
Write the particular solution of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (1 + x)⁴, y = 0 when x = -1.
(a) y = \(\frac{(1+x)^2}{5}\)
(b) y = \(\frac{(2+x)^5}{5}\)
(c) y = \(\frac{(1-x)^5}{5}\)
(d) y = \(\frac{(1+x)^5}{5}\)
Solution:
(d) y = \(\frac{(1+x)^5}{5}\)

Question 16.
Evaluate the integral ∫2x cosec² x² dx?
(a) cot x² + C
(b) -cot x² + C
(c) -cot 2x² + C
(d) cot 2x² + C
Solution:
(b) -cot x² + C

Question 17.
What is the value of \(\frac{d}{d x} \int_{250}^{300}\left(x^4+5 x^3\right)^2\) dx
(a) 0
(b) 1
(c) -1
(d) 2
Solution:
(a) 0

Question 18.
Write down the integral of ∫\(e^{x^2}\) 2x dx.
(a) \(e^{2 x^2}\)
(b) 2\(e^{2 x^2}\)
(c) \(e^{x^2}\)
(d) None of the above
Solution:
(c) \(e^{x^2}\)

Question 19.
What is the integral of ∫log e^x dx?
(a) \(\frac{2 x^2}{2}\) + C
(b) \(\frac{2 x^2}{3}\) + C
(c) \(\frac{x^2}{2}\) + C
(d) None of the above
Solution:
(c) \(\frac{x^2}{2}\) + C

Question 20.
What is the value of \(\int_{-2}^2\)|x| dx?
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0

Question 21.
\(\int_{-1}^1\)|1 – x| dx = ______.
(a) 0
(b) 1
(c) 2
(d) -1
Solution:
(c) 2

Question 22.
If ∫x³\(e^{c x^4}\)dx = \(\frac{1}{20} \mathrm{e}^{\mathrm{cx}}\) then C = ______.
(a) 0
(b) 2
(c) 4
(d) 5
Solution:
(d) 5

Question 23.
\(\int_a^b\)f(x) dx = 1 ⇒ \(\int_a^b\)k f(t)dt ______.
(a) k
(b) -k
(c) 2k
(d) None of the above
Solution:
(b) -k

Question 24.
\(\int_{-1}^1\)f(x) dx = k and f is an even function then \(\int_{-1}^1\)f(x) = ______.
(a) k
(b) -k
(c) 2k
(d) None of the above
Solution:
(c) 2k

Question 25.
If ∫\(\int_0^1\)f(x) dx = 4, \(\int_0^2\)f(t) dt and \(\int_4^2\)f(u) du = 1 then \(\int_1^4\)f(x) dx = ______.
(a) 0
(b) 1
(c) 3
(d) -3
Solution:
(d) -3

Question 26.
I(f) = \(\int_a^x\)f(t) dt and Df = f'(x) then (ID – DI) f = ______.
(a) -f(a)
(b) 2f(a)
(c) f(a)
(d) None of the above
Solution:
(a) -f(a)

Question 27.
\(\int_0^\pi\)cos¹⁰¹ x dx = ______.
(a) 0
(b) 1
(c) -1
(d) 101
Solution:
(a) 0

Question 28.
Let f satisfies all the conditions of Rolle’s theorem in [1, 6] then \(\int_1^6\)f'(x) dx = ______.
(a) 0
(b) 1
(c) -1
(d) 6
Solution:
(a) 0

Question 29.
\(\int_{-2}^2\)|x| dx = ______.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) 4

Question 30.
Integrate ∫log x dx
(a) x. log x + x + C
(b) x. log x – x + C
(c) log x – x + C
(d) None of these
Solution:
(b) x. log x – x + C

Question 31.
Evaluate \(\int_0^2\)[x – 1] dx
(a) 0
(b) 1
(c) -1
(d) 2
Solution:
(b) 1

Question 32.
What is the value of: ∫\(\frac{f^{\prime}(x)-f(x)}{e^x}\) dx?
(a) e^x f(x) + C.
(b) e^2x f(x) + C.
(c) e^-x f(x) + C.
(d) None of the above
Solution:
(c) e^-x f(x) + C.

Question 33.
What is the value of \(\int_0^1\)x(1 – x)⁹⁹ dx?
(a) \(\frac{1}{100}\)
(b) \(\frac{1}{10}\)
(c) \(\frac{1}{1010}\)
(d) \(\frac{1}{10100}\)
Solution:
(d) \(\frac{1}{10100}\)

Question 34.
Solution of \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = xy + x + y + 1 is ______.
(a) 2x + \(\frac{x^2}{2}\) + C
(b) x + \(\frac{x}{2}\) + C
(c) x + \(\frac{2 x^2}{2}\) + C
(d) x + \(\frac{x^2}{2}\) + C
Solution:
(d) x + \(\frac{x^2}{2}\) + C

Question 35.
f(x) = \(\int_0^x\)t sin t dt then f ‘(x) = ______.
(a) x cos x
(b) x sin t
(c) x sin x
(d) x tan x
Solution:
(c) x sin x

Question 36.
What is the value of the integral \(\int_a^b \frac{|x|}{x}\)dx?
(a) |b| – |a|
(b) |a| – |b|
(c) |b| + |a|
(d) |a| + |b|
Solution:
(a) |b| – |a|

Question 37.
What is the value of ∫x^x (1 + ln x) dx?
(a) x^2x + C
(b) x^x + C
(c) 2x^x + C
(d) x² + C
Solution:
(b) x^x + C

Question 38.
Evaluate: \(\int_0^{\mathrm{p} / 2}\)ln(cot x) dx.
(a) 0
(b) 1
(c) cot x
(d) sin x
Solution:
(a) 0

Question 39.
Evaluate: \(\int_{-3}^4\)|x| dx
(a) \(\frac{2}{25}\)
(b) \(\frac{25}{2}\)
(c) \(\frac{25}{4}\)
(d) \(\frac{25}{-3}\)
Solution:
(b) \(\frac{25}{2}\)

Question 40.
Evaluate: \(\int_0^{\frac{\pi}{2}}\)(cos x – sin x) dx
(a) 0
(b) 1
(c) -1
(d) π
Solution:
(a) 0

Question 41.
Evaluate: \(\int_0^{\frac{\pi}{2}}\)log tan x dx.
(a) 1
(b) -1
(c) 0
(d) π
Solution:
(c) 0

Question 42.
Integrate: \(\frac{d x}{3 e^x-1}\)
(a) \(\ln \left(\frac{e^{3 x}-1}{e^x}\right)\) + C
(b) \(\ln \left(\frac{3 e^x+1}{e^x}\right)\) + C
(c) \(\ln \left(\frac{3 e^x-1}{e^x}\right)\) + C
(d) \(\ln \left(\frac{3 e^x+1}{e^{3 x}}\right)\) + C
Solution:
(c) \(\ln \left(\frac{3 e^x-1}{e^x}\right)\) + C

Question 43.
Evaluate: \(\int_0^1 \ln \left(\frac{1}{x}-1\right)\)dx
(a) 1
(b) 2
(c) 0
(d) -1
Solution:
(c) 0

Question 44.
Evaluate: ∫e^x\(\left(\frac{1-\sin x}{1-\cos x}\right)\)dx
(a) -e^x cot\(\frac{x}{2}\) + C
(b) e^x tan\(\frac{x}{2}\) + C
(c) e^x cot\(\frac{x}{2}\) + C
(d) -e^x sin\(\frac{x}{2}\) + C
Solution:
(a) -e^x cot\(\frac{x}{2}\) + C

Question 45.
Evaluate: \(\int_0^1\)x log(1 + x) dx
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{2}{3}\)
Solution:
(b) \(\frac{1}{4}\)

Question 46.
What is the integrating factor of the equation y’ + y cot x = cosec x?
(a) cot x
(b) sin x
(c) cos x
(d) cosec x
Solution:
(b) sin x

(B) Very Short Type Questions With Answers

Question 1.
Write the order of the differential equation whose solution is given by
y = (c₁ + c₂) cos (x + c₃) + c₄\(e^{x+c_5}\) where c₁, c₂, c₄ and c₅ are arbitrary constants.
Solution:
y = (c₁ + c₂) cos (x + c₃) + c₄\(e^{x+c_5}\)
y = (c₁ + c₂) cos (x + c₃) + c₄\(e^{c_5}\).e^x
= A cos(x + c₃) + Be^x
Where c₁ + c₂ = A, c₄\(e^{c_5}\) = B
As there are 3 independent constants the order of the differential equation is 3.

Question 2.
If p and q are respectively degree and order of the differential equation y = e^dy/dx, then write the relation between p and q.
Solution:
Given differential equation is
y = \(e^{\frac{d y}{d x}}\) ⇒ \(\frac{d y}{d x}\) = ln y
Whose order = 1 = p
Degree = 1 = q
∴ p = q

Question 3.
Write the value of \(\int_0^1\){x} dx where {x} stands for fractional part of x.
Solution:

Question 4.
Write the order of the differential equation of the family of circles
ar² + ay² + 2gx + 2fy + c = 0
ax² + ay² + 2gx + 2fy + c = 0
Solution:
As there are 3 independent constants, the order of the differential equation is 3.

Question 5.
If p and q are the order and degree of the differential equation
y\(\left(\frac{d y}{d x}\right)^2\) + x² \(\frac{d^2 y}{d x^2}\) + xy = sin x, then choose the correct statement out of (i) p > q, (ii) p = q, (iii) p < q.
Solution:
Order of the given differential = p = 2
Degree of the given differential equation = q = 1
∴ p > q

Question 6.
Write the order of the differential equation of the system of ellipses:
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
Solution:
As there are two unknown constants in the system of ellipses \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 the order of the differential equation is 2.

Question 7.
What do you mean by integration? Write your answer in one sentence.
Solution:
Integration is the antiderivative of a function.

Question 8.
Write the differential equation of the family of straight lines parallel to the y-axis.
Solution:
\(\frac{d x}{d y}\) = 0 is the differential equation of family of lines parallel to y-axis.

Question 9.
Write the value of ∫\(\int_{-\pi / 4}^{\pi / 4}\)sin⁵ x cos x dx.
Solution:
Let f(x) = sin⁵ x cos x
f(-x) = sin⁵ (-x) cos (-x)
= -sin⁵ x cos x = -f(x)
i.e. f is an odd function.
Thus \(\int_{-\pi / 4}^{\pi / 4}\)sin⁵ x cos x dx = 0

Question 10.
Write the degree of the differential equation ln\(\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right)\) = y
Solution:
The degree of the differential equation ln\(\left(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\right)\) = y is 1.

Question 11.
What is F'(t) if F(t) = \(\int_a^t\)e^3x .cos 2x dx ?
Solution:
F(t) = \(\int_a^t\)e^3x .cos 2x dx
⇒ F'(t) = e^3x cos 2t

Question 12.
Write the order and degree of the following differential equation:
\(\frac{d^2 y}{d x^2}\) = \(\frac{2 y^3+\left(\frac{d y}{d x}\right)^4}{\sqrt{\frac{d^2 y}{d x^2}}}\)
Solution:
Order = 2, Degree = 3

Question 13.
∫\(\frac{\cot x d x}{\ln \sin x}\) = ?
Solution:
∫\(\frac{\cot x d x}{\ln \sin x}\) = ln(ln sin x) + C

Question 14.
What is F'(x) if F(x) = \(\int_0^{\mathbf{x}}\)e^2t sin 3t dt?
Solution:
If F(x) = \(\int_0^{\mathbf{x}}\)e^2t sin 3t dt then F'(x) = e^2x sin 3x

Question 15.
∫\(\frac{d x}{\cos ^2 x \sin ^2 x}\) = ?
Solution:
∫\(\frac{d x}{\cos ^2 x \sin ^2 x}\) = 4∫\(\frac{d x}{\sin ^2 2 x}\)
= 4∫cosec² 2x dx = -2 cot 2x + C

Question 16.
What is the value of ∫\(\frac{d}{d x}\)f(x) dx – \(\frac{d}{d x}\)(∫f(x) dx)?
Solution:
∫\(\frac{d}{d x}\)f(x) dx – \(\frac{d}{d x}\)(∫f(x) dx)
= f(x) + C – f(x) = C (constant)

Question 17.
If \(\int_1^2\)f(x) dx = λ, then what is the value \(\int_1^2\)f(3 – x) dx?
Solution:
If \(\int_1^2\)f(x) dx = λ, then \(\int_1^2\)f(3 – x) dx = λ

Question 18.
What is the value of \(\int_{-1}^1 \frac{d x}{1+x^2}\)?
Solution:
\(\int_{-1}^1 \frac{d x}{1+x^2}\) = \(\left[\tan ^{-1} x\right]_{-1}^1\)
= tan^-1 1 – tan^-1 (-1)
= tan^-1 1 + tan^-1 1
= 2tan^-1 (1) = 2 . \(\frac{\pi}{4}\) = \(\frac{\pi}{2}\)

Question 19.
Write the order and the degree of the following differential equation:
\(\frac{d^3 y}{d x^3}\) = \(\left(\frac{d^2 y}{d x^2}\right)^2\) + \(\left(\frac{d y}{d x}\right)^4\) + y
Solution:
Order = 3
Degree = 1

Question 20.
Write the particular solution of \(\frac{d y}{d x}\) = (1 + x)⁴, y = 0 when x = -1.
Solution:
\(\frac{d y}{d x}\) = (1 + x)⁴ ⇒ \(\frac{(1+x)^5}{5}\) + C
Given y = 0 for x = -1
⇒ o = o + c ⇒ c = o
∴ The particular solution is y = \(\frac{(1+x)^5}{5}\)

(C) Short Type Questions With Answers

Question 1.
Evaluate: ∫\(\frac{2 x+1}{\sqrt{x^2+10 x+29}}\)dx
Solution:

Question 2.
Evaluate: \(\int_0^{\pi / 2} \frac{\cos x d x}{(2-\sin x)(3+\sin x)}\)
Solution:

Question 3.
Evaluate: ∫\(\frac{d x}{(1+x) \sqrt{1-x^2}}\)
Solution:

Question 4.
Solve: cosec x \(\frac{d^2 y}{d x^2}\) = x.
Solution:
cosec x \(\frac{d^2 y}{d x^2}\) = x => \(\frac{d^2 y}{d x^2}\) = x sin x
⇒ \(\frac{d y}{d x}\) = ∫x sin x dx + A
= x (-cos x) – ∫(-cos x) dx + A
= -x cos x + sin x + A
⇒ y = -∫x cos x dx + ∫sin x dx + A∫dx + B
= [x sin x – ∫sin x dx] – cos x + Ax = B
⇒ y = -x sin x – 2 cos x + Ax + B is the solution.

Question 5.
Find the particular solution of the following differential equation:
\(\frac{d y}{d x}\) = \(\frac{1+y^2}{1+x^2}\) given that y = √3 when x = 1
Solution:

Question 6.
Evaluate: \(\int_0^a x^2\left(a^2-x^2\right)^{5 / 2}\) dx
Solution:

Question 7.
Evaluate: \(\int_0^a \frac{d x}{e^{4 x}-5}\)
Solution:

Question 8.
Evaluate: ∫x² tan^-1 x dx.
Solution:

Question 9.
If f(x) = e^x + \(\frac{1}{1+x^2}\) and f(0) = 1, then find f(x).
Solution:
f(x) = e^x + \(\frac{1}{1+x^2}\)
⇒ f(x) = ∫\(\left(e^x+\frac{1}{1+x^2}\right)\)dx + C
= e^x + tan^-1 x + C
f(0) = 1
⇒ 1 = 1 + 0 + C => C = 0
Thus f(x) = e^x + tan^-1 x

Question 10.
Evaluate: ∫(log x)² dx
Solution:
I = ∫(log x)² dx
= (log x)². x – 2∫(log x) . \(\frac{1}{x}\) . x . dx
= x (log x)² – 2 ∫log x. dx
= x (log x)² – 2 {(log x) x – ∫dx}
= x (log x)² – 2x log x + 2x + C

Question 11.
Evaluate: ∫\(\frac{2 x+9}{(x+3)^2}\)dx
Solution:

Question 12.
Solve: ydy + e^-y x sin x dx = 0
Solution:
ydy = e^-y x sin x dx = 0
⇒ y e^y dy + x sin x dx = 0
⇒ ∫y e^y dy + ∫x sin x dx =C
⇒ y e^y – e^y + (-x cos x) + sin x = C
⇒ e^y (y – 1) – x cos x + sin x = C is the general solution.

Question 13.
Evaluate: ∫\(\frac{d x}{x \ln x \sqrt{(\ln x)^2-4}}\)
Solution:

Question 14.
Find the particular solution of the differential equation \(\frac{d^2 y}{d x^2}\) = 6x given that y = 1 and \(\frac{d y}{d x}\) = 2 when x = 0.
Solution:
\(\frac{d^2 y}{d x^2}\) = 6x ⇒ \(\frac{d y}{d x}\) = 6 . \(\frac{x^2}{2}\) + A
\(\frac{d y}{d x}\) = 3x² + A ⇒ y = x³ + Ax + B
Using the givne conditions x = 0, \(\frac{d y}{d x}\) = 2, y = 1, we get
2 = 0 + A ⇒ A = 2
and 1 = 0 + 0 + B ⇒ B = 1
The particular solution is y = x³ + 2x + 1

Question 15.
Evaluate: \(\int_0^{\frac{3}{2}}\)[x²] dx
Solution:

Question 16.
Find the differential equation whose general solution is ax² + by = 1, where a and b are arbitrary constants.
Solution:

Question 17.
Integrate: ∫\(\frac{\sin 6 x+\sin 4 x}{\cos 6 x+\cos 4 x}\) dx.
Solution:

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Ex 12(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Exercise 12(a)

Question 1.
Each question given below has four possible answers out of which only one is correct. Choose the correct one.
(i) \(\vec{a}\) = î + 2ĵ + k̂, \(\vec{b}\) = 2î – 2ĵ + 2k̂ and \(\vec{c}\) = -î + 2 ĵ + k̂ then
(a) \(\vec{a}\) and \(\vec{b}\) have the same direction
(b) \(\vec{a}\) and \(\vec{c}\) have opposite directions.
(c) \(\vec{b}\) and \(\vec{c}\) have opposite directions
(d) no pair of vectors have same direction
Solution:
(d) no pair of vectors have same direction

(ii) If the vectors \(\vec{a}\) = 2î + 3ĵ – 6k̂ and \(\vec{b}\) = -α î – ĵ + 2k̂ are parallel, then α = ______.
(a) 2
(b) \(\frac{2}{3}\)
(c) –\(\frac{2}{3}\)
(d) \(\frac{1}{3}\)
Solution:
(c) –\(\frac{2}{3}\)

(iii) If the position vectors of two points A and B are 3î + k̂, and 2î + ĵ – k̂, then the vector \(\overrightarrow{BA}\) is
(a) -î + ĵ – 2k̂
(b) î + ĵ
(c) î – ĵ + 2k̂
(d) î – ĵ – 2k̂
Solution:
(c) î – ĵ + 2k̂

(iv) If \(|k \vec{a}|\) = 1, then
(a) \(\vec{a}=\frac{1}{k}\)
(b) \(\vec{a}=\frac{1}{|k|}\)
(c) \(k=\frac{1}{|\vec{a}|}\)
(d) \(k=\frac{+1}{|\vec{a}|}\)
Solution:
(d) \(k=\frac{+1}{|\vec{a}|}\)

(v) The direction cosines of the vectors \(\overrightarrow{PQ}\) where \(\overrightarrow{OP}\) = (1, 0, -2) and \(\overrightarrow{OQ}\) = (3, -2, 0) are
(a) 2, -2, 2
(b) 4, -2, -2
(c) \(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
(d) \(\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}}\)
Solution:
(c) \(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)

Question 2.
Rectify the mistakes, if any
(i) \(\vec{a}-\vec{a}\) = 0
Solution:
\(\overrightarrow{0}\)

(ii) The vector \(\overrightarrow{0}\) has unique direction.
Solution:
indefinite direction

(iii) All unit vectors are equal.
Solution:
equal magnitude

(iv) \(|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\vec{b}\)
Solution:
\(\vec{a}=\vec{b} \Rightarrow|\vec{a}|=|\vec{b}|\)

(v) Subtraction of vectors is not commutative.
Solution:
true

Question 3.
(i) If \(\vec{a}\) = (2, 1), \(\vec{b}\) = (-1, 0), find \(3 \vec{a}+2 \vec{b}\).
Solution:
\(3 \vec{a}+2 \vec{b}\) = 3 (2, 1) + 2 (-1, 0)
= (6 – 2, 3 + 0)
= (4, 3 )

(ii) If \(\vec{a}\) = (1, 1, 1) , \(\vec{b}\) = (-1, 3, 0) and \(\vec{c}\) =(2, 0, 2), find \(\vec{a}+2 \vec{b}-\frac{1}{2} \vec{c}\).
Solution:
\(\vec{a}+2 \vec{b}-\frac{1}{2} \vec{c}\)
= (1, 1, 1) + 2 (-1, 3, 0) – \(\frac{1}{2}\)(2, 0, 2)
= (1 – 2 – 1, 1 + 6 – 0, 1 + 0 – 1)
= (-2, 7, 0)

Question 4.
If A, B, C and D are the vertices of a square, find \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}\).
Solution:
Let ABCD be a square.

Question 5.
The given points A, B, C are the vertices of a triangle. Determine the vectors \(\overrightarrow{A B}, \overrightarrow{B C} \text { and } \overrightarrow{C A}\) and the lengths of these vectors in the following cases.
(i) A (4, 5, 5), B (3, 3, 3), C (1, 2, 5)
Solution:

(ii) A (8, 6, 1), B (2, 0, 1), C (-4, 0, -5)
Solution:

Question 6.
Find the vector from origin to the midpoint of the vector \(\overrightarrow{{P}_1 {P}_2}\) joining the points P₁(4, 3) and P₂(8, -5).
Solution:
P₁ = (4, 3) and P₂ = (8, -5)
If P is the mid-point of P₁P₂ then P = (6, -1).
Position vector of P = \(\overrightarrow{{OP}}\) = 6î – ĵ

Question 7.
Find the vectors from the origin to the points of trisection the vector \(\overrightarrow{{P}_1 {P}_2}\) joining P₁ (-4, 3) and P₂ (5, -12).
Solution:

Question 8.
Find the vector from the origin to the intersection of the medians of the triangle whose vertices are A (5, 2, 1), B(-4, 7, 0) and C (5, -3, 5).
Solution:

Question 9.
Prove that the sum of all the vectors drawn from the centre of a regular octagon to its vertices is the null vector.
Solution:

Question 10.
Prove that the sum of the vectors represented by the sides of a closed polygon taken in order is a zero vector.
Solution:
Consider a closed polygon ABCDEFA.

Question 11.
(a) Prove that:
(i) \(|\overrightarrow{a}+\overrightarrow{{b}}| \leq|\overrightarrow{a}|+|\overrightarrow{b}|\)
State when the equality will hold;
Solution:

(ii) \(|\overrightarrow{a}-\overrightarrow{b}| \geq|\overrightarrow{a}|-|\overrightarrow{b}|\)
Solution:

(b) What is the geometrical significance of the relation \(|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|\)?
Solution:

Question 12.
Find the magnitude of the vector \(\overrightarrow{PQ}\), its scalar components and the component vectors along the coordinate axes, if P and Q have the coordinates.
(i) P (-1, 3), Q (1, 2)
Solution:

(ii) P (-1, -2), Q (-5, -6)
Solution:

(iii) P (1, 4, -3), Q (2, -2, -1).
Solution:

Question 13.
In each of the following find the vector \(\overrightarrow{PQ}\), its magnitude and direction cosines, if P and Q have co-ordinates.
(i) P (2, -1, -1), Q (-1, -3, 2);
Solution:

(ii) P (3, -1, 7), Q (4, -3, -1).
Solution:

Question 14.
If \(\vec{a}\) = (2, -2, 1), \(\vec{b}\) = (2, 3, 6) and \(\vec{c}\) = (-1, 0, 2), find the magnitude and direction of
\(\vec{a}-\vec{b}+2 \vec{c}\).
Solution:

Question 15.
Determine the unit vector having the direction of the given vector in each of the following problems:
(i) 5î – 12ĵ
Solution:

(ii) 2î + ĵ
Solution:

(iii) 3î + 6ĵ – k̂
Solution:

(iv) 3î + ĵ – 2k̂
Solution:

Question 16.
Find the unit vector in the direction of the vector \(\overrightarrow{r_1}-\overrightarrow{r_2}\), where \(\vec{r}_1\) = î + 2ĵ + k̂ and \(\vec{r}_2\) = 3î + ĵ – 5k̂.
Solution:

Question 17.
Find the unit vector parallel to the sum of the vectors \(\vec{a}\) = 2î + 4ĵ – 5k̂ and \(\vec{b}\) = î + 2ĵ + 3k̂. Also find its direction cosines.
Solution:

Question 18.
If the sum of two unit vectors is a unit vector, show that the magnitude of their difference is √3.
Solution:

Question 19.
The position vectors of the points A, B, C and D are 4î + 3ĵ – k̂, 5î + 2ĵ + 2k̂, 2î – 2ĵ – 3k̂ and 4î – 4ĵ + 3k̂ respectively. Show that AB and CD are parallel.
Solution:
Given that the
position vector of A = 4î + 3ĵ – k̂
position vector of B = 5î + 2ĵ + 2k̂
position vector of C = 2î – 2ĵ – 3k̂
position vector of D = 4î – 4ĵ + 3k̂

Question 20.
In each of the following problems, show by vector method that the given points are collinear.
(i) A (2, 6, 3), B (1, 2, 7) and C (3, 10, -1)
Solution:
Given that A = (2, 6, 3), B = (1, 2, 7) and C = (3, 10, -1)
Then

(ii) P (2, -1, 3), Q (3, -5, 1) and R (-1, 11, 9).
Solution:
Given that P = (2, -1, 3) Q = (3, -5, 1) and R = (-1, 11, 9)

Hence the points P, Q, R are collinear. (Proved)

Question 21.
Prove that the vectors 2î – ĵ + k̂, î – 3ĵ – 5k̂, 3î – 4ĵ – 4k̂ are the sides of a right angled triangle.
Solution:
Let A, B and C be the points whose position vectors are 2î – ĵ – k̂, î – 3ĵ – 5k̂ and 3î – 4ĵ – 4k̂ respectively.

Question 22.
Prove by vector method that:
(a) the medians of a triangle are concurrent;
Solution:

The symmetry of the result shows that the point G also lies on the other two medians.
Hence the medians are concurrent. (Proved)

(b) the diagonals of a parallelogram bisect each other;
Solution:

(c) the line segment joining the midpoints of two sides of a triangle is parallel to the third and half of it;
Solution:

(d) the lines joining the midpoints of consecutive sides of a quadrilateral is a parallelogram;
Solution:

⇒ SR = PQ and SR || PQ
Hence PQRS is a parallelogram.
(Proved)

(e) in any triangle ABC, the point P being on the side \(\overrightarrow{B C} \text {; if } \overrightarrow{P Q}\) is the resultant of the vectors \(\overrightarrow{A P}, \overrightarrow{P B}\) and \(\overrightarrow{P C}\) then ABQC is a parallelogram;
Solution:

Hence ABQC is parallelogram. (Proved)

(f) In a parallelogram, the line joining a vertex to the midpoint of an opposite side trisects the other diagonal.
Solution:

⇒ P divides BD into the ratio 1 : 2.
Similarly we can show that Q divides BD into the ratio 2 : 1.
Hence P, Q are the points of trisection of the diagonal BD. (Proved)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Ex 11(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Exercise 11(b)

Solve the following differential equations.
Question 1.
\(\frac{d y}{d x}\) + y = e^-x
Solution:
Given equation is \(\frac{d y}{d x}\) + y = e^-x … (1)
This is a linear differential equation.
Here P = 1, Q = e^-x
So the integrating factor
I.F. = e^{∫P dx} = e^∫dx = e^x
The solution of (1) is given by
ye^x = ∫e^-x . e^x dx = ∫dx = x + C
⇒ y – xe^-x + Ce^-x

Question 2.
(x² – 1)\(\frac{d y}{d x}\) + 2xy = 1
Solution:
Given equation is (x² – 1)\(\frac{d y}{d x}\) + 2xy = 1

Question 3.
(1 – x²)\(\frac{d y}{d x}\) + 2xy = x \(\sqrt{1-x^2}\)
Solution:
Given equation is

Question 4.
x log x \(\frac{d y}{d x}\) + y = 2 log x
Solution:
Given equation is

Question 5.
(1 + x²)\(\frac{d y}{d x}\) + 2xy = cos x
Solution:

Question 6.
\(\frac{d y}{d x}\) + y sec x = tan x
Solution:
Given equation is
\(\frac{d y}{d x}\) + y sec x = tan x
This is a linear equation where
P = sec x, Q = tan x
I.F. = e^{∫sec dx}
= e^{(sec x + tan x)} = sec x + tan x
The solution is y . (sec x + tan x)
= ∫(sec x + tan x) tan x dx
= ∫(sec x tan x + tan² x) dx
= ∫(sec x . tan x + sec² x – 1) dx
= ∫(sec x + tan x) – x + C
⇒ (y – 1) (sec x + tan x) + x = C

Question 7.
(x + tan y) dy = sin 2y dx
Given equation can be written as
Solution:

Question 8.
(x + 2y³)\(\frac{d y}{d x}\) = y
Solution:
Given equation can be written as

Question 9.
sin x\(\frac{d y}{d x}\)+ 3y = cos x
Solution:

Question 10.
(x + y + 1)\(\frac{d y}{d x}\) = 1
Solution:

Question 11.
(1 + y²) dx + (x – \(e^{-\tan ^{-1} y}\)) dy = 0
Solution:
Given equation can be written as

Question 12.
x\(\frac{d y}{d x}\) + y = xy²
Solution:
Given equation can be written as

⇒ z = -x ln x + Cx
⇒ \(\frac{1}{y}\) = -x ln x + Cx
⇒ 1 = -xy ln x + Cxy
∴ The solution is (C – ln x) xy = 1

Question 13.
\(\frac{d y}{d x}\) + y = y² log x
Solution:

Question 14.
(1 + x²)\(\frac{d y}{d x}\) = xy – y²
Solution:
The given equation can be written as

Question 15.
\(\frac{d y}{d x}\) + \(\frac{y}{x-1}\) = \(x y^{\frac{1}{2}}\)
Solution:
The given equation can be written as

Question 16.
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x², y(1) = 1
Solution:
The given equation can be written as
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x², y(1) = 1 … (1)
This is a linear equation.

Question 17.
\(\frac{d y}{d x}\) + 2y tan x = sin x, y\(\left(\frac{\pi}{3}\right)\) = 0.
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Ex 11(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Exercise 11(a)

Question 1.
Determine the order and degree of each of the following differential equations.
(i) y sec² x dx + tan x dy = 0
Solution:
Order: 1, Degree: 1

(ii) \(\left(\frac{d y}{d x}\right)^4\) + y⁵ = \(\frac{d^3 y}{d x^3}\)
Solution:
Order: 3, Degree: 1

(iii) a\(\frac{d^2 y}{d x^2}\) = \(\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^{\frac{3}{2}}\)
Solution:
Order: 2, Degree: 2

(iv) tan^-1\(\sqrt{\frac{d y}{d x}}\) = x
Solution:
Order: 1, Degree: 1

(v) ln\(\left(\frac{d^2 y}{d x^2}\right)\) = y
Solution:
Order: 2, Degree: 1

(vi) \(\frac{\frac{d y}{d t}}{y+\frac{d y}{d t}}\) = \(\frac{y t}{d y}\)
Solution:
Order: 1, Degree: 2

(vii) \(\frac{d^2 y}{d u^2}\) = \(\frac{3 y+\frac{d y}{d u}}{\sqrt{\frac{d^2 y}{d u^2}}}\)
Solution:
Order: 2, Degree: 3

(viii) \(e^{\frac{d z}{d x}}\) = x²
Solution:
Order: 1, Degree: 1

Question 2.
Form the differential equation by eliminating the arbitrary constants in each of the following cases.
(i) y = A sec x
Solution:
y = A sec x
Then \(\frac{d y}{d x}\) = A sec x tan x = y tan x

(ii) y = C tan^-1 x
Solution:

(iii) y = Ae^t + Be^2t
Solution:

(iv) y = Ax² + Bx
Solution:

(v) y = -acos x + b sin x
Solution:

(vi) y = a sin^-1 x + b cos^-1 x
Solution:

(vii) y = at + be^t
Solution:

(viii) y = a sin t + be^t
Solution:

(ix) ax² + by = 1
Solution:

Question 3.
Find the general solution ofthe following differential equations.
(i) \(\frac{d y}{d x}\) = \(\frac{e^{2 x}+1}{e^x}\)
Solution:
\(\frac{d y}{d x}\) = \(\frac{e^{2 x}+1}{e^x}\)
⇒ y = ∫(e^x + e^-x) dx = e^x – e^-x + C

(ii) \(\frac{d y}{d x}\) = x cos x
Solution:
\(\frac{d y}{d x}\) = x cos x
⇒ y = ∫x cos x dx
= x . sin x – ∫sin x dx – x sin x + cos x + C

(iii) \(\frac{d y}{d x}\) = t⁵ log t
Solution:

(iv) \(\frac{d y}{d x}\) = 3t² + 4t + sec² t
Solution:
\(\frac{d y}{d x}\) = 3t² + 4t + sec² t
⇒ y = t³ + 2t² + tan t + C

(v) \(\frac{d y}{d x}\) = \(\frac{1}{x^2-7 x+12}\)
Solution:

(vi) \(\frac{d y}{d u}\) = \(\frac{u+1}{\sqrt{3 u^2+6 u+5}}\)
Solution:

(vii) (x² + 3x + 2) dy – dx = 0
Solution:

(viii) \(\frac{d y}{d t}\) = \(\frac{\sin ^{-1} t e^{\sin ^{-1} t}}{\sqrt{1-t^2}}\)
Solution:

Question 4.
Solve the following differential equations.
(i) \(\frac{d y}{d x}\) = y + 2
Solution:

(ii) \(\frac{d y}{d t}\) = \(\sqrt{1-y^2}\)
Solution:
\(\frac{d y}{d t}\) = \(\sqrt{1-y^2}\)
⇒ \(\frac{d y}{\sqrt{1-y^2}}\) = dt
⇒ sin^-1 y = t + C

(iii) \(\frac{d y}{d z}\) = sec y
Solution:
\(\frac{d y}{d z}\) = sec y
⇒ cos y dy = dz
⇒ sin y = z + C

(iv) \(\frac{d y}{d x}\) = e^y
Solution:

(v) \(\frac{d y}{d x}\) = y² + 2y
Solution:

(vi) dy + (y² + 1) dx = 0
Solution:

(vii) \(\frac{d y}{d x}\) + \(\frac{e^y}{y}\) = 0
Solution:

(viii) dx + cot x dt = 0
Solution:
dx + cot x dt = 0
⇒ tan x dx + dt = 0
⇒ ∫tan x dx + ∫dt = C₁
⇒ In sec x + t = C₁
⇒ In sec x = C₁ – t
⇒ sec x = \(e^{C_1}\) . e^-t
⇒ cos x = \(e^{-C_1}\) . e^t
⇒ cos x = Ce^t where C = \(e^{-C_1}\)

Question 5.
Obtain the general solution of the following differential equations.
(i) \(\frac{d y}{d x}\) = (x² + 1) (y² + 1)
Solution:

(ii) \(\frac{d y}{d t}\) = e^2t+3y
Solution:

⇒ 2e^-3y + 3e^2t + 6C₁ = 0
⇒ 2e^-3y + 3e^2t = C
where C = -6C₁

(iii) \(\frac{d y}{d z}\) = \(\frac{\sqrt{1-y^2}}{\sqrt{1-z^2}}\)
Solution:

(iv) \(\frac{d y}{d z}\) = \(\frac{x \log x}{3 y^2+4 y}\)
Solution:

(v) x²\(\sqrt{y^2+3}\) dx + y\(\sqrt{x^3+1}\) dy = 0
Solution:

(vi) tan y dx + cot x dy = 0
Solution:
tan y dx + cot x dy = 0
⇒ tan x . dx + cot y dy = 0
⇒ ∫tan x dx + ∫cot y dy = 0
⇒ -ln cos x + ln siny = ln C
⇒ ln\(\frac{\sin y}{\cos x}\) = ln C
⇒ \(\frac{\sin y}{\cos x}\) = C
⇒ sin y = C cos x

(vii) (x² + 7x + 12) dy + (y² – 6y + 5) dx = 0
Solution:

(viii) y dy + e^-y x sin x dx = 0
Solution:
y dy + e^-y x sin x dx = 0
⇒ ye^y dy + x sin x dx = 0
⇒ ∫ye^y dy + ∫x sin dx = C
[Integrating by parts.
⇒ ye^y – ∫e^y dy + x(-cos x) – ∫(-cos x) dx = C
⇒ ye^y – e^y – x cos x + sin x = C
⇒ (y – 1) e^y – x cos x + sin x = C

Question 6.
Solve the following second order equations.
(i) \(\frac{d^2 y}{d x^2}\) = 12x² + 2x
Solution:

(ii) \(\frac{d^2 y}{d t^2}\) =e^2t +e^-t
Solution:

(iii) \(\frac{d^2 y}{d \vartheta^2}\) = -sin υ + cos υ + sec² υ
Solution:
\(\frac{d^2 y}{d \vartheta^2}\) = -sin υ + cos υ + sec² υ
Integrating we get
\(\frac{d y}{d υ}\) = ∫sin υ dυ + ∫cos υ dυ + ∫sec² υ dυ
= cos υ + sin υ + tan υ + A
Again integratingwe get
y = ∫(cos υ + sin υ + tan υ + A)dυ + B
where A, B are arbritrary constants.
⇒ y = sin υ – cos υ + ln |sec υ| + A.υ. + B

(iv) cosec x \(\frac{d^2 y}{d x^2}\) = x
Solution:
cosec x \(\frac{d^2 y}{d x^2}\) = x
\(\frac{d^2 y}{d x^2}\) = x sin x
Integrating we get
\(\frac{d y}{d x}\) = ∫x sin x dx + A
= x . (-cos x) – ∫(-cos x) dx + A
= -x cos x + ∫cos x dx + A
= -x cos x + sin x + A
Again integrating we get
y = -∫x cos x dx + ∫sin x + ∫A dx + B
= -{x sin x -∫1 . sin x dx} – cos x + Ax + B
= -x sin x – 2cos x + Ax + B

(v) x²\(\frac{d^2 y}{d x^2}\) + 2 = 0
Solution:

(vi) sec x \(\frac{d^2 y}{d x^2}\) = sec 3x
Solution:

(vii) \(\frac{d^2 y}{d x^2}\) = sec² x + cos² x
Solution:

(viii) e^-x\(\frac{d^2 y}{d x^2}\) = x
Solution:
e^–x\(\frac{d^2 y}{d x^2}\) = x
⇒ \(\frac{d^2 y}{d x^2}\) = xe^x
Integrating we get
\(\frac{d y}{d x}\) = ∫xe^x dx = ∫e^x dx + Ax + B
= xe^x – e^x – e^x + Ax + B
= (x – 2)e^x + Ax + B

Question 7.
Find the particular solutions of the following equations subject to the given conditions.
(i) \(\frac{d y}{d x}\) = cos x, given that y = 2 when x = 0.
Solution:
\(\frac{d y}{d x}\) = cos x
Integrating we get
y = ∫cos x dx = sin x + C
Given that when x = 0, y = 2
So 2 = C
∴ The particular solution is y = sin x + 2

(ii) \(\frac{d y}{d t}\) = cos² y subject to y = \(\frac{\pi}{4}\) when t = 0.
Solution:
\(\frac{d y}{d t}\) = cos² y
⇒ sec² y dy = dt
∫sec² dy = ∫dt
⇒ tan y = t + C
When t = 0, y = \(\frac{\pi}{4}\)
So tan \(\frac{\pi}{4}\) = C ⇒ C = 1
∴ The particular solution is tan y = t + 1

(iii) \(\frac{d y}{d x}\) = \(\frac{1+y^2}{1+x^2}\) given that y = √3 when x = 1.
Solution:

(iv) \(\frac{d^2 y}{d x^2}\) = 6x given that y = 1 and \(\frac{d y}{d x}\) = 2 when x = 0.
Solution:
\(\frac{d^2 y}{d x^2}\) = 6x ⇒ \(\frac{d y}{d x}\) = 3x² + 2
When x = 0, \(\frac{d y}{d x}\) = 2
So 2 = A
∴ \(\frac{d y}{d x}\) = 3x² + 2
Again integrating we get
y = x³ + 2x + B
When x = 0, y = 1
So B = 1.
∴ The particular solution is y = x³ + 2x + 1

Question 8.
(i) Solve : \(\frac{d y}{d x}\) = sec (x + y)
Solution:

(ii) Solve : \(\frac{d y}{d x}\) = sin(x + y) + cos(x + y)
Solution:

(iii) Solve : \(\frac{d y}{d x}\) = cos (x + y)
Solution:

(iv) Solve : \(\frac{d y}{d x}\) + 1 = e^x+y
Solution:

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Ex 12(d) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Exercise 12(d)

Question 1.
Each question given below has four possible answers out of which only one is correct. Choose the correct one.
(i) \(\vec{a} \cdot \vec{b} \times \vec{a}\) = _______.
(a) \(\overrightarrow{0}\)
(b) 0
(c) 1
(d) \(\vec{a}^2 \vec{b}\)
Solution:
\(\vec{a} \cdot(\vec{b} \times \vec{a})\) = \((\vec{b} \times \vec{a}) \cdot \vec{a}\)
= \(\vec{b} \cdot(\vec{a} \times \vec{a})\) = \(\vec{b} \cdot \overrightarrow{0}\)
= 0 [∴ Dot product is commutative and in the scalar triple product the dot and cross can be interchanged.]

(ii) \((-\vec{a}) \cdot \vec{b} \times(-\vec{c}))\) = _______.
(a) \(\vec{a} \times \vec{b} \cdot \vec{c}\)
(b) \(-\vec{a} \cdot(\vec{b} \times \vec{c})\)
(c) \(\vec{a} \times \vec{c} \cdot \vec{b}\)
(d) \(\vec{a} \cdot(\vec{c} \times \vec{b})\)
Solution:
\((-\vec{a}) \cdot \vec{b} \times(-\vec{c})\) = \(\vec{a} \cdot(\vec{b} \times \vec{c})\)

(iii) For the non-zero vectors \(\vec{a}, \vec{b}\) and \(\vec{c}, \vec{a} \cdot(\vec{b} \times \vec{c})\) = 0 if
(a) \(\vec{b} \perp \vec{c}\)
(b) \(\vec{a} \perp \vec{b}\)
(c) \(\vec{a} \| \vec{c}\)
(d) \(\vec{a} \perp \vec{c}\)
Solution:
\(\vec{a} \cdot(\vec{b} \times \vec{c})\) = \((\vec{a} \times \vec{b}) \cdot \vec{c}\)
\(\vec{c} \perp(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})\)
but \(\vec{a} \times \vec{b}\) is perpendicular to \(\vec{a}\) and \(\vec{b}\)
∴ \(\vec{a} \| \vec{b}\)

Question 2.
Find the scalar triple product \(\vec{b} \cdot(\vec{c} \times \vec{a})\) where \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are respectively.
(i) î + ĵ, î – ĵ, 5î + 2ĵ + 3k̂
Solution:

= 1 (0 – 3) + 1 (0 – 3) + 0 (5 – 2)
= 3 – 3 = -6

(ii) 5î – ĵ + 4k̂, 2î + 3ĵ + 5k̂, 5î – 2ĵ
Solution:

= 5 (18 + 10) + 1 (12 – 25) + 4 (- 4 – 15)
= 140 – 13 – 76 = 140 – 89 = 51

Question 3.
Find the volume of the parallelopiped whose sides are given by the vectors.
(i) î + ĵ + k̂, k̂, 3î – ĵ + 2k̂
Solution:

= 1 (0 + 1) – 1 (0 – 3) + 1 (0 – 0)
= 1 + 3 = 4 cube units.

(ii) (1, 0, 0), (0, 1, 0), (0, 0, 1).
Solution:

Question 4.
Show that the following vector are co-planar
(i) î – 2ĵ + 2k̂, 3î + 4ĵ + 5k̂, -2î + 4ĵ – 4k̂
Solution:

(ii) î + 2ĵ + 3k̂, -2î – 4ĵ + 5k̂, 3î + 6
Solution:

Question 5.
Find the value of λ so that the three vectors are co-planar.
(i) î + 2ĵ + 3k̂, 4î + ĵ + λk̂ and λî – 4ĵ + k̂
Solution:

(ii) (2, -1, 1), (1, 2, -3) and (3, λ, 5)
Solution:

⇒ 2 (10 + 3λ) + 1 (5 + 9) + 1 (λ – 6) = 0
⇒ 20 + 6λ +14 + λ – 6 = 0
⇒ 7λ + 28 = 0 ⇒ λ = -4

Question 6.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) mutually perpendiculars, show that \([\vec{a} .(\vec{b} \times \vec{c})]^2\) = a²b²c²
Solution:

Question 7.
Show that \([\vec{a}+\vec{b} \vec{b}+\vec{c} \vec{c}+\vec{a}]\) = 2\([\vec{a} \vec{b} \vec{c}]\)
Solution:

Question 8.
Prove that \([\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]\) = \([\vec{a} \vec{b} \vec{c}]^2\)
Solution:

Question 9.
For \(\vec{a}\) = î + ĵ, \(\vec{b}\) = -î + 2k̂, \(\vec{c}\) = ĵ + k̂ obtain \(\vec{a} \times(\vec{b} \times \vec{c})\) and also verify the formula \(\vec{a} \times(\vec{b} \times \vec{c})\) = \((\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}\).
Solution:

Question 10.
Prove that \(\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b})\) and hence prove that \(\vec{a} \times(\vec{b} \times \vec{c}), \vec{b} \times(\vec{c} \times \vec{a}), \vec{c} \times(\vec{a} \times \vec{b})\) are coplanar.
Solution:

Question 11.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) unit vectors and \(\hat{a} \times(\hat{b} \times \hat{c})=\frac{1}{2} \hat{b}\) find the angles that â makes with b̂ and ĉ, where b̂, ĉ are not parallel.
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Additional Exercise Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Additional Exercise

Question 1.
∫\(\sqrt{1-\sin 2 x}\) dx
Solution:
I = ∫\(\sqrt{1-\sin 2 x}\) dx
= ∫\(\sqrt{(\cos x-\sin x)^2}\) dx
= ∫(cos x – sin x) dx
= sin x + cos x + c

Question 2.
∫\(\frac{d x}{1+\sin x}\)
Solution:
I = ∫\(\frac{d x}{1+\sin x}\)
= ∫\(\frac{1-\sin x}{\cos ^2 x}\)
= ∫sec² x – sec x tan x dx
= tan x – sec x + c

Question 3.
∫\(\frac{\sin x}{1+\sin x}\) dx
Solution:

Question 4.
∫\(\frac{\sec x}{\sec x+\tan x}\) dx
Solution:
I = ∫\(\frac{\sec x}{\sec x+\tan x}\) dx
= ∫\(\frac{\sec x(\sec x-\tan x)}{\sec ^2 x-\tan ^2 x}\) dx
= ∫sec² x – sec x tan x dx
= tan x – sec x + c

Question 5.
∫\(\frac{1+\sin x}{1-\sin x}\) dx
Solution:
I = ∫\(\frac{1+\sin x}{1-\sin x}\) dx
= ∫\(\frac{(1+\sin x)^2}{\cos ^2 x}\) dx
= ∫[sec² x+ tan² x+ 2sec x tan x) dx
= ∫[2sec² x – 1 + 2sec x tan x) dx
= 2tan x – x + 2sec x + c

Question 6.
∫tan^-1 (sec x + tan x) dx
Solution:

Question 7.
∫\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx
Solution:
I = ∫\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx
= ∫\(\frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha}\) dx
= 2 ∫(cos x + cos α) dx
= 2 sin x + 2x cos α + c

Question 8.
∫tan-1\(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\) dx
Solution:

Question 9.
∫\(\frac{d x}{\sqrt{x+1+} \sqrt{x+2}}\)
Solution:

Question 10.
∫\(\frac{2+3 x}{3-2 x}\) dx
Solution:

Question 11.
∫\(\frac{d x}{\sqrt{x}+x}\)
Solution:

Question 12.
∫\(\frac{d x}{1+\tan x}\)
Solution:

Question 13.
∫\(\frac{x+\sqrt{x+1}}{x+2}\) dx (Hints put : \(\sqrt{x+1}\) = t)
Solution:

Question 14.
∫sin^-1\(\sqrt{\frac{x}{a+x}}\) dx (Hints put : x = a tan² t)
Solution:

Question 15.
∫e^x\(\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right)\) dx
Solution:

Question 16.
∫\(\frac{\left(x^2+1\right) e^x}{(x+1)^2}\) dx
Solution:

Question 17.
∫\(\frac{x^2-1}{x^4+x^2+1}\) dx
Solution:

Question 18.
∫\(\frac{x^2 d x}{x^4+x^2+1}\)
Solution:

Question 19.
∫\(\sqrt{\cot x}\) dx
Solution:

Question 20.
∫\((\sqrt{\tan x}+\sqrt{\cot x})\) dx
Solution:

Question 21.
∫\(\frac{\mathrm{dx}}{x\left(x^4+1\right)}\)
Solution:

Question 22.
∫\(\frac{\mathrm{dx}}{e^x-1}\)
Solution:

Question 23.
∫\(\frac{(x-1)(x-2)(x-3)}{(x+4)(x-5)(x-6)}\) dx
Solution:

Question 24.
∫\(\frac{d x}{\left(e^x-1\right)^2}\)
Solution:

Question 25.
∫\(\frac{d x}{\sin x \cos ^2 x}\)
Solution:

Question 26.
\(\int_2^4 \frac{\left(x^2+x\right) d x}{\sqrt{2 x+1}}\)
Solution:

Question 27.
\(\int_{-a}^a \sqrt{\frac{a-x}{a+x}}\) dx
Solution:

Let a² – x² = t²
-2x dx = 2t dt
x = -a ⇒ 0 t = 0
x = a ⇒ t = 0
= 0
I = aI₁ – I₂ = aπ

Question 28.
\(\int_0^{\pi / 2}(\sqrt{\tan x}+\sqrt{\cot x})\) dx
Solution:

Question 29.
\(\int_0^{\pi / 2} \frac{\cos x d x}{1+\cos x+\sin x}\)
Solution:

Question 30.
\(\int_0^1\)x (1 – x)ⁿ dx
Solution:

Question 31.
\(\int_0^{\pi / 2}\)sin 2x log (tan x) dx
Solution:

Question 32.
\(\int_0^{\pi / 2} \frac{\sin ^2 x d x}{\sin x+\cos x}\)
Solution:

Question 33.
\(\int_0^{\pi / 2} \frac{\sin ^2 x d x}{1+\sin x \cos x}\)
Solution:

Question 34.
\(\int_0^{\pi / 2} \frac{x d x}{\sin x+\cos x}\)
Solution:

Question 35.
Prove that \(\int_0^\pi\) x sin³ x dx = \(\frac{2 \pi}{3}\)
Solution:

Question 36.
\(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x d x}{\sin x+\cos x}\)
Solution:

Question 37.
\(\int_0^\pi\)|cos x| dx
Solution:

Question 38.
\(\int_1^4\)(|x – 1| + |x – 2| + |x – 3|) dx
Solution:

Question 39.
\(\int_{-\pi / 2}^{\pi / 2}\)(sin |x| + cos |x|) dx
Solution:

Question 40.
\(\int_0^\pi\)log (1 + cos x) dx
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(l) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(l)

Question 1.
\(\int_0^{\frac{\pi}{2}}\)sin¹⁰ θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)sin¹⁰ θ dθ = \(\frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\) = \(\frac{405 \pi}{7680}\)

Question 2.
\(\int_0^{\frac{\pi}{2}}\)cos¹² θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)cos¹² θ dθ = \(\frac{11}{12} \cdot \frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\) = \(\frac{4455 \pi}{92160}\)

Question 3.
\(\int_0^{\frac{\pi}{2}}\)sin¹¹ θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)sin¹¹ θ dθ = \(\frac{10}{11} \cdot \frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}\) = \(\frac{3840}{4455}\)

Question 4.
\(\int_0^{\frac{\pi}{2}}\)cos⁹ θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)cos⁹ θ dθ = \(\frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}\) = \(\frac{384}{405}\)

Question 5.
\(\int_0^1 \frac{x^7}{\sqrt{1-x^2}}\) dx
Solution:

Question 6.
\(\int_0^1 \frac{x^5\left(4-x^2\right)}{\sqrt{1-x^2}}\) dx
Solution:

Question 7.
\(\int_0^a x^3\left(a^2-x^2\right)^{\frac{5}{2}}\) dx
Solution:

Question 8.
\(\int_0^1 x^5 \sqrt{\frac{1+x^2}{1-x^2}}\) dx
Solution:

Question 9.
\(\int_0^{\infty} \frac{x^2}{\left(1+x^6\right)^n}\) dx
Solution:

Question 10.
\(\int_0^\pi\)sin⁸ θ dθ
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(k) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(k)

Evaluate the following Integrals:
Question 1.
(i) \(\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan x}\)dx
Solution:

(ii) \(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\)dx
Solution:

(iii) \(\int_0^1 \frac{\ln (1+x)}{2+x^2}\)dx (x = tan θ)
Solution:

(iv) \(\int_0^\pi \frac{x d x}{1+\sin x}\)
Solution:

Question 2.
(i) \(\int_{-a}^a\)x⁴ dx
Solution:

(ii) \(\int_{-a}^a\)(x⁵ + 2x² + x) dx
Solution:

(iii) \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\)cos² x dx
Solution:

(iv) \(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\)sin⁵ x dx
Solution:
Let f(x) = sin⁵ x
Then f(-x) = sin⁵ (-x)
= -sin⁵ x = -f(x)
So f(x) is an odd function.
Thus \(\int_{-a}^a\)f(x) dx = 0
\(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\)sin⁵ x dx = 0

Question 3.
(i) \(\int_0^\pi\)cos³ x dx
Solution:

(ii) \(\int_0^\pi\)cos² x dx
Solution:

(iii) \(\int_0^\pi\)sin³ x cos x dx
Solution:
\(\int_0^\pi\)sin³ x cos x dx
[Put sin x = t, then cos x dx = dt
When x = 0, t = 0, when x = π, t = 0
\(\int_0^\pi\)t³ dt = 0

(iv) \(\int_0^\pi\)sin x cos² x dx
Solution:

Question 4.
Show that
(i) \(\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}\) dx = \(\frac{\pi}{2} \ln \frac{1}{2}\)
Solution:

(ii) \(\int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x}\) dx = 0
Solution:

(iii) \(\int_0^\pi\)x ln sin x dx = \(\frac{\pi^2}{2} \ln \frac{1}{2}\)
Solution:

Question 5.
(i) \(\int_0^{\pi / 2}\)ln (tan x + cot x) dx
Solution:

(ii) \(\int_0^\pi \frac{x \tan x-\sin x}{1+\sin x \cos x}\) dx
Solution:

(iii) \(\int_1^3 \frac{\sqrt{x} d x}{\sqrt{4-x}+\sqrt{x}}\)
Solution:

(iv) \(\int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x}\)
Solution:

(v) \(\int_0^1\)x (1 – x)¹⁰⁰ dx
Solution:

(vi) \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}}\)
Solution:

(vii) \(\int_0^{50}\)e^x-[x] dx
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Ex 11(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Exercise 11(c)

Find the solutions of the following differential equations:
Question 1.
(x + y) dy + (x – y) dx = 0
Solution:
Given equation can be written as

Question 2.
\(\frac{d y}{d x}\) = \(\frac{1}{2}\left(\frac{y}{x}+\frac{y^2}{x^2}\right)\)
Solution:

Question 3.
(x² – y²) dx + 2xy dy = 0
Solution:
Given equation can be written as

Question 4.
x\(\frac{d y}{d x}\) + \(\sqrt{x^2+y^2}\) = y
Solution:
Given equation can be written as

Question 5.
x (x + y) dy = (x² + y²) dx
Solution:


This is the required solution.

Question 6.
y² + x² \(\frac{d y}{d x}\) = xy \(\frac{d y}{d x}\)
Solution:
Given equation can be written as

This is the required solution.

Question 7.
x sin\(\frac{y}{x}\) dy = \(\left(y \sin \frac{y}{x}-x\right)\)dx
Solution:
Given equation can be written as

Question 8.
x dy – y dx= \(\sqrt{x^2+y^2}\) dx
Solution:

This is the required solution.

Question 9.
\(\frac{d y}{d x}\) = \(\frac{y-x+1}{y+x+5}\)
Solution:
Given equation is


This is the required solution.

Question 10.
(x – y) dy = (x + y + 1) dx
Solution:
Given equation can be written as


This is the required solution.

Question 11.
(x – y – 2) dx + (x – 2y – 3) dy = 0
Solution:
Given equation can be written as


This is the required solution.

Question 12.
\(\frac{d y}{d x}\) = \(\frac{3 x-7 y+7}{3 y-7 x-3}\)
Solution:
Given equation can be written as


This is the required solution.

Question 13.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0
Solution:
Given equation can be written as

⇒ 2z + ln (z – 1) = 3x + C
⇒ 2 (2x + y) + ln (2x + y – 1 ) = 3x + C
⇒ (x + 2y) + ln (2x + y – 1 ) = C
This is the required solution.

Question 14.
(2x + 3y – 5)\(\frac{d y}{d x}\) + 3x + 2y – 5 = 0
Solution:
Given equation can be written as

Question 15.
(4x + 6y + 5) dx – (2x + 3y + 4) dy = 0
Solution:
Given equation can be written as


This is the required solution.