Ch 1Sets, Relations and Functions
5-Mark Questions
Write the following in roaster form. (i) {x ∈ N: x 2 < 121 and x is a prime}
Let A = { x ∈ N: x 2 < 121 and x is a prime } A = { 2, 3, 5, 7 } (ii) The set of positive roots of the equation (x – 1) ( x + 1) (x – 1 ) = 0 The set of positive roots of the equations (x – 1) (x + 1) (x 2 – 1) = 0 (x – 1 ) (x + 1 ) (x + 1) (x – 1) = 0 (x + 1 ) 2 (x – 1) 2 = 0 (x + 1) 2 = 0 or (x – 1) 2 = 0 x + 1 = 0 or x – 1 = 0 x = -1 or x = 1 A = { 1 } (iii) {x ∈ N: 4x + 9 < 52} 4x + 9 < 52 4x + 9 – 9 < 52 – 9 4x < 43 x < \(\frac{43}{4}\) (i.e.) x < 10.75 4 But x ∈ N ∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (iv) Let A = ⇒ \(\frac{x-4}{x+2}\) = 3 ⇒ x – 4 = 3(x + 2) ⇒ x – 4 = 3x + 6 ⇒ 3x – x = – …
State whether the following sets are finite or infinite. (i) {x ∈ N: x is an even prime number }
Let A = { x ∈ N: x is an even prime number ) A = {2} A is a finite set. (ii) {x ∈ N: x is an odd prime number } Let B = {x ∈ N: x is an odd prime number} B = {1, 3, 5, 7, 11, …………….. } B is an infinite set. (iii) {x ∈ Z: x is even and < 10 } C = {x ∈ Z: x is even and< 10} C = { ……….. -8, -6, -4, -2, 0, 2, 4, 6, 8} C is an infinite set. (iv) {x ∈ R: x is a rational number } D = { x ∈ R: x is a rational number } D is an infinite set. (v) {x ∈ N: x is a rational number } E = { x ∈ N: x is a rational number ) E = {1, 2, 3, 4, 5, 6, …………..) Every integer is a rational number. …
2-Mark Questions
Write the set {-1, 1} in set builder form.
A = {x: x 2 – 1 = 0, x ∈ R}
Justify the trueness of the statement: “An element of a set can never be a subset of itself”.
A set itself can be a subset of itself (i.e.) A ⊆ A. But it cannot be a proper subset.
If n (A ∩ B ) = 3 and n(A ∪ B ) = 10, then find n(P(A ∆ B)).
n(A ∪ B) = 10; n(A ∩ B) = 3 n(A ∆ B) = 10 – 3 = 7 and n(P(A ∆ B)) = 27 = 128
1-Mark Questions (MCQ)
Let A and B be two sets such that n (A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.
n(A) = 3 ⇒ set A contains 3 elements n(B) = 2 ⇒ set B contains 2 elements – we are given (x, 1), (y, 2), (z, 1) are elements in A × B ⇒ A = {x, y, z} and B = {1, 2}
Ch 2Basic Algebra
5-Mark Questions
Classify each element of {√7, \(-\frac{1}{4}\), 0, 3.14, 4, \(\frac{22}{7}\)} as a member of N, Q, R – Q or Z.
√7 is an irrational number. ∴ √7 ∈ R – Q \(-\frac{1}{4}\) is a negative rational number. ∴ \(-\frac{1}{4}\) ∈ Q 0 is an integer. ∴ 0 ∈ Z, Q 3.14 is a rational number. ∴ 3.14 ∈ Q 4 is a positive integers. ∴ 4 ∈ Z, N, Q \(\frac{22}{7}\) is an rational number. ∴ \(\frac{22}{7}\) ∈ Q
Prove that √3 is an irrational number.
Suppose that √3 is rational. Let √3 = \(\frac{\mathrm{m}}{\mathrm{n}}\) where m and n are positive integers with no common factors greater than 1. √3 = \(\frac{\mathrm{m}}{\mathrm{n}}\) ⇒ √3n = m ⇒ 3n 2 = m 2 ——– (1) By assumption n is an integer ∴ n 2 is an integer. Hence 3n 2 is an integral multiple of 3. ∴ From equation (1) m 2 is an integral multiple of 3 ⇒ m is an intergral multiple of 3 [Here m is an integer and m 2 is an integral multiple of 3. That m 2 is cannot take all integral multiples of 3. For example suppose m 2 = 3 = 1 × 3 which is an integral multiple of 3. …
2-Mark Questions
Solve – 3 |x| + 5 ≤ – 2 and graph the solution set in a number line.
-3|x| + 5 ≤ – 2 ⇒ -3 |x| ≤ – 2 – 5 (= -7) -3|x| ≤ – 7 ⇒ 3 |x| ≥ 7
SoIve 2|x + 1| – 6 ≤ 7 and graph the solution set in a number line.
Given 2|x + 1| – 6 ≤ 7 2|x + 1| ≤ 7 + 6 2|x + 1| ≤ 13 |x + 1| ≤ \(\frac{13}{2}\)
Simplify and hence find the value of n: \(\frac{3^{2 n} \cdot 9^{2} \cdot 3^{-n}}{3^{3 n}}\) = 27
Given 4 – 2n = 3 2n = 4 – 3 2n = 1 n = \(\frac { 1 }{ 2 }\)
1-Mark Questions (MCQ)
Find the condition that one of the roots of ax 2 + bx + c may be (a) negative of the other (b) thrice the other (c) reciprocal of the other.
The given quadratic equation is ax 2 + bx + c = 0 ——- (1) Let α and β be the roots of the equation (1) then Sum of the roots α + β = ——- (2) Product of the roots αβ = ——- (3) (a) Given one root is the negative of the other β = – α (2) ⇒ α + (-α) = – \(\frac{b}{a}\) 0 = – \(\frac{b}{a}\) ⇒ b = 0 (3) ⇒ α(-α) = \(\frac{c}{a}\) – α 2 = \(\frac{c}{a}\) Hence the required condition is b = 0 (b) Given that one root is thrice the other β = 3α When is the required condition? (c) One root is reciprocal of the other When is the required condition?
Ch 3Trigonometry
5-Mark Questions
Identify the quadrant in which an angle of each given measure lies, (i) 25° (ii) 825° (iii) – 55° (iv) 328° (v) – 230°
(i) 25° 25° First quadrant (ii) 825° 825° = 9 × 90° + 15° 825° = 2 × 360° + 105° ∴ 825° lies in the second quadrant. iii) -55° -55° lies in the fourth quadrant iv) 328° 328° = 270° + 58° lies in the fourth quadrant. v) -230° – 230° = – 180° + (- 50°) lies in the second quadrant.
For each given angle, find a co-terminal angle with a measure of 9 such that 0 ≤ θ < 360°. (i) 395° (ii) 525° (iii) 1150° (iv) – 270° (v) – 450°
(i) 395° 395° = 360° + 35° 395° – 35° = 360° ∴ Coterminal angle for 395° is 35° (ii) 525° 525° = 360° + 165° 360° – 165° = 360° ∴Coterminal angle for 525° is 165° (iii) 1150° 1150° = 360° + 360° + 360° + 70° 1150° = 3 × 360° + 70° 1150° – 70° = 3 × 360° ∴ Coterminal angle for 1150° is 70°. (iv) – 270° – 270° = 360° + 90° – 270° – 90° = 360° ∴ Coterminal angle for -270° is 90° (v) – 450° – 450° = – 720° + 270° – 450° – 270° = – 2 × 360° ∴ Coterminal angle for – 450° is 270°
2-Mark Questions
If sin θ + cos θ = m, show that cos 6 θ + sin 6 θ = \(\frac{4-3\left(m^{2}-1\right)^{2}}{4}\) where m 2 ≤ 2.
sin θ + cos θ = m (sin θ + cos θ) 2 = m 2
Express each of the following in radian measure. (i) 30° (ii) 135° (iii) -205° (iv) 150° (v) 330°
(i) 30° (ii) 135° (iii) – 205° (iv) 150° (v) 330°
Find the degree measure corresponding to the following radian measures. (i) \(\frac{\pi}{3}\) (ii) \(\frac{\pi}{9}\) (iii) \(\frac{2 \pi}{5}\) (iv) \(\frac{7 \pi}{3}\) (v) \(\frac{10 \pi}{9}\)
(i) \(\frac{\pi}{3}\) radians (ii) \(\frac{\pi}{9}\) radians (iii) \(\frac{2 \pi}{5}\) radians (iv) \(\frac{7 \pi}{3}\) radians (v) \(\frac{10 \pi}{9}\) radians
1-Mark Questions (MCQ)
Which of the following is not true? (1) sin θ = – \(\frac{3}{4}\) (2) cos θ = – 1 (3) tan θ = 25 (4) sec θ = \(\frac{1}{4}\)
(4) sec θ = \(\frac{1}{4}\) Explaination: We know |cos θ| < 1 sec θ = \(\frac{1}{4}\) ⇒ \(\frac{1}{\cos \theta}\) = \(\frac{1}{4}\) ⇒ cos θ = 4 which is not possible.
Ch 4Combinatorics and Mathematical Induction
5-Mark Questions
(i) A person went to a restaurant for dinner. In the menu card, the person saw 10 Indian and 7 Chinese food items. In how many ways the person can select either an Indian or Chinese food?
Selecting an Indian food item from the given 10 can be done in 10 ways. Selecting a Chinese food item from the given 7 can be done in 7 ways. ∴ Selecting an Indian or Chinese food can be done in 10 + 7 = 17 ways. (ii) There are 3 types of a toy car and 2 types of toy train are available in a shop. Find the number of ways a baby can buy a toy car and a toy train? Number of types of Toy car = 3 Number of types of Toy Train = 2 Number of ways of buying a Toy car = 3 ways Number of ways of buying a toy train = 2 ways ∴ By fundamental principle of multiplication, number of ways of buying a toy car …
(i) A mobile phone has a passcode of 6 distinct digits. What is the maximum number of attempts one makes to retrieve the passcode?
Number of distinct digit in a passcode of a mobile phone = 6 First digit can be tried in 10 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Second digit can be tried in 9 ways Third digit can be tried in 8 ways Fourth digit can be tried in 7 ways Fifth digit can be tried in 6 ways Sixth digit can be tried in 5 ways Therefore, the maximum number of attempts made to retrieve the passcode = 10 × 9 × 8 × 7 × 6 × 5 = 151200 (ii) Given four flags of different colours, how many different signals can be generated if each signal requires the use of three flags, one below the other? Number of flags = …
2-Mark Questions
Evaluate \(\frac{\mathbf{n} !}{\mathbf{r} !(\mathbf{n}-\mathbf{r}) !}\) when (i) n = 6, r = 2 (ii) n = 10, r = 3 (iii) For any n with r = 2.
(i) n = 6, r = 2 (ii) n = 10, r = 3 (iii) For any n with r = 2.
If (n – 1)P 3: nP 4 = 1: 10 find n.
Given (n – 1)P 3: nP 4 = 1: 10
In 3 fingers, the number of ways four rings can be worn is ————- ways. (1) 4 3 – 1 (2) 3 4 (3) 68 (4) 64
(4) 64 Explanation: Each letter can be ported in 3 ways ∴ 4 letter is 3 4 ways
1-Mark Questions (MCQ)
A test consists of 10 multiple choice questions. In how many ways can the test be answered if (i) Each question has four choices ? (ii) The first four questions have three choices and the remaining have five choices? (iii) Question number n has n + 1 choices ?
Each question has 4 choices. So each question can be answered in 4 ways. Number of Questions = 10 So they can be answered in 410 ways (ii) The first four questions have 3 choices. So they can be answered in 3 4 ways. The remaining 6 questions have 5 choices. So they can be answered in 5 6 ways. So all 10 questions can be answered in 3 4 × 5 6 ways. (iii) Given question n has n + 1 choices
Ch 5Binomial Theorem, Sequences and Series
5-Mark Questions
Compute (i) 102 4 (ii) 99 4 (iii) 9 7
(i) 102 4 102 4 = (100 + 2) 4 = 100000000 + 8 × 1000000 + 24 × 10000 + 3200 + 16 = 100000000 + 8000000 + 240000 + 3216 = 108243216 (ii) 99 4 99 4 = (100 – 1) 4 = 100000000 – 4 × 1000000 + 6 × 10000 – 400 + 1 = 100000000 – 4000000 + 60000 -400+1 = 96059601 (iii) 9 7 9 7 = (10 – 1) 7 = 10 3 (10 4 – 7 × 10 3 + 21 × 10 2 – 35 × 10 + 35) – 21 × 100 + 70 – 1 = 10 3 (10000 – 7000 + 2100 – 350 + 35 ) – 2100 + 70 – 1 = 10 3 (12135 – 7350) – 2031 = 10 3 × 4785 – 2031 = 4785000 – 2031 = 4782969
Find the coefficient of x 15 in \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)
General term T r+1 = nC r x n-r. a r ∴ In the expansion \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\) To find the coefficient of x 15, Put 20 – 5r = 15 20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 1 ∴ T 1 + 1 = 10C 1 x 20 – 5 ⇒ T 2 = 10. x 15 ∴ The coefficient of x 15 is 10
2-Mark Questions
Expand (i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\) (ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)
(i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\) (ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)
Find the constant term of \(\left(x^{3}-\frac{1}{3 x^{2}}\right)^{5}\)
To get the constant term, Put 15 – 5r = 0 ⇒ 5r = 15 ⇒ r = 3
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y) n are equal.
Given (x + y) n If n is odd the middle term in the expansion of (x + y) n are \(\frac{T_{n-1}}{2}+1\) and \(\mathrm{T}_{\frac{\mathrm{n}+1}{2}+1}\)
Ch 6Two Dimensional Analytical Geometry
5-Mark Questions
Find the locus of P, if for all values of a, the coordinates of a moving point P is (i) (9 cos α, 9 sin α) (ii) (9 cos α, 6 sin α)
(i) (9 cos α, 9 sin α) Let P (h, k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 9 sin α The locus of P (h, k) is obtained by replacing h by x and k by y. ∴ The required locus becomes x 2 + y 2 = 81 (ii) ( 9 cos α, 6 sin α) Let P (h, k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 6 sin α The locus of p(h, k) is obtained by replacing h by x and k by y ∴ The required locus is \(\frac{x^{2}}{81}+\frac{y^{2}}{36}\) = 1
Find the locus of a point P that moves a constant distant of (i) two units from the x-axis (ii) three units from the y-axis.
(i) Two units from x-axis: Let P (h, k) be any point on the required path. From the given data, we have k = 2 The locus of P (h, k) is obtained by replacing h by x and k by y. ∴ The required locus is y = 2 (ii) Three units from y-axis: Let (h, k) be any point on the required path. From the given data, we have h = 3. The locus of P (h, k) is obtained by replacing h by x and k by y. ∴ The required locus is x = 3
2-Mark Questions
If 6 is a parameter, find the equation of the locus of a moving point, whose coordinates are x = a cos 3 θ, y = a sin 3 θ.
The given moving points is (a cos 3 θ, a sin 3 θ)
If the point (8, – 5) lies on the focus \(\frac{x^{2}}{16}-\frac{y^{2}}{25}\) = k, then the value of k is (1) 0 (2) 1 (3) 2 (4) 3
(4) 3 Explaination: Given the point (8, – 5) lies on the locus
The line (p + 2q)x + (p – 3q)y = p – q for different values of p and q passes through the point (1) \(\left(\frac{3}{2}, \frac{5}{2}\right)\) (2) \(\left(\frac{2}{5}, \frac{2}{5}\right)\) (3) \(\left(\frac{3}{5}, \frac{3}{5}\right)\) (4) \(\left(\frac{2}{5}, \frac{3}{5}\right)\)
(4) \(\left(\frac{2}{5}, \frac{3}{5}\right)\) Explaination: The equation of the given line is (p + 2q)x + (p – 3q)y = p – q ……….. (1)
1-Mark Questions (MCQ)
A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time shown in the following table.
Choose the weight along the x-axis and Length along the y-axis. (a) (b) The points are(2, 3), (4, 4), (5, 4.5), (8, 6) The relation connecting weight and Length is the equation of the straight line joining the points (2, 3) and (4, 4) x – 2 = 2(y – 3) x – 2 = 2y – 6 x – 2y + 6 – 2 = 0 x – 2y + 4 = 0 —– (1) which the required relation connecting weight and length. (c) To find the actual length of the spring, put weight x = 0 in equation (1) 0 – 2y + 4 = 0 ⇒ 2y = 4 ⇒ y = 2 ∴ The actual length of the spring is 2 cm. …
Ch 7Matrices and Determinants
5-Mark Questions
Construct an m × n matrix A = [a ij ], where a ij is given by (i) a ij = \(\frac{(\mathbf{i}-2 \mathbf{j})^{2}}{2}\) with m = 2, n = 3 (ii) a ij = \(\frac{|3 \mathbf{i}-4 \mathbf{j}|}{4}\) with m = 3, n = 4
(i) a ij = \(\frac{(\mathbf{i}-2 \mathbf{j})^{2}}{2}\) with m = 2, n = 3 To construct 2 × 3 matrices. (ii) a ij = \(\frac{|3 \mathbf{i}-4 \mathbf{j}|}{4}\) with m = 3, n = 4 To construct a 3 × 4 matrices.
Find the value of p, q, r and s if
Equating the corresponding entries ⇒ p 2 – 1 = 1 ⇒ p 2 = 1 + 1 = 2 p = ± \(\sqrt{2}\) -31 – q 3 = -4 -q 3 = -4 + 31 = 27 q 3 = -27 = (-3) 3 ⇒ q = -3 r + 1 = \(\frac{3}{2}\) ⇒ r = \(\frac{3}{2}\) – 1 = \(\frac{3-2}{2}\) = \(\frac{1}{2}\) s – 1 = π ⇒ s = – π + 1 (i.e.,) s = 1 – π So, p = ± \(\sqrt{2}\), q = -3, r = 1/2 and s = 1 – π
2-Mark Questions
Consider the matrix A α = \(\left[ \begin{matrix} cos\quad α & -\quad sin\quad α \\ sin\quad α & cos\quad α \end{matrix} \right] \) (i) Show that A α A β = A (α+β)
(ii) Find all possible real values α satisfying the condition A α + A α T = I
If A = \(\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right]\) and A 3 – 6A 2 + 7A + kI = 0, find the value of k.
Equating the corresponding entries – 2 + k = 0 ⇒ k = 2 ∴ The required value of k is k = 2
without expanding the determinant,
= s (a 2 + b 2 + c 2 ) × 0 since two columns are equal. = 0
1-Mark Questions (MCQ)
Which one of the following is not true about the matrix \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) ?. (1) a scalar matrix (2) a diagonal matrix (3) an upper triangular matrix (4) a lower triangular matrix
(1) a scalar matrix Explaination: Let A = \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) (1) a scalar matrix – not true (2) a diagonal matrix – true (3) an upper triangular matrix – true (4) a lower triangular matrix – true [(1) A square matrix A = [a ij ] m × n is called a diagonal matrix if a ij = 0 whenever i ≠ j (2) A diagonal matrix whose entries along the principle diagonal are equal is called a scalar matrix. (3) A square matrix is said to be an upper triangular matrix if all the elements below the main diagonal are zero. …
Ch 8Vector Algebra – I
5-Mark Questions
Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side whose length is half of the length of the third side.
ABC be the triangle and D, E, F is the midpoints of the sides AB, BC, and AC respectively. Let O be the origin and let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are the position vectors of the points A, B and C respectively. ∴ \(\overrightarrow{\mathbf{D E}}\) is parallel to \(\overrightarrow{\mathbf{B C}}\) and length of DE is half the length of BC. Similarly we can prove \(\overrightarrow{\mathbf{E F}}\) = \(\frac{1}{2} \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{EF}}\) parallel to \(\overrightarrow{\mathbf{A B}}\) and length of E is half the length of AB. …
Prove that the line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Let ABCD be any quadrilateral D, E, F, G are the midpoints of the sides AB, BC, CD, AD respectively. ∴ EF and DG are parallel and equal. ∴ In the quadrilateral, DEFG’s opposite sides are parallel and equal. Therefore, DEFG is a parallelogram.
2-Mark Questions
Represent graphically the displacement of (i) 45cm, 30° north of east (ii) 80 km, 60° south of west
(i) 45cm, 30° north of east (ii) 80 km, 60° south of west
If \(\vec{a}\) and \(\vec{b}\) represent a side and a diagonal of a parallelogram, find the other sides and the other diagonal.
Let ABCD be a parallelogram Let \(\overrightarrow{\mathbf{A B}}\) = \(\vec{a}\) \(\overrightarrow{\mathbf{A C}}\) = \(\vec{b}\) Since ABCD is a parallelogram, we have
If D is the midpoint of the aide BC of i triangle ABC, prove that \(\overrightarrow{\mathbf{A B}}\) + \(\overrightarrow{\mathbf{A C}}\) = 2 \(\overrightarrow{\mathbf{A D}}\)
Given D is the mid point of the side BC of a triangle ABC
1-Mark Questions (MCQ)
If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are the position vectors of three collinear points, then which of the following is true? (1) \(\overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\) (2) \(2 \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\) (3) \(\overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}}\) (4) \(4 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=0\)
(2) \(2 \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}\) Explaination:
Ch 9Limits and Continuity
5-Mark Questions
\(\lim _{x \rightarrow 1}\) sin πx
\(\lim _{x \rightarrow 1}\) sin πx From the graph x = 1, the curve y = f(x) intersects the line x = 1 at x – axis. ∴ y = f(1) = 0 Hence \(\lim _{x \rightarrow 1}\) sin πx = 0
\(\lim _{x \rightarrow 0}\) sec x
To find \(\lim _{x \rightarrow 0}\) sec x Let y = f(x) = sec x From the graph at x = 0 the curve intersect the y – axis. At x = 0 we have y = 1 ∴ \(\lim _{x \rightarrow 0}\) sec x = 1
2-Mark Questions
Sketch the graph of a function f that satisfies the given values: (i) f(0) is defined \(\lim _{x \rightarrow 0}\) f(x) = 4 f(2) = 6 \(\lim _{x \rightarrow 2}\) f(x) = 3
(ii) f(-2) = 0 f(2) = 0 \(\lim _{x \rightarrow-2}\) f(x) = 0 \(\lim _{x \rightarrow 2}\) f(x) does not exist.
Write a brief description of the meaning of the notation \(\lim _{x \rightarrow 8}\) f(x) = 25
Given \(\lim _{x \rightarrow 8}\) f(x) = 25 By the definition of limit ∴ f(8 – ) = f(8 + ) = 25
Verify the existence of \(\lim _{x \rightarrow 0}\) f(x), where
From equations (1) and (2) we get f(1 – ) ≠ f(1 + ) ∴ The limit of f(x) does not exist.
Ch 10Differentiability and Methods of Differentiation
5-Mark Questions
Find the derivatives of the following functions using the first principle. (i) f(x) = 6 (ii) f(x) = -4x + 7 (iii) f(x) = -x 2 + 2
(i) f(x) = 6 (ii) f(x) = – 4x + 7, f(x + Δx) = -4(x + Δx) + 7 f(x + Δx) – f(x) = [-4(x + Δx) + 7] – [-4x + 7] f(x + Δx) – f(x) = [-4(x + Δx) + 7] + 4x – 7 f(x + Δx) – f(x) = -4 Δx (iii) f(x) = -x 2 + 2 f (x + Δx) = – (x + Δx) 2 + 2 f (x + Δx) – f(x) = – [x 2 + 2x Δx + (Δx) 2 ] + 2 – [- x 2 + 2]
Determine whether the following function is differentiable at the indicated values. (i) f(x) = x |x| at x = 0
(ii) f(x) = |x 2 – 1|at x = 1 (iii) f(x) = |x| + |x – 1| at x = 0, 1 To find the limit at x = 0 First we find the left limit of f(x) at x = 0 When x = 0 – |x| = -x and |x – 1| = -(x – 1) ∴ When x = 0 we have f(x) = -x – (x – 1) f(x) = -x – x + 1 = -2x + 1 f(0) = 2 × 0 + 1 = 1 f'(0 – = – 2 ……… (1) ∴When x = 0 + we have |x| = x and |x – 1| = – (x – 1) ∴ f(x) = x – (x – 1) f(x) = x – x + 1 f(x) = 1 f(0) = 1 From equations (1) and (2), we get f'(0 – ) ≠ f’(0 + ) ∴ f(x) is not differentiable at x = 0. …
2-Mark Questions
Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1? (i) f (x) = |x – 1|
(ii) f (x) = \(\sqrt{1-x^{2}}\) (iii)
f(x) = x – 3 sin x
f(x) = x – 3 sinx = f'(x) = 1 – 3 (cos x) = 1 – 3 cos x
y = sin x + cos x
y = sin x + cos x \(\frac{d y}{d x}\) = cos x – sin x
1-Mark Questions (MCQ)
If f(x) = then which one of the following is true? (1) f(x) is not differentiable at x = a (2) f(x) is discontinuous at x = a (3) f(x) is continuous for all x in R (4) f(x) is differentiable for all x ≥ a
(1) f(x) is not differentiable at x = a Explaination: f’ (a + ) = 3 ………. (2) From equations (1) and (2) we get f'(a – ) ≠ f'(a + ) ∴ f’ (x) does not exist at x = a ∴ f(x) is not differentiable at x = a
Ch 11Integral Calculus
5-Mark Questions
(i) \(\frac{1}{\sin ^{2} x}\) (ii) \(\frac{\tan x}{\cos x}\) (iii) \(\frac{\cos x}{\sin ^{2} x}\) (iv) \(\frac{1}{\cos ^{2} x}\)
(i) \(\frac{1}{\sin ^{2} x}\) \(\frac{1}{\sin ^{2} x}\) = ∫cosec 2 x dx = – cot x + c (ii) \(\frac{\tan x}{\cos x}\) \(\frac{\tan x}{\cos x}\) = ∫sec x tan x dx = sec x + c (iii) \(\frac{\cos x}{\sin ^{2} x}\) (iv) \(\frac{1}{\cos ^{2} x}\) \(\frac{1}{\cos ^{2} x}\) = ∫sec 2 x dx = tan x + c
(i) 12 3 (ii) \(\frac{x^{24}}{x^{25}}\) (iii) e x
(i) 12 3 ∫12 3 dx = 12 3 ∫dx = 12 3 x + c (ii) \(\frac{x^{24}}{x^{25}}\) (iii) e x ∫e x dx = e x + c
2-Mark Questions
(i) x 11 (ii) \(\frac{1}{x^{7}}\) (iii) \(\sqrt[3]{x^{4}}\) (iv) (x 5 ) 1/8
(i) x 11 (ii) \(\frac{1}{x^{7}}\) (iii) \(\sqrt[3]{x^{4}}\) (iv) (x 5 ) 1/8
(i) (1 + x 2 ) -1 (ii) (1 – x 2 ) -1/2
(i) (1 + x 2 ) -1 (ii) (1 – x 2 ) -1/2 Read More: JUBLFOOD Pivot Point Calculator
(i) (x + 5) 6 (ii) \(\frac{1}{(2-3 x)^{4}}\) (iii) \(\sqrt{3 x+2}\)
(i) (x + 5) 6 (ii) \(\frac{1}{(2-3 x)^{4}}\) (iii) \(\sqrt{3 x+2}\)
Ch 12Introduction to Probability Theory
5-Mark Questions
If two coins are tossed simultaneously, then find the probability of getting (i) one head and one tail (ii) at most two tails
Two coins are tossed simultaneously = one coin is tossed two times. The sample space S = { H, T } × { H, T } S = {HH, HT, TH, TT } n(S) = 4 Let A be the event of getting one head and one tail, B be the event of getting atmost two tails. A = {HT, TH} n(A) = 2 B = {HH, HT, TH, TT} n (B) = 4 (i) P (getting one head and one tail) (ii) P (getting atmost two tails)
Five mangoes and 4 apples are in a box. If two fruits are chosen at random, find the probability that (i) one is a mango and the other is an apple (ii) both are of the same variety.
(i) One is a mango and the other is an apple: Let S be the sample space, A be the event of taking one mango and one apple. n(S) = 9C 2 n (A) = 5C 1 × 4C 1 = 5 × 4 = 20 (ii) Both are of the same variety: Let S be the sample space, A be the event of taking 2 mangoes and B be the event of taking 2 apples ∴ n(s) = 9C 2 P(taking 2 fruits are of the same colour) = P(A or B) = P(A ∪ B) = P(A) + P(B)
2-Mark Questions
Can two events be mutually exclusive and independent simultaneously?
When A and B are independent P(A ∩ B) = P(A) P(B) But when A and B are mutually Exclusive P(A ∩ B) = 0
If X and Y be two events such that P(X/Y) = \(\frac{1}{2}\), P(Y/X) = \(\frac{1}{3}\) and P( X ∩ Y) = \(\frac{1}{6}\),then (1) \(\frac{1}{3}\) (2) \(\frac{2}{5}\) (3) \(\frac{1}{6}\) (4) \(\frac{2}{3}\)
(4) \(\frac{2}{3}\) Explaination:
If two events A and B are such that P(A̅) = \(\frac{3}{10}\) and P(A ∩ B̅) = \(\frac{1}{2}\) then P(A ∩ B) is (1) \(\frac{1}{2}\) (2) \(\frac{1}{3}\) (3) \(\frac{1}{4}\) (4) \(\frac{1}{5}\)
(4) \(\frac{1}{5}\) Explaination:
1-Mark Questions (MCQ)
An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C, and D. Check whether the following assignments of probability are permissible. (i) P
P(A) = 0.15, P(B) = 0.30, P(C) = 0.43, P(D) = 0.12 Now P(A) + P(B) + P(C) + P(D) = 0.15 + 0.30 + 0.43 + 0.12 = 1 0.15 + 0.30 + 0.43 + 0.12 = 1 ∴ The assignment of probability is permissible. (ii) P (A) = 0.22, P (B) = 0.38, P (C) = 0.16, P (D) = 0.34 Given that P (A) = 0.22 ≥ 0, P (B) = 0.38 ≥ 0, P(C) = 0.16 ≥ 0, P (D) = 0.34 ≥ 0 P(S) = P (A) + P(B) + P(C) + P(D) = 0.22 + 0.38 + 0.16 + 0.34 = 1.1 > 1 Therefore the assignment of probability isn’t permissible (iii) P(A) = \(\frac{2}{5}\), P(B) = \(\frac{3}{5}\), P(C) = – \(\frac{1}{5}\), P(D) = \(\frac{1}{5}\) P(A) = \(\frac{2}{5}\), P(B) = \(\f …
Frequently asked questions
- Write the following in roaster form. (i) {x ∈ N: x 2 < 121 and x is a prime}
- Let A = { x ∈ N: x 2 < 121 and x is a prime } A = { 2, 3, 5, 7 } (ii) The set of positive roots of the equation (x – 1) ( x + 1) (x – 1 ) = 0 The set of positive roots of the equations (x – 1) (x + 1) (x 2 – 1) = 0 (x – 1 ) (x + 1 ) (x + 1) (x – 1) = 0 (x + 1 ) 2 (x – 1) 2 = 0 (x + 1) 2 = 0 or (x – 1) 2 = 0 x + 1 = 0 or x – 1 = 0 x = -1 or x = 1 A = { 1 } (iii) {x ∈ N: 4x + 9 < 52} 4x + 9 < 52 4x + 9 – 9 < 52 – 9 4x < 43 x < \(\frac{43}{4}\) (i.e.) x < 10.75 4 But x ∈ N ∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (iv) Let A = ⇒ \(\frac{x-4}{x+2}\) = 3 ⇒ x – 4 = 3(x + 2) ⇒ x – 4 = 3x + 6 ⇒ 3x – x = – …
- State whether the following sets are finite or infinite. (i) {x ∈ N: x is an even prime number }
- Let A = { x ∈ N: x is an even prime number ) A = {2} A is a finite set. (ii) {x ∈ N: x is an odd prime number } Let B = {x ∈ N: x is an odd prime number} B = {1, 3, 5, 7, 11, …………….. } B is an infinite set. (iii) {x ∈ Z: x is even and < 10 } C = {x ∈ Z: x is even and< 10} C = { ……….. -8, -6, -4, -2, 0, 2, 4, 6, 8} C is an infinite set. (iv) {x ∈ R: x is a rational number } D = { x ∈ R: x is a rational number } D is an infinite set. (v) {x ∈ N: x is a rational number } E = { x ∈ N: x is a rational number ) E = {1, 2, 3, 4, 5, 6, …………..) Every integer is a rational number. …
- Write the set {-1, 1} in set builder form.
- A = {x: x 2 – 1 = 0, x ∈ R}
- Justify the trueness of the statement: “An element of a set can never be a subset of itself”.
- A set itself can be a subset of itself (i.e.) A ⊆ A. But it cannot be a proper subset.
These important questions are selected from the Samacheer Kalvi Class 11 Maths textbook book-back exercises to help you revise the most useful questions. Mark weightage (5/2/1) follows the usual exam pattern and may vary by exam — always check your latest syllabus and question pattern. Open each chapter for the complete set of questions and answers.