Category: Class 11 Mathematics

NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-6-linear-inequalities/

NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities

Linear Inequalities Class 11 Extra Questions NCERT

Q1. \(\frac { 4 }{ x+1 } \le 3\le \frac { 6 }{ x+1 } \)

NCERT Exemplar Class 11 Maths Chapter 6 Solutions

Q2. \(\frac { |x-2|-1 }{ |x-2|-2 } \le 0  \)

Sol:

Linear Inequalities Class 11 Questions And Answers NCERT

Q3. \(\frac { 1 }{ |x|-3 } \le \frac { 1 }{ 2 }   \)
Sol: 

NCERT Exemplar Linear Inequalities Class 11

Q4. |x-1|≤ 5, |x| ≥ 2
Sol:

Linear Inequalities Class 11 Exemplar Solutions NCERT

Q5. \(-5\le \frac { 2-3x }{ 4 } \le 9   \)
Sol:

Linear Inequalities Class 11 Important Questions NCERT

Q6. 4x + 3≥2x + 17, 3x – 5 < -2.
Sol: 

NCERT Exemplar Class 11 Maths Linear Inequalities

Q7. A company manufactures cassettes. Its cost and revenue functions are C(x) = 26000 + 30x and R(x) = 43x, respectively, where x is the number of cassettes produced and sold in a week How many cassettes must be sold by the company to realise some profit?

Sol. Cost function: C(x) = 26000 + 3Ox Revenue function: R(x) = 43x For profit, R(x) > C(x)
⟹    26000 + 30x < 43x
⟹43x – 30x > 26000
⟹ 13x> 26000
⟹     x > 2000
Hence, more than 2000 cassettes must be produced to get profit.

Linear Inequalities Class 11 Questions NCERT

Q8. The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH readings are 8.48 and 8.35, find the range of pH value for the third reading that will result in the acidity level being normal.

Sol: Given, first pH value = 8.48
And second pH value = 8.35
Let third pH value be x.
Since it is given that average pH value lies between 8.2 and 8.5, we get

Class 11 Maths Chapter 6 Extra Questions NCERT

Q9. A solution of 9% acid is to be diluted by adding 3% acid solution to it. The resulting mixture is to be more than 5% but less than 7% acid. If there is 460 litres of the 9% solution, how many litres of 3% solution will have to be added?

Sol:

Extra Questions Of Linear Inequalities Class 11 NCERT

Q10. A solution is to be kept between 40°C and 45°C. What is the range of temperature in degree Fahrenheit, if the conversion formula is F= 9/5 C + 32?
Sol.

Linear Inequalities Class 11 Exemplar NCERT

Q11. The longest side of a triangle is twice the shortest side and the third side is 2 cm longer than the shortest side. If the perimeter of the triangle is more than 166 cm then find the minimum length of the shortest side.
Sol: Let the length of shortest side be x cm.
According to the given information,
Longest side = 2 x Shortest side = 2x cm
And third side = 2 + Shortest side = (2 + x) cm
Perimeter of triangle = x + 2x + (x + 2) = 4x + 2
But it is given that,
Perimeter > 166 cm
=> 4x + 2 > 166 => 4x> 166-2 => 4x>164
x>164/4 =41 cm
Hence, the minimum length of shortest side is 41 cm.

Q12. In drilling world’s deepest hole it was found that the temperature T in degree Celsius, jc km below the earth’s surface was given by T = 30 + 25 (x – 3), 3 ≤ x≤ At what depth will the temperature be between 155°C and 205°C?

Q13. \(\frac { 2x+1 }{ 7x-1 } >5,\frac { x+7 }{ x-8 } >2 \)
Solution:

Q14. Find the linear inequalities for which the shaded region in the given figure is the solution set.

Sol: We observe that the shaded region and the origin are on the same side of the line 3x + 2y = 48.
For (0, 0), we have 3(0) + 2(0) – 48 < 0. So, the shaded region satisfies the inequality 3x + 2y≤ 48.
Also, the shaded region and the origin are on the same side of the line x+y = 20.
For (0,0), we have 0 + 0 – 20 < 0. So, the shaded region satisfies the inequality x +y ≤ 20.
Also, the shaded region lies in the first quadrant. So, x ≥0,y≥0.
Thus, the linear inequation corresponding to the given solution set are 3x + 2y ≤ 48, x + y ≤ 20 and x ≥ 0, y ≥0.

Q15Find the linear inequalities for which the shaded region in the given figure is the solution set.

Sol: We observe that the shaded region and the origin are on the same side of the linex+y = 8.
For (0,0), we have 0 + 0 – 8 < 0. So, the shaded region satisfies the inequality x + 2≤8.
The shaded region and the origin are on the opposite side of the line x+y = 4.
For (0,0), we have 0 + 0 – 4 < 0. So, the shaded region satisfies the inequality x + 2≥4.
Further, the shaded region and the origin are on the same side of the lines x = 5 andy = 5.
So, it satisfies the inequality x ≤ 5 andy < 5.
Also, the shaded region lies in the first quadrant. So, x > 0, y > 0.
Thus, the linear inequation comprising the given solution set are: x+y≥4;x+y≤ 8;x≤ 5;y < 5;x≥ 0 andy ≥ 0.

Q16. Show that the following system of linear inequalities has no solution: x + 2y≤3, 3x + 4y> 12,x≥0,y≥ 1
Sol: We have x + 2y ≤ 3, 3x + 4y > 12, x > 0, y ≥ 1
Now let’s plot lines x + 2y = 3, 3x + 4y = 12, x = 0 and y = 1 in coordinate plane.
Line x + 2y = 3 passes through the points (0, 3/2) and (3, 0).
Line 3jc + 4y = 12 passes through points (4, 0) and (0, 3).
For (0, 0), 0 + 2(0) – 3 < 0.
Therefore, the region satisfying the inequality x + 2y ≤ 3 and (0,0) lie on the same side of the line x + 2y = 3.
For (0, 0), 3(0) + 4(0)- 12 ≤0.
Therefore, the region satisfying the inequality 3x + 4y ≥ 12 and (0, 0) lie on the opposite side of the line 3x + 4y = 12.
The region satisfying x > 0 lies to the right hand side of the y-axis.
The region satisfying y > 1 lies above the line y = 1.
These regions are plotted as shown in the following figure

It is clear from the graph that the Shaded portions do not have common region. So, solution set is null set.

Q17. Solve the following system of linear inequalities:
3x+2y≥24,3x+y≤ 15,x≥4
Sol: We have, 3x + 2y ≥24,
3x +y ≤ 15, x ≥ 4
Now let’s plot lines 3x + 2y = 24, 3x + y = 15 and x = 4 on the coordinate plane.
Line 3x + 2y = 24 passes through the points (0, 12) and (8, 0).
Line 3x+y = 15 passes through points (5,0) and (0, 15).
Also line x = 4 is passing through the point (4, 0) and vertical.
For (0, 0), 3(0) + 2(0) – 24 < 0.
Therefore, the region satisfying the inequality 3x + 2y≥ 24 and (0, 0) lie on the opposite of the line 3x + 2y = 24.
For (0), 3(0) + (0) – 15 ≤ 0.
Therefore, the region satisfying the inequality 3x +y ≤ 15 and (0,0) lie on the same side of the line 3x +y = 15.
The region satisfying x ≥ 4 lies to the right hand side of the line x = 4.
These regions are plotted as shown in the following figure

It is clear from the graph that there is no common region corresponding to these inequalities.

Hence, the gi ven system of inequalities has no solution.

Q18. Show that the solution set of the following system of linear inequalities is an unbounded region:
2x +y ≥ 8, x + 2y > 10, x ≥ 0, y ≥ 0
Sol: We have 2x+y≥8,x + 2y> 10, x ≥ 0, y ≥ 0
Line 2x + y = 8 passes through the points (0, 8) and (4, 0).
Line x + 2y = 10 passes through points (10, 0) and (0, 5).\
For (0, 0), 2(0) + (0) – 8 < 0.
Therefore, the region satisfying the inequality 2x+y ≥ 8 and (0, 0) lie on the opposite side of the line 2x +y = 8.
For (0,0), (0) + 2(0)- 10 <0.
Therefore, the region satisfying the inequality x + 2y≥ 10 and (0,0) lie on the opposite side of the line x + 2y = 10.
Also, for x ≥ 0, y ≥0, region lies in the first quadrant.
The common region is plotted as shown in the following figure.

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We hope the NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 6 Linear Inequalities, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-5-complex-numbers-quadratic-equations/

NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Complex Numbers Class 11 Questions And Answers NCERT

Short Answer Type Questions

NCERT Exemplar Class 11 Complex Numbers

Questions On Complex Numbers Class 11 NCERT

NCERT Exemplar Class 11 Maths Ch 5

Q6. If a = cos θ + i sin θ, then find the value of (1+a/1-a) 
Sol:  a = cos θ + i sin θ


Q10. Show that the complex number z, satisfying the condition arg lies on arg (z-1/z+1) = π/4 lies on a  circle.

Sol: Let z = x + iy

Q11. Solve the equation |z| = z + 1 + 2i.
Sol: We have |z| = z + 1 + 2i
Putting z = x + iy, we get
|x + iy| = x + iy + 1+2i

Long Answer Type Questions
Q12. If |z + 1| = z + 2( 1 + i), then find the value of z.
Sol: We have |z + 11 = z + 2(1+ i)
Putting z = x + iy, we get
Then, |x + iy + 11 = x + iy + 2(1 + i)
⟹|x + iy + l|=x + iy + 2(1 +i)

Q13. If arg (z – 1) = arg (z + 3i), then find (x – 1) : y, where z = x + iy.
Sol: We have arg (z – 1) = arg (z + 3i), where z = x + iy
=>  arg (x + iy – 1) = arg (x + iy + 3i)
=> arg (x – 1 + iy) = arg [x + i(y + 3)]

Q14. Show that | z-2/z-3| = 2 represents a circle . Find its center and radius .
Sol: We have | z-2/z-3| = 2
Puttingz=x + iy, we get

 

Q15. If z-1/z+1 is a purely imaginary number (z ≠1), then find the value of |z|.

Sol: Let   z = x + iy

Q17. If |z₁ | = 1 (z₁≠ -1) and z₂ = z₁ – 1/ z₁ + 1 , then show that real part of z₂ is zero .

Q18. If Z₁, Z₂ and Z₃, Z₄ are two pairs of conjugate complex numbers, then find arg (Z_1/ Z₄) + arg (Z_2/ Z₃)
Sol. It is given that z₁ and z₂ are conjugate complex numbers.

Q20. If for complex number z₁ and z₂, arg (z₁) – arg (z₂) = 0, then show that |z₁ – z₂| = | z₁|- |z₂ |

Q21. Solve the system of equations Re (z²) = 0, |z| = 2.

Sol: Given that, Re(z²) = 0, |z| = 2

Q22. Find the complex number satisfying the equation z + √2 |(z + 1)| + i = 0.

Fill in the blanks 

True/False Type Questions

Q26. State true or false for the following.
(i) The order relation is defined on the set of complex numbers.
(ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction.
(iii) For any complex number z, the minimum value of |z| + |z – 11 is 1.
(iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1).
(v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z²) = 0.
(vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z₁ and Z₂ be two complex numbers such that |z, + z₂| = |z₁ j + |z₂|, then arg (z₁ – z₂) = 0.
(viii) 2 is not a complex number.

Sol:(i) False
We can compare two complex numbers when they are purely real. Otherwise comparison of complex numbers is not possible or has no meaning.

Matching Column Type Questions
Q24. Match the statements of Column A and Column B.

Column A			Column B
(a)	The polar form of i + √3 is	(i)	Perpendicular bisector of segment joining (-2, 0) and (2,0)
(b)	The amplitude of- 1 + √-3 is	(ii)	On or outside the circle having centre at (0, -4) and radius 3.
(c)	It |z + 2| = |z – 2|, then locus of z is	(iii)	2/3
(d)	It |z + 2i| = |z – 2i|, then locus of z is	(iv)	Perpendicular bisector of segment joining (0, -2) and (0,2)
(e)	Region represented by |z + 4i| ≥ 3 is	(v)	2(cos /6 +I sin /6)
(0	Region represented by |z + 4| ≤ 3 is	(Vi)	On or inside the circle having centre (-4,0) and radius 3 units.
(g)	Conjugate of 1+2i/1-I  lies in	(vii)	First quadrant
(h)	Reciprocal of 1 – i lies in	(viii)	Third quadrant

Q28. What is the conjugate of 2-i / (1 – 2i)²

Q29. If |Z₁| = |Z₂|, is it necessary that Z₁ = Z₂?
Sol: If |Z₁| = |Z₂| then z₁ and z₂ are at the same distance from origin.
But if arg(Z₁) ≠arg(z₂), then z₁ and z₂ are different.
So, if (z₁| = |z₂|, then it is not necessary that z₁ = z₂.
Consider Z₁ = 3 + 4i and Z₂ = 4 + 3i

Q30.If  (a²+1)² / 2a –i = x + iy, then what is the value of x² + y²?
Sol: (a²+1)² / 2a –i = x + iy

Q31. Find the value of z, if |z| = 4 and arg (z) = 5π/6

Q34. Where does z lies, if | z – 5i / z + 5i  |  = 1?
Sol: We have | z – 5i / z + 5i  |

Instruction for Exercises 35-40: Choose the correct answer from the given four options indicated against each of the Exercises.

Q35. sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for

Question 40.

Q41. Which of the following is correct for any two complex numbers z₁ and z₂?

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We hope the NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations help you. If you have any query regarding NCERT Exemplar Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-4-principle-mathematical-induction/

NCERT Exemplar Class 11 Maths Chapter 4 Principle of Mathematical Induction

Short Answer Type Questions
Q1. Give an example of a statement P(n) which is true for all n≥ 4 but P(l), P(2) and P(3) are not true. Justify your answer.

Sol. Consider the statement P(n): 3n < n!

For n = 1, 3 x 1 < 1!, which is not true
For n = 2, 3 x 2 < 2!, which is not true
For n = 3, 3 x 3 < 3!, which is not true
For n = 4, 3 x 4 < 4!, which is true
For n = 5, 3 x 5 < 5!, which is true

Q2. Give an example of a statement P(n) which is true for all Justify your answer.

Q3. 4ⁿ – 1 is divisible by 3, for each natural number
Sol: Let P(n): 4ⁿ – 1 is divisible by 3 for each natural number n.
Now, P(l): 4¹ – 1 = 3, which is divisible by 3 Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 4^k – 1 is divisible by 3
or               4^k – 1 = 3m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 4^k+1 – 1
= 4^k-4-l
= 4(3m + 1) – 1  [Using (i)]
= 12 m + 3
= 3(4m + 1), which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q4. 2³ⁿ – 1 is divisible by 7, for all natural numbers
Sol: Let P(n): 2³ⁿ – 1 is divisible by 7
Now, P( 1): 2³ — 1 = 7, which is divisible by 7.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 2^3k – 1 is divisible by 7.
or               2^3k -1 = 7m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 2^3(k+1)– 1
= 2^3k.2³– 1
= 8(7 m + 1) – 1
= 56 m + 7
= 7(8m + 1), which is divisible by 7.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q5. n³ – 7n + 3 is divisible by 3, for all natural numbers
Sol: Let P(n): n³ – 7n + 3 is divisible by 3, for all natural numbers n.
Now P(l): (l)³ – 7(1) + 3 = -3, which is divisible by 3.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) = K³ – 7k + 3 is divisible by 3
or K³ – 7k + 3 = 3m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1 ):(k + l)³ – 7(k + 1) + 3
= k³ + 1 + 3k(k + 1) – 7k— 7 + 3 = k³ -7k + 3 + 3k(k + l)-6
= 3m + 3[k(k+l)-2]  [Using (i)]
= 3[m + (k(k + 1) – 2)], which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q6. 3²ⁿ – 1 is divisible by 8, for all natural numbers
Sol: Let P(n): 3²ⁿ – 1 is divisible by 8, for all natural numbers n.
Now, P(l): 3² – 1 = 8, which is divisible by 8.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): 3^2k – 1 is divisible by 8
or               3^2k -1 = 8m, m ∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 3^2(k+1)– l
= 3^2k • 3² — 1
= 9(8m + 1) – 1     (using (i))
= 72m + 9 – 1
= 72m + 8
= 8(9m +1), which is divisible by 8 Thus P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q7. For any natural number n, 7ⁿ – 2ⁿ is divisible by 5.
Sol: Let P(n): 7ⁿ – 2ⁿ is divisible by 5, for any natural number n.
Now, P(l) = 7¹-2¹ = 5, which is divisible by 5.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.

.’.  P(k) = 7^k -2^k is divisible by 5
or  7^k – 2^k = 5m, m∈ N                                                                           (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 7^k+1 -2^k+1
= 7^k-7-2^k-2
= (5 + 2)7^k -2^k-2
= 5.7^k + 2.7^k-2-2^k
= 5.7^k + 2(7^k – 2^k)
= 5 • 7^k + 2(5 m)     (using (i))
= 5(7^k + 2m), which divisible by 5.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q8. For any natural number n, xⁿ -yⁿ is divisible by x -y, where x and y are any integers with x ≠y
Sol: Let P(n) : xⁿ – yⁿ is divisible by x – y, where x and y are any integers with x≠y.
Now, P(l): x¹ -y¹ = x-y, which is divisible by (x-y)
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): x^k -y^k is divisible by (x – y)
or   x^k-y^k = m(x-y),m ∈ N …(i)
Now, we have to prove that P(k + 1) is true.
P(k+l):x^k+l-y^k+l
= x^k-x-x^k-y + x^k-y-y^ky
= x^k(x-y) +y(x^k-y^k)
= x^k(x – y) + ym(x – y)  (using (i))
= (x -y) [x^k+ym], which is divisible by (x-y)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q9. n³ -n is divisible by 6, for each natural number n≥
Sol: Let P(n): n³ – n is divisible by 6, for each natural number n> 2.
Now, P(2): (2)³ -2 = 6, which is divisible by 6.
Hence, P(2) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): k³ – k is divisible by 6
or    k³ -k= 6m, m∈ N       (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): (k+ l)³-(k+ 1)
= k³+ 1 +3k(k+ l)-(k+ 1)
= k³+ 1 +3k² + 3k-k- 1 = (k³-k) + 3k(k+ 1)
= 6m + 3 k(k +1)  (using (i))
Above is divisible by 6.   (∴ k(k + 1) is even)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 2.

 

Q10. n(n² + 5) is divisible by 6, for each natural number
Sol: Let P(n): n(n² + 5) is divisible by 6, for each natural number.
Now P(l): 1 (l² + 5) = 6, which is divisible by 6.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): k( k² + 5) is divisible by 6.
or K (k²+ 5) = 6m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(K+l):(K+l)[(K+l)² + 5]
= (K + l)[K² + 2K+6]
= K³ + 3 K² + 8K + 6
= (K² + 5K) + 3 K² + 3K + 6 =K(K² + 5) + 3(K² + K + 2)
= (6m) + 3(K² + K + 2)        (using (i))
Now, K² + K + 2 is always even if A is odd or even.
So, 3(K² + K + 2) is divisible by 6 and hence, (6m) + 3(K² + K + 2) is divisible by 6.
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q11. n² < 2ⁿ, for all natural numbers n ≥
Sol: Let P(n): n² < 2ⁿ for all natural numbers n≥ 5.
Now P(5): 5² < 2⁵ or 25 < 32, which is true.
Hence, P(5) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): k² < 2^k  (i)
Now, to prove that P(k + 1) is true, we have to show that P(k+ 1): (k+ l)² <2^k+1
Using (i), we get
(k + l)² = k² + 2k + 1 < 2^k + 2k + 1         (ii)
Now let, 2^k + 2k + 1 < 2^k+1     (iii)
∴ 2^k + 2k + 1 < 2 • 2^k
2k + 1 < 2^k, which is true for all k > 5 Using (ii) and (iii), we get (k + l)² < 2^k+1 Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 5.

Q12. 2n<(n + 2)! for all natural numbers
Sol: Let P(n): 2n < (n + 2)! for all natural numbers n.
P( 1): 2 < (1 + 2)! or 2 < 3! or 2 < 6, which is true.
Hence,P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) :2k<(k + 2)!  (i)
To prove that P(k + 1) is true, we have to show that
P(k + 1): 2(k+ 1) < (k + 1 + 2)!
or 2(k+ 1) < (k + 3)!
Using (i), we get
2(k + 1) = 2k + 2<(k+2) !  +2  (ii)
Now let, (k + 2)! + 2 < (k + 3)!  (iii)
=>  2 < (k+ 3)! – (k+2) !
=> 2 < (k + 2) ! [k+ 3-1]
=>2<(k+ 2) ! (k + 2), which is true for any natural number.
Using (ii) and (iii), we get 2(k + 1) < (k + 3)!
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q14. 2 + 4 + 6+… + 2n = n² + n, for all natural numbers
Sol: Let P(n) :2 + 4 + 6+ …+2 n = n² + n
P(l): 2 = l² + 1 = 2, which is true
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): 2 + 4 + 6 + .,.+2k = k² + k  (i)
Now, we have to prove that P(k + 1) is true.
P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)
= k² + k + 2(k+ 1)  [Using (i)]
= k² + k + 2k + 2
= k² + 2k+1+k+1
= (k + 1)² + k+ 1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q15. 1 + 2 + 2² + … + 2ⁿ = 2^{n +1} – 1 for all natural numbers
Sol: Let P(n): 1 + 2 + 2² + … + 2ⁿ = 2^{n +1} – 1, for all natural numbers n
P(1): 1 =2^{0 + 1} — 1 = 2 — 1 = 1, which is true.
Hence, ,P(1) is true.
Let us assume that P(n) is true for some natural number n = k.

P(k): l+2 + 2²+…+2^k = 2^k+1-l              (i)

Now, we have to prove that P(k + 1) is true.

P(k+1): 1+2 + 2²+ …+2^k + 2^k+1
= 2^{k +1} – 1 + 2^k+1  [Using (i)]
= 2.2^k+l– 1 = 1
₌ 2^(k+1)+1-1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q16. 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers
Sol: Let P(n): 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers n.
P(1): 1 = 1(2 x 1 – 1) = 1, which is true.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k):l+5 + 9 +…+(4k-3) = k(2k-1)  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 1 + 5 + 9 + … +  (4k- 3) + [4(k+ 1) – 3]
= 2k² -k+4k+ 4-3
= 2k² + 3k + 1
= (k+ 1)( 2k + 1)

= (k+l)[2(k+l)-l]

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.

Long Answer Type Questions
Q17. A sequence a_x, a₂, a₃, … is defined by letting a₁=3 and a_k = 7a_k–₁ for all natural numbers k≥ Show that a_n = 3 • 7 ^n-1 for all natural numbers.

Q18. A sequence b₀, b₁, b₂, … is defined by letting b₀ = 5 and b_k = 4 + b_k–₁, for all natural numbers Show that b_n = 5 + 4n, for all natural number n using mathematical induction.
Sol. We have a sequence b₀, b₁, b₂,… defined by letting b₀ = 5 and b_k = 4 + b_k–₁,, for all natural numbers k.

So, by the principle of mathematical induction P(n) is true for any natural number rt,n> 1.

Q25. Prove that number of subsets of a set containing n distinct elements is 2″, for all n ∈
Sol: Let P(n): Number of subset of a set containing n distinct elements is 2″, for all ne N.
For n = 1, consider set A = {1}. So, set of subsets is {{1}, ∅}, which contains 2¹ elements.
So, P(1) is true.
Let us assume that P(n) is true, for some natural number n = k.
P(k): Number of subsets of a set containing k distinct elements is 2^k To prove that P(k + 1) is true,
we have to show that P(k + 1): Number of subsets of a set containing (k + 1) distinct elements is 2^k+1
We know that, with the addition of one element in the set, the number of subsets become double.
Number of subsets of a set containing (k+ 1) distinct elements = 2×2^k = 2^k+1
So, P(k + 1) is true. Hence, P(n) is true.

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NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-3-trigonometric-functions/

NCERT Exemplar Class 11 Maths Chapter 3  Trigonometric Functions

Class 11 Trigonometry Questions Chapter 3

Trigonometry Class 11 Questions Chapter 3

Trigonometry Class 11 Extra Questions Chapter 3

NCERT Exemplar Class 11 Trigonometry Chapter 3

Q4. If cos (α + ) =4/5 and sin (α- )=5/13 , where α lie between 0 and π/4, then find the value of tan 2α.

Solution:

Trigonometry Questions For Class 11 Chapter 3

Trigonometry Questions Class 11 Chapter 3

Q6. Prove that cos cos /2- cos 3 cos 9/2 = sin 7/2 sin 4 .

Trigonometry Class 11 Important Questions Chapter 3

Q7. If a cos θ + b sin θ =m and a sin θ -b cosθ = n, then show that a² + b²-m² + n²

Sol: We have, a cos θ + b sin θ = m (i)
and a sin θ -bcos θ = n (ii)

Q8. Find the value of tan 22°30′

NCERT Exemplar Class 11 Trigonometry Solutions Chapter 3

Q9. Prove that sin 4A = 4 sin A cos³A – 4 cos A sin³ A.

Sol: L.H.S. = sin 4A
= 2 sin 2A- cos 2A = 2(2 sin A cosA)(cos² A – sin² A)
= 4 sin A • cos³ A – 4 cos A sin³ A = R.H.S.

Q10. If tan + sin = m and tan – sin = n, then prove that m²-n² = 4 sin tan

Sol:We have, tan + sin = m   (i)
And tan -sin =n  (ii)
Now,         m + n = 2 tan
And          m – n = 2 sin.
(m + n)(m -n) = 4 sin 6
tan m² -n² = 4 sin -tan

Trigonometric Functions Class 11 Extra Questions Chapter 3

Q11. If tan (A + B) =p and tan (A – B) = q, then show that tan 2A = p+q / 1 – pq

Sol: We have tan (A + B) =p and tan (A – B) = q
tan2A = tan [(A + B) + (A-B)]

Q12. If cos + cos = 0 = sin + sin β, then prove that cos 2 + cos 2β = -2 cos (α + ).

NCERT Exemplar Class 11 Maths Trigonometry Solutions Chapter 3

Q15.  If sin θ+ cos θ =1, then find the general value of θ

Q16. Find the most general value of θ satisfying the equation tan θ = -1 and cos θ = 1/√2 .
Sol: We have tan θ = -1 and cos θ =1/√2 .
So, θ lies in IV quadrant.
θ = 7/4
So, general solution is θ = 7π/4 + 2 n π, n∈ Z

Q17. If cot θ + tan θ = 2 cosec θ, then find the general value of θ
Sol: Given that, cot θ + tan θ = 2 cosec θ

Q18. If 2 sin² θ =3 cos θ, where O≤θ≤2, then find the value of θ

Q19. If sec x cos 5x + 1 = 0, where 0 < x <π/2 , then find the value of x.

Long Answer Type Questions

Q20. If sin(θ + α) = a and sin(θ + β) = b , then prove that cos2(α – β) – 4abcos(α – β) = 1-2a² -2b²

Sol: We have sin(θ + α) = a —(i)
sin(θ + β) = b ——-(ii)

Q22. Find the value of the expression

Q23. If a cos 2+b sin 2 = c has α and β as its roots, then prove that tan α +tan β = 2b/a+c

Q24. If x = sec ϕ-tanϕandy = cosec ϕ + cot ϕ then show that xy + x -y +1=0.

Q25. If lies in the first quadrant and cos =8/17 , then find the value of cos (30° + ) + cos (45° – ) + cos (120° – ).

Q26. Find the value of the expression cos⁴ π/8 + cos⁴ 3π/8  + cos⁴ 5π/8  + cos⁴7π/8

Q27. Find the general solution of the equation 5 cos² +7 sin² -6 = 0.

Q28. Find the general solution of‘the equation sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x.
Sol: We have, (sin x + sin 3x) – 3 sin 2x = (cos x + cos 3x) – 3 cos 2x
=> 2 sin 2x cos x – 3 sin 2x = 2 cos 2x.cos x – 3 cos 2x
=> sin 2x(2 cos x – 3) = cos 2x(2 cos x – 3)
=> sin 2x = cos 2x (As cos x ≠ 3/2)
=>              tan 2x = 1    => tan 2x = tan π/4
=>              2x = nπ + π/4 , n∈Z
x = nπ/2 +π/8 , n∈Z

Q29. Find the general solution of the equation (√3- l)cos + (√3+ 1)sin = 2.

Objective Type Questions

Q30. If sin + cosec =2, then sin² + cosec² is equal to
(a) 1
(b) 4                          
(c) 2                         
(d) None of these

Q31. If f(x) = cos² x + sec² x, then ‘
(a) f(x) <1             
(b) f(x) = 1              
(c) 2 <f(x) < 1      
(d) fx) ≥ 2

Q32. If tan θ = 1/2 and tan ϕ = 1/3, then the value of θ + ϕ is

Q33. Which of the following is not correct?

(a) sin θ = – 1/5 (b) cos θ = 1                 (c) sec θ = -1/2         (d) tan θ = 20
Sol: (c)
We know that, the range of sec θ is R – (-1, 1).
Hence, sec θ cannot be equal to -1/2

Q34. The value of tan 1° tan 2° tan 3° … tan 89° is
(a) 0
(b) 1
(c) 1/2
(d) Not defined

Sol: (b)
tan 1° tan 2° tan 3° … tan 89°
= [tan 1° tan 2° … tan 44°] tan 45°[tan (90° – 44°) tan (90° – 43°)… tan (90° – 1°)]
= [tan 1° tan 2° … tan 44°] [cot 44° cot 43°……. cot 1°]
= 1-1… 1-1 = 1

Q36. The value of cos 1° cos 2° cos 3° … cos 179° is
(a) 1/√2
(b) 0
(c) 1
(d) -1

Sol: (b)
Since cos 90° = 0, we have
cos 1° cos 2° cos 3° …cos 90°… cos 179° = 0

Q37. If tan θ = 3 and θ lies in the third quadrant, then the value of sin θ is

Q38. The value of tan 75° – cot 75° is equal to

Q39. Which of the following is correct?
(a) sin 1° > sin 1                                     
(b) sin 1° < sin 1
(c) sin l° = sin l
(d) sin l° = π/18° sin 1

 

Sol: We know that, in first quadrant if θ is increasing, then sin θ is also increasing.
∴sin 1° < sin 1 [∵ 1 radian = 57◦30′]

Q41. The minimum value of 3 cos x + 4 sin x + 8 is
(a) 5
(b) 9
(c) 7
(d) 3
Sol: (d)
3 cos x + 4sin x + 8 = 5 (3/5 cos x + 4/5sin x) + 8
= 5(sin α cos x + cos α sin x) + 8
= 5 sin(α + x) + 8, where tan α = 3/4

Q42. The value of tan 3A – tan 2A – tan A is
(a) tan 3A . tan 2A . tan A
(b) -tan 3A .tan 2A . tan A
(c) tan A . tan 2A – tan 2A . tan 3A – tan 3A . tan A
(d) None of these
Sol: (a)
3A= A+ 2A
=> tan 3A = tan (A + 2A)
=> tan 3 A = tanA + tan2A/ 1 – tan A . tan 2A
=> tan A + tan 2A = tan 3A – tan 3A• tan 2A . tan A
=> tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A

Q43. The value of sin (45° + )- cos (45° – ) is
(a) 2 cos              
(b) 2 sin              
(c) 1                         
(d) 0
Sol: (d)
sin (45° + ) – cos (45° – ) = sin (45° + ) – sin (90° – (45° – ))
= sin (45° + ) – sin (45°+ ) = 0

Q44. The value of (π/4+ ) cot (π/4- ) is
(a) -1                       
(b)  0  
(c)  1                     
(d)   Not defined

Q46. The value of cos 12° + cos 84° + cos 156° + cos 132° is
(a) 1/2            
(b) 1                       
(c) -1/2            
(d) 1/8

Q47. If tan A = 1/2 and tan B = 1/3 then tan (2A + B) is equal to
(a) 1
(b) 2
(c) 3
(d) 4

Q49. The value of sin 50° – sin 70° + sin 10° is equal to
(a) 1                       
(b) 0                       
(c) 1
(d) 2

Q50. If sin + cos =1, then the value of sin 2 is
(a) 1                      
(b) 1   
(c) 0                        
(d) -1

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NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions are part of NCERT Exemplar Class 11 Maths. Here we have given NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions. https://www.cbselabs.com/ncert-exemplar-problems-class-11-mathematics-chapter-2-relations-functions/

NCERT Exemplar Class 11 Maths Chapter 2 Relations and Functions

Relations and Functions

Short Answer Type Questions
Q1. If A = {-1, 2, 3 } and B = {1, 3}, then determine
(i) AxB (ii) BxC (c) BxB (iv) AxA
Sol: We have A = {-1,2,3} and B = {1,3}
(i) A x B = {(-1, 1), (-1, 3), (2, 1), (2, 3), (3,1), (3, 3)}
(ii) BxA = {( 1, -1), (1, 2), (1,3), (3,-1), (3,2), (3, 3)}
(iii) BxB= {(1,1), (1,3), (3,1), (3, 3)}
(iv) A xA = {(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3,3)}

Q2. If P = {x : x < 3, x e N}, Q= {x : x≤2,x ∈ W}. Find (P∪ Q) x (P∩ Q), where W is the set of whole numbers.
Sol: We have, P={x: x<3,x ∈ N} = {1,2}
And Q = {x :x≤ 2,x∈ W] = {0,1,2}
P∪Q= {0, 1,2} and P ∩ Q= {1,2}
(P ∪ Q) x (P ∩ Q) = {0,1, 2} x {1,2}
= {(0,1), (0, 2), (1,1), (1,2), (2,1), (2, 2)}

Q3. lfA={x:x∈ W,x < 2}, 5 = {x : x∈N, 1 <.x < 5}, C= {3, 5}. Find
(i) Ax(B∩Q) (ii) Ax(B∪C)
Sol: We have, A = {x :x∈ W,x< 2} = {0, 1};
B = {x : x ∈ N, 1 <x< 5} = {2, 3,4}; and C= {3, 5}

(i) B∩ C = {3}
A x (B ∩ C) = {0, 1} x {3} = {(0, 3), (1, 3)}

(ii) (B ∪ C) ={2,3,4, 5}
A x (B ∪ C) = {0, 1} x {2, 3,4, 5}
= {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1, 5)}

Q4. In each of the following cases, find a and b. (2a + b, a – b) = (8, 3) (ii) {a/4, a – 2b) = (0, 6 + b)
Sol: (i) We have, (2a + b,a-b) = (8,3)
=> 2a + b = 8 and a – b = 3
On solving, we get a = 11/3 and b = 2/3

 

Q5. Given A = {1,2,3,4, 5}, S= {(x,y) :x∈ A,y∈ A}.Find the ordered pairs which satisfy the conditions given below
x+y = 5 (ii) x+y<5 (iii) x+y>8
Sol: We have, A = {1,2, 3,4, 5}, S= {(x,y) : x ∈ A,y∈ A}
(i) The set of ordered pairs satisfying x + y= 5 is {(1,4), (2,3), (3,2), (4,1)}
(ii) The set of ordered pairs satisfying x+y < 5 is {(1,1), (1,2), (1,3), (2, 1), (2,2), (3,1)}
(iii) The set of ordered pairs satisfying x +y > 8 is {(4, 5), (5,4), (5, 5)}.

Q6. Given R = {(x,y) : x,y ∈ W, x² + y² = 25}. Find the domain and range of R
Sol: We have, R = {(x,y):x,y∈ W, x² + y² = 25}
= {(0,5), (3,4), (4, 3), (5,0)}
Domain of R = Set of first element of ordered pairs in R = {0,3,4, 5}
Range of R = Set of second element of ordered pairs in R = {5,4, 3, 0}

Q7. If R₁ = {(x, y)| y = 2x + 7, where x∈ R and -5 ≤ x ≤ 5} is a relation. Then find the domain and range of R₁.
Sol: We have, R₁ = {(x, y)|y = 2x + 7, where x∈ R and -5 ≤x ≤ 5}
Domain of R₁ = {-5 ≤ x ≤ 5, x ∈ R} = [-5, 5]
x ∈ [-5, 5]
=> 2x ∈ [-10,10]
=>2x + 7∈ [-3, 17]
Range is [-3, 17]

Q8. If R₂ = {(x, y) | x and y are integers and x² +y² = 64} is a relation. Then find R₂
Sol: We have, R₂ = {(x, y) | x and y are integers and x² + y² – 64}
Clearly, x² = 0 and y² = 64 or x² = 64 andy² = 0
x = 0 and y = ±8
or x = ±8 and y = 0
R₂ = {(0, 8), (0, -8), (8,0), (-8,0)}

Q9. If R₃ = {(x, |x|) | x is a real number} is a relation. Then find domain and range
Sol: We have, R₃ = {(x, |x)) | x is real number}
Clearly, domain of R₃ = R
Now, x ∈ R and |x| ≥ 0 .
Range of R₃ is [0,∞)

Q10. Is the given relation a function? Give reasons for your answer.
(i) h={(4,6), (3,9), (-11,6), (3,11)}
(ii) f = {(x, x) | x is a real number}
(iii) g = {(n, 1 In)| nis a positive integer}
(iv) s= {(n, n²) | n is a positive integer}
(v) t= {(x, 3) | x is a real number}
Sol: (i) We have, h = {(4,6),(3,9), (-11,6), (3,11)}.
Since pre-image 3 has two images 9 and 11, it is not a function.
(ii) We have, f = {(x, x) | x is a real number}
Since every element in the domain has unique image, it is a function.
(iii) We have, g= {(n, 1/n) | nis a positive integer}
For n, it is a positive integer and 1/n is unique and distinct. Therefore,every element in the domain has unique image. So, it is a function.
(iii) We have, s = {(n, n²) | n is a positive integer}
Since the square of any positive integer is unique, every element in the domain has unique image. Hence, ibis a function.
(iv) We have, t = {(x, 3)| x is a real number}.
Since every element in the domain has the image 3, it is a constant function.

Q11. If f and g are real functions defined byf( x) = x² + 7 and g(x) = 3x + 5, find each of the following

Q12. Let f and g be real functions defined by f(x) = 2x+ 1 and g(x) = 4x – 7.
(i) For what real numbers x,f(x)= g(x)?
(ii) For what real numbers x,f (x) < g(x)?
Sol: We have,f(x) = 2x + 1 and g(x) = 4x-7
(i) Now f (x) = g(x)
=> 2x+l=4x-7
=> 2x = 8 =>x = 4
(ii) f (x) < g(x)
=> 2x + 1 < 4x – 7
=> 8 < 2x
=> x > 4

Q13. If f and g are two real valued ftmctions defined as f(x) = 2x + 1, g(x) = x² + 1, then find.

Q14. Express the following functions as set of ordered pairs and determine their range.
f:X->R,f{x) =  x³ + 1, where X= {-1,0, 3, 9, 7}
Sol: We have, f:X→ R,flx) = x³ + 1.
Where X = {-1, 0, 3, 9, 7}
Now f (-l) = (-l)³+1 =-l + 1 =0
f(0) = (0)³+l=0+l = l
f(3) = (3)³ + 1 = 27 + 1 = 28
f(9) = (9)³ + 1 = 729 + 1 = 730
f(7) = (7)³ + 1 = 343 + 1 = 344
f= {(-1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}
Range of f= {0, 1, 28, 730, 344}

Q15. Find the values of x for which the functions f(x) = 3x² -1 and g(x) = 3+ x are equal.
Sol: f(x) = g(x)
=> 3x²-l=3+x => 3x²-x-4 = 0 => (3x – 4)(x+ 1) – 0
x= -1,4/3

Q16. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g(x) = x +, then what values should be assigned to and ?
Sol:We have, g = {(1, 1), (2, 3), (3, 5), (4,7)}
Since, every element has unique image under g. So, g is a function.
Now, g(x) = x + For (1,1), g(l) = a(l) + P
=>             l = +            (i)
For (2, 3), g(2) = (2) +
=>             3 = 2 +         (ii)
On solving Eqs. (i) and (ii), we get = 2, = -l
f(x) = 2x-1
Also, (3, 5) and (4, 7) satisfy the above function.

Q17. Find the domain of each of the following functions given by

Q18. Find the range of the following functions given by

 

Q19. Redefine the function f(x) = |x-2| + |2+x| , -3 ≤x ≤3

Q21. Let f (x) = √x and g(x) = xbe two functions defined in the domain R⁺ ∪ {0}. Find
(i) (f+g)(x)        
(ii) (f-g)(x)
(iii) (fg)(x)
(iv) f/g(x)

 

Q23. If f( x)= y = ax-b/ cx-a then prove that f (y) = x

Objective Type Questions
Q24. Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is
(a) mⁿ                     
(b) n^m– 1                  
(c) mn – 1               
(d) 2^mn– 1

Sol: (d) We have, n(A) = m and n(B) = n
n(A xB) = n(A). n(B) = mn
Total number of relation from A to B = Number of subsets of AxB = 2^mn
So, total number of non-empty relations = 2^mn – 1

Q25. If [x]² – 5[x] + 6 = 0, where [. ] denote the greatest integer function, then
(a) x ∈ [3,4]
(b) x∈ (2, 3]            
(c) x∈ [2, 3]           
(d) x ∈ [2, 4)
Sol: (d) We have [x]² – 5[x] + 6 = 0 => [(x – 3)([x] – 2) = 0
=> [x] = 2,3 .
For [x] = 2, x ∈ [2, 3)
For [x] = 3, x ∈ [3,4)
x ∈ [2, 3) u [3,4)
Or x ∈ [2,4)

Q29. If fx) ax+ b, where a and b are integers,f(-1) = -5 and f(3) – 3, then a and b are equal to
(a) a = -3, b =-1
(b) a = 2, b =-3
(c) a = 0, b = 2
(d) a = 2, b = 3

Fill in the Blanks Type Questions

Q36. Let f and g be two real functions given by  f= {(0, 1), (2,0), (3,.-4), (4,2), (5, 1)}
g= {(1,0), (2,2), (3,-1), (4,4), (5, 3)}  then the domain of f  x g is given by________ .
Sol: We have, f = {(0, 1), (2, 0), (3, -4), (4, 2), (5,1)} and g= {(1, 0), (2, 2), (3, 1), (4,4), (5, 3)}
Domain of  f = {0,2, 3, 4, 5}
And Domain of g= {1, 2, 3,4, 5}
Domain of (f x g) = (Domain of f) ∩ (Domain of g) = {2, 3,4, 5}

Matching Column Type Questions

Q37. Let f= {(2,4), (5,6), (8, -1), (10, -3)} andg = {(2, 5), (7,1), (8,4), (10,13), (11, 5)} be two real functions. Then match the following:

True/False Type Questions

Q38. The ordered pair (5,2) belongs to the relation R ={(x,y): y = x – 5, x,y∈Z}
Sol: False
We have, R = {(x, y): y = x – 5, x, y ∈ Z}
When x = 5, then y  = 5-5=0 Hence, (5, 2) does not belong to R.

Q39. If P = {1, 2}, then P x P x P = {(1, 1,1), (2,2, 2), (1, 2,2), (2,1, 1)}
Sol:False
We have, P = {1, 2} and n(P) = 2
n(P xPxP) = n(P) x n(P) x n(P) = 2 x 2 x 2
= 8 But given P x P x P has 4 elements.

Q40. If A= {1,2, 3}, 5= {3,4} and C= {4, 5, 6}, then (A x B) ∪ (A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3, 5), (3,6)}.
Sol: True
We have.4 = {1,2, 3}, 5= {3,4} andC= {4,5,6}
AxB= {(1, 3), (1,4), (2, 3), (2,4), (3, 3), (3,4)}
And A x C =  {(1,4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6)}
(A x B)∪(A xC)= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3,3), (3,4), (3, 5), (3,6)}

Q42. If Ax B= {(a, x), (a, y), (b, x), (b, y)}, thenM = {a, b},B= {x, y}.
Sol: True
We have, AxB= {{a, x), {a, y), (b, x), {b, y)}
A = Set of first element of ordered pairs in A x B = {a, b}
B = Set of second element of ordered pairs in A x B = {x, y}

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