Haryana Board (HBSE) Class 12 Maths Question Paper 2024 with a fully solved answer key. Students can use this HBSE Class 12 Mathematics Solved Paper to match their responses and understand the question pattern. This BSEH Math Answer Key 2024 is based on the latest syllabus and exam format to support accurate preparation and revision for the board exams.
HBSE Class 12 Maths Question Paper 2024 Answer Key
1. Let N be the set of natural numbers and the function f : N → N be defined by f(n) = 2n + 3 ∀ n ∈ N, then f is :
(A) Surjective
(B) Injective
(C) Bijective
(D) None of these
Answer : (B) Injective
For f(n) = 2n + 3
Assume f(n1) = f(n2) for some n1, n2 ∈ N
2n1 + 3 = 2n2 + 3
2n1 = 2n2
n1 = n2
Since f(n1) = f(n2) implies n1 = n2, the function is injective.
2. The domain of sin–12x is :
(A) [0, 1]
(B) [–1, 1]
(C) [–1/2, 1/2]
(D) [–2, 2]
Answer : (C) [–1/2, 1/2]
sin–12x = a
sina = 2x
–1 ≤ sina ≤ 1
–1 ≤ 2x ≤ 1
–1/2 ≤ x ≤ 1/2
3. Total numbers of possible matrices of order 3 × 3 with each entry 2 or 0, are :
(A) 9
(B) 27
(C) 81
(D) 512
Answer : (D) 512
A 3 × 3 matrix has 3 × 3 = 9 independent entries. Each entry can take one of 2 values (either 2 or 0).
Total matrices = CN = 29 = 512
4. If A = |x 2| = |6 2|
. |18 x| = |18 6|, then x is equal to :
(A) 6
(B) ± 6
(C) – 6
(D) 0
Answer : (B) ± 6
x(x) – 2(18) = 6(6) – 2(18)
x2 – 36 = 36 – 36
x2 = 36
x = √36 = ± 6
5. Let A be a non-singular square matrix of order 3 × 3. Then |adj A| is equal to …………
Answer : |adj A| = |A|n–1 = |A|3–1 = |A|2
6. The derivative of sin(logx) w.r.t. x is :
(A) cos(logx)/x2
(B) cos(logx)
(C) cos(logx)/x
(D) None of these
Answer : (C) cos(logx)/x
Using chain rule,
Derivative of sin(logx) = cos(logx) × 1/x = cos(logx)/x
7. The derivative of sinx w.r.t. cos x is ……………
Answer : Let y = sinx, dy/dx = cosx
z = cosx, dz/dx = – sinx
dy/dz = (dy/dx) / (dz/dx) = cosx / (–sinx) = –cotx
8. ∫ex secx(1+tanx)dx is equal to :
(A) ex cosx + c
(B) ex secx + c
(C) ex sinx + c
(D) ex tanx + c
Answer : (B) ex secx + c
Here f(x) = secx and f'(x) = secx.tanx
Using ∫ex [f(x) + f'(x)]dx = ex f(x) + c
∫ex secx(1+tanx)dx = ex secx + c
9. The value of –π/2 ∫ π/2 sin5xdx is equal to ………….
Answer : The function f(x) = sin5x is an odd function because f(–x) = sin5(–x) = (–sin x)5 = –sin5x = –f(x)
For any odd function, the integration of f(x)dx limits [–π/2, π/2] is 0.
Hence, –π/2 ∫ π/2 sin5xdx = 0
10. The area enclosed by circle x2 + y2 = 2 is equal to :
(A) 4π sq. units
(B) 2√2π sq. units
(C) 4π2 sq. units
(D) 2π sq. units
Answer : (D) 2π sq. units
The standard equation of a circle centered at the origin is given by x2 + y2 = r2, where r is the radius.
Compare x2 + y2 = 2 with x2 + y2 = r2
r2 = 2
r = √2
Area of circle = πr2 = π(√2)2 = π(2) = 2π sq. units
11. The order and degree of the differential equation [1 + (dy/dx)2]2 = d2y/dx2 respectively are :
(A) 1, 2
(B) 2, 2
(C) 2, 1
(D) 4, 2
Answer : (C) 2, 1
12. Integrating factor of the differential equation x dy/dx – y = sinx is ………..
Answer : Standard linear form dy/dx + P(x)y = Q(x)
x dy/dx – y = sinx
dy/dx – y/x = sinx/x
Here, P(x) = –1/x
Integrating factor (IF) = e ∫P(x)dx = e ∫–1/xdx = e–logx = 1/x
13. Let the vectors a and b such that | a | = 3 and | b |= √2/3, then a × b is a unit vector, if angle between vectors a and b is :
(A) π/6
(B) π/4
(C) π/3
(D) π/2
Answer : (B) π/4
Given, |a × b| = 1
|a × b| = | a | | b | sinθ
| a | | b | sinθ = 1
(3)(√2/3)sinθ = 1
sinθ = 1/√2 = sin45° = sin(π/4)
θ = π/4
Angle between vectors a and b is π/4.
14. If the vectors a = 2î + λĵ + k̂ and b = î + 2ĵ + 3k̂ are orthogonal, then the value of λ is …………..
Answer : a.b = 0
(2î + λĵ + k̂).(î + 2ĵ + 3k̂) = 0
2(1) + λ(2) + 1(3) = 0
2 + 2λ + 3 = 0
2λ = – 5
λ = – 5/2
15. The equations of x-axis in space are :
(A) x = 0, y = 0
(B) x = 0, z = 0
(C) x = 0
(D) y = 0, z = 0
Answer : (D) y = 0, z = 0
16. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is …………..
Answer : The only even prime number is 2, with probability (1/6) per die.
For two dice, probability is = 1/6 × 1/6 = 1/36
17. A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, the probability of getting exactly one red ball is …………….
Answer : Probability of getting exactly one red ball if 3 balls are randomly drawn without replacement
= P(R).P(B).P(B) + P(B).P(R).P(B) + P(B) .P(B).P(R)
= (5/8 × 3/7 × 2/6) + (3/8 × 5/7 × 2/6) + (3/8 × 2/7 × 5/6)
= 30/336 + 30/336 + 30/336
= 90/336
= 15/56
18. Two cards are drawn from a well shuffled deck of 52 playing cards with replacement. The probability that both cards are queens is …………..
Answer : P(Queens) = Number of Queens / Total Cards = 4/52 = 1/13
The drawing is done with replacement, meaning the first card is returned to the deck and the deck is re-shuffled. This makes the two draws independent events. The probability of both independent events occurring is the product of their individual probabilities:
P(Both Queens) = 1/13 × 1/13 = 1/169
19. Assertion (A) : A relation R = {(a, b); |a – b| < 2} defined on the set A = {1, 2, 3, 4, 5} is reflexive.
Reason (R) : A relation R on the set A is said to be reflexive if for (a, b) ∈ R and (b, c) ∈ R, we have (a, c) ∈ R.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Answer : (C) Assertion (A) is true, but Reason (R) is false.
(The reason is false because this is actually the definition of transitive relation, not reflexive. Reflexive means (a, a) ∈ R for all a ∈ A).
20. Assertion (A) : The angle between the straight lines (x+1)/2 = (y–2)/5 = (z+3)/4 and (x–1)/1 = (y+2)/2 = (z–3)/(–3) is 90°.
Reason (R) : Skew lines are lines in different planes which are parallel and intersecting.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Answer : (C) Assertion (A) is true, but Reason (R) is false.
(The reason is incorrect because skew lines are lines in different planes which are neither parallel nor intersecting).
21. Let f : R → R be defined by f(x) = sinx and g : R → R be defined by g(x) = x2 then find fog and gof. Show that :
fog ≠ gof
Answer : Given f(x) = sinx and g(x) = x2
fog = f(g(x)) = f(x2) = sinx2
gof = g(f(x)) = g(sinx) = (sinx)2 = sin2x
so, sinx2 ≠ sin2x
Hence, fog ≠ gof
OR
Find the value of :
cos–1(1/2) + 2sin–1(1/2)
Answer : cos–1(1/2) + 2sin–1(1/2)
= π/3 + 2(π/6)
= π/3 + π/3
= 2π/3
22. If A = [cosθ sinθ]
. [–sinθ cosθ]
then verify that : A’A = I
Answer : The transpose (A’) is found by interchanging the rows and columns of A.
A’ = [cosθ –sinθ]
. [sinθ cosθ]
A’A = [cosθ –sinθ] [cosθ sinθ]
. [sinθ cosθ] [–sinθ cosθ]
First row, first column = (cosθ)(cosθ) + (–sinθ)(–sinθ) = cos2θ + sin2θ = 1
First row, second column = (cosθ)(sinθ) + (–sinθ)(cosθ) = sinθ.cosθ – sinθ.cosθ = 0
Second row, first column = (sinθ)(cosθ) + (cosθ)(–sinθ) = sinθ.cosθ – sinθ.cosθ = 0
Second row, second column = (sinθ)(sinθ) + (cosθ)(cosθ) = sin2θ + cos2θ = 0
A’A = [1 0]
. [0 1]
A’A = I
Therefore, we have verified that A’A = I.
23. Differentiate w.r.t. x :
tan–1(√(1–cosx)/(1+cosx)), π/4 < x < π/4
Answer : y = tan–1(√(1–cosx)/(1+cosx)
= tan–1(√(2sin2x/2)/(2cos2x/2))
= tan–1(√(tan2x/2)
= x/2
dy/dx = d/dx(x/2) = 1/2
24. Verify that the function xy = log y + c is a solution of the differential equation dy/dx = y2/(1–xy), xy ≠ 1.
Answer : d/dx (xy) = d/dx (log y + c)
x dy/dx + y(1) = 1/y dy/dx + 0
y = dy/dx (1/y – x)
y = dy/dx [(1–xy)/y]
dy/dx = y2/(1–xy)
Hence Proved.
OR
Find the general solution of differential equation dy/dx =(1+x2)(1+y2).
Answer : dy/dx =(1+x2)(1+y2)
1/(1+y2) dy = (1+x2) dx
Integrating both side,
∫1/(1+y2) dy = ∫(1+x2) dx
tan–¹y = x + x3/3 + c
y = tan(x + x3/3 + c)
25. Evaluate P(A∪B), if 2P(A) = P(B) = 5/13 and P(A/B) = 2/5.
Answer : 2P(A) = 5/13
P(A) = 5/26
P(B) = 5/13
P(A/B) = 2/5
Using, P(A/B) = P(A∩B)/P(B)
P(A∩B) = P(A/B) × P(B) = 2/5 × 5/13 = 2/13
Using, P(A∪B) = P(A) + P(B) – P(A∩B) = 5/26 + 5/13 – 2/13 = 11/26
The probability P(A∪B) is 11/26.
26. Show that the relation R defined in the set A of all triangles as R = {(T1, T2); T1 is similar to T2}, is an equivalence relation.
Answer : A relation R is Reflexive if every element is related to itself.
That is, for every triangle T1 ∈ A, we must have (T1, T1) ∈ R.
Proof: Every triangle is always similar to itself because all its corresponding angles and sides are in the same ratio. Therefore, T1~T1
Thus, R is reflexive.
A relation R is symmetric if:
(T1 ,T2) ∈ R→(T2, T1) ∈ R
Proof: If triangle T1 is similar to triangle T2 then the equality of corresponding angles and the proportionality of corresponding sides hold in both directions. Hence T1~T2→T2~T1.
Thus, R is symmetric.
A relation R is transitive if:
(T1, T2) ∈ R and (T2, T3) ∈ R→(T1, T3) ∈ R.
Proof: If T1 is similar to T2, and T2 is similar to T3, then they all have equal corresponding angles and proportional sides. This implies that T1~T3.
Thus, R is transitive.
Conclusion : Since the relation “is similar to” on the set of all triangles satisfies reflexivity, symmetry, and transitivity, the relation R is an equivalence relation.
OR
Write the following function in simplest form :
tan–1[(3a2x–x3)/(a3–3ax2)], a > 0; –a/√3 < x < a/√3
Answer : To simplify the expression inside the inverse tangent function, we divide the numerator and the denominator by a3 to get an expression in terms of x/a:
= (3a2x–x3)/a3 / (a3–3ax2)/a3
= [3(x/a) – (x/a)3] / [1 – 3(x/a)2]
Substituting x/a = tanθ into the expression:
= [3tanθ – tan3θ] / [1 – 3tan2θ]
= tan3θ
so, tan–1(tan3θ) = 3θ
Range, –a/√3 < x < a/√3
Dividing by a gives, –1/√3 < x/a < 1/√3
–1/√3 < tanθ < 1/√3
–π/6 < θ < π/6
–π/2 < 3θ < π/2
Since 3θ falls within the principal range of the inverse tangent function, tan–1(tan3θ) = 3θ
Substituting θ = tan–1(x/a), the simplified form is 3tan–1(x/a).
tan–1[(3a2x–x3)/(a3–3ax2)] = 3tan–1(x/a)
27. If A = [3 –2], I = [1 0]
. [4 –2] [0 1],
find K, so that A2 = KA – 2I.
Answer :
A2 = A.A = [3 –2] [3 –2] = [1 –2]
. [4 –2] [4 –2] [4 –4]
KA–2I=K[3 –2] – 2[1 0] = [3K–2 –2K]
. [4 –2] [0 1] [4K –2K–2]
Use A2 = KA – 2I
[1 –2] = [3K–2 –2K]
[4 –4] [4K –2K–2]
4 = 4K
K = 1
28. Find the value of K, so that the function
f(x) = {Kcosx/(π–2x), if x ≠ π/2
. {3, if x = π/2
is continuous at x = π/2.
Answer : To take f(x) continuous at x = π/2, we need limx→π/2 f(x) = f(π/2) = 3
Numerator derivative: d/dx [Kcosx] = –Ksinx
Denominator derivative: d/dx [π–2x] = –2
As x→π/2, cosx→0, π – 2x→0, so it is a 0/0 form.
so, x ≠ π/2
Use L’Hospital’s Rule:
= limx→π/2 Kcosx/(π–2x)
= limx→π/2 –Ksinx/(–2)
= K/2 sinπ/2
= K/2
For continuity, this limit must equal 3:
K/2 = 3
K = 6
29. A balloon, which always remains spherical, has a variable diameter 3/2(2x+1). Find the rate of change of its volume with respect to x.
Answer : The diameter of the spherical balloon is given by 3/2(2x+1).
Diameter (D) = 3/2(2x+1)
Radius (r) = D/2 = 3/4(2x+1)
Volume (V) = 4/3πr3 = 4/3π[3/4(2x+1)]3 = 9/16π(2x+1)3
Differentiate the volume expression using the chain rule to find the rate of change of the volume with respect to x,
dV/dx = d/dx [9/16π(2x+1)3] = 9/16π × 3(2x+1)2 × 2 = 27/8π(2x+1)2
The rate of change of its volume with respect to x is 27/8π(2x+1)2.
30. Integrate: ∫ √tanx / sinx.cosx dx
Answer : ∫ √tanx / sinx.cosx.(cosx/cosx) dx
= ∫ √tanx / tanx.cos2x dx
= ∫ sec2x / √tanx dx
Put t = tanx
dt/dx = sec2x
dt = sec2x dx
= ∫ 1/√t dt
= ∫ t–1/2 dt
= t1/2 / 1/2 + C
= 2√t + C
= 2√tanx + C
The integral is 2√tanx + C.
OR
Integrate: ∫1 / x(x4–1) dx
Answer : Let I = ∫1 / x(x4–1) dx
= 1/4 ∫4x3 / x4(x4–1) dx
Put x4 = t
4x3 dx = dt
I = 1/4 ∫1/t(t–1) dt
Let 1/t(t–1) = A/t + B/(t–1)
1 = A(t–1) + Bt ……..(i)
Putting t = 0 in eqn.(i), we get
A = –1
Putting t = 1 in eqn.(i), we get
B = 1
so, 1/t(t–1) = –1/t + 1/(t–1)
I = 1/4 ∫(–1/t + 1/(t–1)) dt
= 1/4 [–log | t | + log |t–1| + C
= 1/4 log | (t–1)/t | + C
= 1/4 log | (x4–1)/x4 | + C
31. Find λ and µ, if (2î + 6ĵ + 27k̂) × (î + λĵ + µk̂) = 0.
Answer : The cross product of the two vectors A = 2î + 6ĵ + 27k̂ and B = î + λĵ + µk̂ must equal the zero vector, 0. We can express this condition using the determinant formula:
. |î ĵ k̂|
A × B = |2 6 27| = 0
. |1 λ µ|
(6µ – 27λ)î – (2µ – 27)ĵ + (2λ – 6)k̂ = 0î + 0ĵ + 0k̂
From the equation,
2λ – 6 = 0 then λ = 3
2µ – 27 = 0 then µ = 27/2
The values are λ = 3 and µ = 27/2.
32. Solve the following system of linear equations, using matrix method :
2x + y + z = 1
x – 2y – z = 3/2
3y – 5z = 9
Answer : The given system of equations can be written in the matrix form AX = B, where:
. [2 1 1] [x] [1]
A = [1 –2 –1], X = [y], B = [3/2]
. [0 3 –5] [z] [9]
| A | = 2(10 + 3) – 1(–5) + 1(3) = 26 + 5 + 3 = 34
Since | A | = 34 ≠ 0, the matrix A is invertible, and a unique solution exists.
The cofactors are calculated as:
A11 = 13, A12 = 5, A13 = 3
A21 = 8, A22 = –10, A23 = –6
A31 = 1, A32 = 3, A33 = –5
The matrix of cofactors is:
[13 5 3]
[8 –10 –6]
[1 3 –5]
The adjoint of A, denoted adj(A), is the transpose of the matrix of cofactors:
[13 8 1]
[5 –10 3]
[3 –6 –5]
The inverse of A is given by A–1 = 1/|A| adj(A)
. [13 8 1]
A–1 = 1/34 [5 –10 3]
. [3 –6 –5]
X = A–1B
. [13 8 1] [1]
X = 1/34 [5 –10 3] [1.5]
. [3 –6 –5] [9]
. [13(1) + 8(1.5) + 1(9)] [34]
X =1/34 [5(1) – 10(1.5) + 3(9)] =[17]
. [3(1) – 6(1.5) – 5(9)] [–51]
. [1]
X = [1/2]
. [–3/2]
The solution to the system of equations is x = 1, y = 1/2 and z = –3/2.
33. Evaluate the following definite integral:
0 ∫ π/4 log(1+tanx) dx
Answer : Let I = 0 ∫ π/4 log(1+tanx) dx
Using, 0 ∫ a f(x) dx = 0 ∫ a f(a – x) dx
I = 0 ∫ π/4 log[1+tan(π/4 – x)] dx ….…(i)
= 0 ∫ π/4 log[1 + (1–tanx)/(1+1.tanx] dx
= 0 ∫ π/4 log[2/(1+tanx)]
= 0 ∫ π/4 [log2 – log(1+tanx)] dx
I = 0 ∫ π/4 log2 dx – 0 ∫ π/4 log(1+tanx)dx ……(ii)
Adding eqn.(i) and (ii), we get
2I = 0 ∫ π/4 log2 dx
I = log2/2 0∫π/4 dx
= log2/2 0[x]π/4
= log2/2 [π/4 – 0]
= log2/2 ×π/4
I = π/8 log2
The value of the definite integral is π/8 log2.
OR
Find the area of the region bounded by the ellipse 9x2 + 4y2 = 36.
Answer : 9x2 + 4y2 = 36
Dividing by 36,
9x2/36 + 4y2/36 = 36/36
x2/4 + y2/9 = 1
Standard form, x2/a2 + y2/b2 = 1
a2 = 4 and a = 2
b2 = 9 and b = 3
The area (A) of an ellipse with semi-major and semi-minor axes a and b is given by the formula,
Area = πab = π(2)(3) = 6π
The area of the region bounded by the ellipse is 6π square units.
34. Find the shortest distance between the lines whose vector equations are :
r = î + 2ĵ + 3k̂ + λ(î – 3ĵ + 2k̂)
and r = 4î + 5ĵ + 6k̂ + µ(2î + 3ĵ + k̂)
Answer : The given equations of the lines are in the form r = a1 + λb1 and r = a2 + µb2. From the given equations:
a1 = î + 2ĵ + 3k̂
b1 = î – 3ĵ + 2k̂
a2 = 4î + 5ĵ + 6k̂
b2 = 2î + 3ĵ + k̂
Calculate the vector connecting a point on the first line to a point on the second line:
a2 – a1 = (4î + 5ĵ + 6k̂) – (î + 2ĵ + 3k̂) = 3î + 3ĵ + 3k̂
. |î ĵ k̂|
b1 × b2 = |1 –3 2|
. |2 3 1|
b1 × b2 = î(–3 × 1 – 2 × 3) – ĵ(1 × 1 – 2 × 2) + k̂(1 × 3 – (–3) × 2)
b1 × b2 = –9î + 3ĵ + 9k̂
Magnitude of (b1 × b2):
|b1 × b2| = |–9î + 3ĵ + 9k̂| = √[(–9)2+(3)2+(9)2] = √81+9+81 = √171
|b1 × b2| = √171 = 3√19
Also,
(a2–a1).(b1×b2) = (3î + 3ĵ + 3k̂).(–9î + 3ĵ + 9k̂) = 3(–9) + 3(3) + 3(9) = –27 + 9 + 27 = 9
(a2–a1).(b1×b2) = 9
Shortest distance (d) = |(a2–a1).(b1×b2)| / |b1×b2| = |9 / (3√19)| = 3/√19
Therefore, the shortest distance between the two lines is 3/√19 units.
OR
Find the shortest distance between the lines whose vector equations are :
r = (1 – t)î + (t – 2)ĵ + (3 – 2t)k̂
and r = (s + 1)î + (2s – 1)ĵ – (2s + 1)k̂
Answer : The given equations of the lines are in the form r = a + λb are:
r1 = (1 – t)î + (t – 2)ĵ + (3 – 2t)k̂ = (î – 2ĵ + 3k̂) + t(–î + ĵ – 2k̂)
r2 = (s + 1)î + (2s – 1)ĵ – (2s + 1)k̂ = (î – ĵ – k̂) + s(î + 2ĵ – 2k̂)
Here,
a1 = î – 2ĵ + 3k̂
b1 = – î + ĵ – 2k̂
a2 = î – ĵ – k̂
b2 = î + 2ĵ – 2k̂
a2 – a1 = (î – ĵ – k̂) – (î – 2ĵ + 3k̂) = 0î + ĵ – 4k̂
. |î ĵ k̂|
b1 × b2 = |–1 1 –2|
. |1 2 –2|
b1 × b2 = î[1 × (–2) – 2 × (–2)] – ĵ[–1 × (–2) – 1 × (–2)] + k̂[–1 × 2 – 1 × 1]
b1 × b2 = 2î – 4ĵ – 3k̂
Magnitude of (b1 × b2):
|b1 × b2| = |2î – 4ĵ – 3k̂| = √[(2)2+(–4)2+(–3)2] = √4+16+9 = √29
|b1 × b2| = √29
Also,
(b1×b2).(a2–a1) = (2î – 4ĵ – 3k̂).(0î + ĵ – 4k̂) = 2(0) + (–4)(1) + (–3)(–4) = 0 – 4 + 12 = 8
(b1×b2).(a2–a1) = 8
Shortest distance (d) = |(b1×b2).(a2–a1)| / |b1×b2| = |8 / √29)| = 8/√29
Therefore, the shortest distance between the two lines is 8/√29 units.
35. Minimize and maximize z = 5x + 10y subject to constraints :
x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, y ≥ 0
Answer : The feasible region is determined by the constraints x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, y ≥ 0. The vertices of this region, found by determining the intersection points of the boundary lines, are (40, 20), (60, 30), (120, 0), and (60, 0).
The objective function z = 5x + 10y is evaluated at each vertex:
At (40, 20): z = 5(40) + 10(20) = 400
At (60, 30): z = 5(60) + 10(30) = 600
At (120, 0): z = 5(120) + 10(0) = 600
At (60, 0): z = 5(60) + 10(0) = 300
Minimum value of z is 300, which occurs at the point (60, 0).
Maximum value of z is 600, which occurs at all points along the line segment connecting the points (60, 30) and (120, 0).
36. Let a cone is inscribed in a sphere of radius R. The height and radius of cone are h and r respectively.
On the basis of above information, answer the following questions :
(i) Write the relation between r and R in terms of x.
Answer : Using the Pythagorean theorem, a right triangle is formed by the radius of the sphere (R), the radius of the cone (r), and the distance from the center of the sphere to the cone’s base (x).
R2 = r2 + x2
r2 = R2 – x2
(ii) Write the volume V of the cone in terms of R and x.
Answer : Height, h = R + x
V = 1/3 πr2h
V = 1/3 π(R2 – x2)(R + x)
V = π/3 (R3 + R2x – Rx2 – x3)
(iii) Show that volume V of the cone is maximum, when x = R/3.
Answer : V = = π/3 (R3 + R2x – Rx2 – x3)
dV/dt = π/3 (R2 – 2Rx – 3x2)
Set, dV/dt = 0
R² – 2Rx – 3x2 = 0
(R – 3x)(R + x) = 0
x = R/3 and x = – R
Since (x) is a physical distance, it must be positive, so we take x = R/3.
(The second derivative test confirms this is a maximum, as d2V/dx2 = π/3 (–2R–6x), which is negative at x = R/3).
OR
If volume V of the cone is maximum at x = R/3, then find the maximum value of V and find the ratio of volume of cone and volume of sphere, when volume of cone is maximum.
Answer : Substitute x = R/3 back into the volume equation from part (ii):
Vmax = π/3 (R2 – x2)(R + x) = π/3 [R2 – (R/3)2] [R + R/3] = π/3 (8R2/9)(4R/3)
Vmax = 32/81 πR3
Vsphere = 4/3 πR3
Vmax : Vsphere = 32/81 πR3 : 4/3 πR3 = 8 : 27
37. Polio drops are delivered to 50 K children in a district. The rate at which polio drops are given is directly proportional to the number of children who have not been administered the drops. By the end of 2nd week, half the children have been given the polio drops. How many will have been given the drops by the end of 3rd week can be estimated using the solution to the differential equation dy/dx = λ(50 – y), where x denotes the number of weeks and the y number of children who have been given the drops.
Based on the above information, answer the following questions :
(i) State the order of the above differential equation.
Answer : Order = 1 (first order)
(ii) Which method of solving a differential equation can be used to solve dy/dx = λ(50 – y) ?
Answer : Variable separation method
(iii) Solve the differential equation: dy/dx = λ(50 – y) ?
Answer : dy/dx = λ(50 – y)
∫ 1/(50–y) dy = ∫ λ dx
– ln |50 – y| = λx + C
ln |50 – y| = –(λx + C)
|50 – y| = e–(λx + C) = e–C e–λx
Let A = ± e–C be an arbitrary constant,
50 – y = Ae–λx
y = 50 – Ae–λx
The general solution is y = 50 – Ae–λx
OR
If λ = 0.049 and y(0) = 0, then find the particular solution of differential equation.
Answer : Take, y = 50 – Ae–λx
0 = 50 – Ae–0.049 x 0
0 = 50 – A(1)
A = 50
Substituting A = 50 and λ = 0.049 back into the general solution givens the particular solution,
y = 50 – 50e–0.049x
y = 50(1 – e–0.049x)
The particular solution is y = 50(1 – e–0.049x).
38. In a school, teacher asks a question to three students Ravi, Mohit and Sonia. The probability of solving the question by Ravi, Mohit and Sonia are 30%, 25% and 45% respectively. The probability of making errors by Ravi, Mohit and Sonia are 1%, 1.2% and 2% respectively.
Based on the above information, answer the following questions :
(i) Find the total probability of committing an error in solving the question.
Answer : Probability that Ravi solves the question = P(R) = 0.30
Probability that Mohit solves the question = P(M) = 0.25
Probability that Sonia solves the question = P(S) = 0.45
Probability of making an error while solving:
Ravi = ER = 0.01
Mohit = EM = 0.012
Sonia = ES = 0.02
By law of total probability:
P(Error) = P(R).ER + P(M).ЕM + P(S).ES P(Error) = 0.30(0.01) + 0.25(0.012) + 0.45(0.02) = 0.003 + 0.003 + 0.009 = 0.015
P(Error) = 0.015 or 1.5%
(ii) If the solution of question is checked by teacher and has some error, then find the probability that the question is not solved by Ravi.
Answer : We want, P(Not Ravi | Error)
P(Ravi and Error) = P(R).ER = 0.30 × 0.01 = 0.003
P(Error) = 0.015
P(R | Error) = P(Not Ravi | Error) / P(Error) = 0.003 / 0.015 = 0.20
P(Not Ravi | Error) = 1 – P(R | Error)
P(Not Ravi | Error) = 1 – 0.20 = 0.80
The probability that the question is not solved by Ravi is 0.80 or 80%.