CBSE Class 12 Maths 2025 Question Paper 65/1/1 with Solutions
All 45 questions from the CBSE Class 12 Mathematics board paper, Set 65/1/1 (2025),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
The corner points of the feasible region in graphical representation of a L.P.P. are (2, 72), (15, 20) and (40, 15). If Z = 18x+9y be the objective function, then
(A)Z is maximum at (2, 72), minimum at (15, 20)
(B)Z is maximum at (15, 20) minimum at (40, 15)
(C)Z is maximum at (40, 15), minimum at (15, 20)
(D)Z is maximum at (40, 15), minimum at (2, 72)
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Answer: (C) Z is maximum at (40, 15), minimum at (15, 20)
Assertion (A): Let Z be the set of integers. A function f : Z → Z defined as f(x) = 3x−5, ∀x∈ Z is a bijective. Reason (R): A function is a bijective if it is both surjective and injective.
(A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): f(x) = {3x−8,2k,x≤5x>5 is continuous at x=5 for k = 25. Reason (R): For a function f to be continuous at x = a, limx→a−f(x)=limx→a+f(x)=f(a).
(A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
limx→5−f(x)=f(5)=15−8=7; limx→5+f(x)=2k.
Continuity needs 2k=7, i.e. k=27, not 25. So A is false.
R is the correct condition for continuity, so R is true.
Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors a=3i^+j^+2k^ and b=2i^−2j^+4k^. Determine the angle formed between the kite strings. Assume there is no slack in the strings.
Verify that lines given by r=(1−λ)i^+(λ−2)j^+(3−2λ)k^ and r=(μ+1)i^+(2μ−1)j^−(2μ+1)k^ are skew lines. Hence, find shortest distance between the lines.
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Answer: Shortest distance =298 units (non-zero, so the lines are skew)
Line 1: a1=i^−2j^+3k^, b1=−i^+j^−2k^.
Line 2: a2=i^−j^−k^, b2=i^+2j^−2k^. The directions are not parallel.
During a cricket match, the position of the bowler, the wicket keeper and the leg slip fielder are in a line given by B=2i^+8j^, W=6i^+12j^ and F=12i^+18j^ respectively. Calculate the ratio in which the wicketkeeper divides the line segment joining the bowler and the leg slip fielder.
Q303 marksShort AnswerProbabilityNot in current syllabus
The probability distribution for the number of students being absent in a class on a Saturday is as follows : X: 0, 2, 4, 5 P(X): p, 2p, 3p, p Where X is the number of students absent. (i) Calculate p. (1) (ii) Calculate the mean of the number of absent students on Saturday. (2)
For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in written test is 0.4 and that a female getting a distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test.
Find a point P on the line 1x+5=4y+3=−9z−6 such that its distance from point Q(2, 4, −1) is 7 units. Also, find the equation of line joining P and Q.
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Answer: P(−4, 1, −3); line PQ: 6x−2=3y−4=2z+1
General point P(λ−5,4λ−3,6−9λ).
PQ2=(λ−7)2+(4λ−7)2+(7−9λ)2=49.
98λ2−196λ+147=49⇒λ2−2λ+1=0⇒λ=1.
P = (−4, 1, −3). Direction PQ = (6, 3, 2) (check: 36+9+4=7).
A school wants to allocate students into three clubs : Sports, Music and Drama, under following conditions : • The number of students in Sports club should be equal to the sum of the number of students in Music and Drama club. • The number of students in Music club should be 20 more than half the number of students in Sports club. • The total number of students to be allocated in all three clubs are 180. Find the number of students allocated to different clubs, using matrix method.
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Answer: Sports 90, Music 65, Drama 25
Let Sports = x, Music = y, Drama = z: x−y−z=0, −x+2y=40, x+y+z=180.
A = 1−11−121−101, X = xyz, B = 040180.
∣A∣=1(2)+1(−1)−1(−3)=4=0.
adj A = 21−302−2211.
X = A−1B=41 adj A ⋅ B =41360260100, so x=90, y=65, z=25.
A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections. Let the length of the side perpendicular to the partition be x metres and with parallel to the partition be y metres. Based on this information, answer the following questions : (i) Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of x and y. (1) (ii) Write the area of the solar panel as a function of x. (1) (iii) (a) Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area. (2) OR (iii) (b) Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition. (2)
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Answer: (i) 2x+3y=300 (ii) A(x)=100x−32x2 (iii) (a) critical point x=75; A′′<0 so maximum; maximum area 3750 m2 (y = 50) OR (iii) (b) maximum area 3750 m2
(i) Two sides of length x and three of length y (two sides plus the partition): 2x+3y=300.
(ii) y=3300−2x, so A=xy=100x−32x2.
(iii)(a) A′(x)=100−34x=0⇒x=75. A′′(x)=−34<0, so maximum at x=75, y = 50; max area =75×50=3750 m2.
(iii)(b) A′(x)=34(75−x) is positive for x<75 and negative for x>75, so A has a maximum at x=75; maximum area =3750 m2.
A class-room teacher is keen to assess the learning of her students the concept of “relations” taught to them. She writes the following five relations each defined on the set A = {1, 2, 3} : R1 = {(2, 3), (3, 2)} R2 = {(1, 2), (1, 3), (3, 2)} R3 = {(1, 2), (2, 1), (1, 1)} R4 = {(1, 1), (1, 2), (3, 3), (2, 2)} R5 = {(1, 1), (1, 2), (3, 3), (2, 2), (2, 1), (2, 3), (3, 2)} The students are asked to answer the following questions about the above relations : (i) Identify the relation which is reflexive, transitive but not symmetric. (ii) Identify the relation which is reflexive and symmetric but not transitive. (iii) (a) Identify the relations which are symmetric but neither reflexive nor transitive. OR (iii) (b) What pairs should be added to the relation R2 to make it an equivalence relation ?
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Answer: (i) R4 (ii) R5 (iii) (a) R1 and R3 OR (iii) (b) (1, 1), (2, 2), (3, 3), (2, 1), (3, 1), (2, 3)
R1: symmetric; not reflexive; not transitive ((2, 3), (3, 2) but (2, 2) missing).
R2: not reflexive, not symmetric; transitive ((1, 3), (3, 2) gives (1, 2), present).
R3: symmetric; not reflexive; not transitive ((2, 1), (1, 2) but (2, 2) missing).
R4: reflexive and transitive, not symmetric ((1, 2) in but (2, 1) not).
R5: reflexive and symmetric, not transitive ((1, 2), (2, 3) but (1, 3) missing).
(i) R4 (ii) R5 (iii)(a) R1 and R3.
(iii)(b) Reflexivity needs (1, 1), (2, 2), (3, 3); symmetry needs (2, 1), (3, 1), (2, 3). The result is A × A, which is transitive, so add these six pairs.
A bank offers loan to its customers on different types of interest namely, fixed rate, floating rate and variable rate. From the past data with the bank, it is known that a customer avails loan on fixed rate, floating rate or variable rate with probabilities 10%, 20% and 70% respectively. A customer after availing loan can pay the loan or default on loan repayment. The bank data suggests that the probability that a person defaults on loan after availing it at fixed rate, floating rate and variable rate is 5%, 3% and 1% respectively. Based on the above information, answer the following : (i) What is the probability that a customer after availing the loan will default on the loan repayment ? (2) (ii) A customer after availing the loan, defaults on loan repayment. What is the probability that he availed the loan at a variable rate of interest ? (2)