CBSE Class 12 Maths 2026 Question Paper 65/4/1 with Solutions
All 45 questions from the CBSE Class 12 Mathematics board paper, Set 65/4/1 (2026),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
In the graph, the feasible region representing the Linear Programming Problem for maximising objective function Z=px+qy, p,q>0 is shaded. If all points on segment AB give max (Z), then which of the following is true ?
(A)p=2q
(B)p=3q
(C)q=3p
(D)q=2p
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Answer: (C) q=3p
At A(0, 5): Z=5q. At B(3, 4): Z=3p+4q.
Maximum at every point of AB means Z(A)=Z(B): 5q=3p+4q.
A box contains 4 red, 5 blue and 1 green marble. A child randomly takes out a marble from the box, notes down the colour and puts it back in the box. If the activity is repeated 3 times, what is the probability that at least one marble is red ?
(A)12527
(B)1258
(C)1252
(D)12598
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Answer: (D) 12598
P(red in one draw) =104=52, so P(not red) =53.
Draws are independent (with replacement): P(no red in 3 draws) =(53)3=12527.
Assertion (A) : If A and B are two square matrices such that AB and BA are defined, then it is not necessary that AB = BA. Reason (R) : Product of two diagonal matrices of same order is commutative.
(A)Both Assertion (A) and Reason (R) are true and 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: (B) Both A and R are true, but R is not the correct explanation of A.
Matrix multiplication is not commutative in general, e.g. A=[0010], B=[1000] give AB=O=BA. So A is true.
For diagonal matrices the product is diagonal with entries aibi=biai, so R is true.
R is about a special case and does not explain why AB need not equal BA in general.
Assertion (A) : A function f:N→N given by f(x)=x3+2,∀x∈N is one-one but not onto. Reason (R) : Since ∀y∈N (Codomain), there does not exist x=(y−2)1/3 in N (Domain) such that f(x)=x3+2=y.
(A)Both Assertion (A) and Reason (R) are true and 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: (C) Assertion (A) is true, but Reason (R) is false.
x13+2=x23+2⇒x1=x2, so f is one-one.
y=1 (or 2) has no pre-image in N, so f is not onto. A is true.
R says that for every y∈N no such x exists, which is false: for y=3, x=1∈N.
Let two rods placed on the ground be represented by vectors 4i^−j^+3k^ and −2i^+j^−2k^. Find a vector representing a flag-post of height 5 m that has to be erected perpendicular to both the rods.
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Answer:±35(−i^+2j^+2k^)
a×b=i^4−2j^−11k^3−2=−i^+2j^+2k^.
Its magnitude is 1+4+4=3, so a unit vector perpendicular to both is 31(−i^+2j^+2k^).
Required vector of length 5: ±35(−i^+2j^+2k^) (the upward one is taken for the flag-post).
Let n be a fixed positive integer. A relation R is defined in set Z such that R={(x,y):(x−y) is divisible by n,x,y∈Z}. Determine if R is an equivalence relation.
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Answer: Yes, R is an equivalence relation.
Reflexive: x−x=0=0⋅n, so (x,x)∈R for all x∈Z.
Symmetric: if x−y=kn then y−x=(−k)n, so (y,x)∈R.
Transitive: if x−y=kn and y−z=mn then x−z=(k+m)n, so (x,z)∈R.
Solve the following Linear Programming Problem graphically : Maximise Z=200x+120y subject to the constraints x+y≤300 3x+y≤600 x−y≥−100 x,y≥0
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Answer: Maximum Z = 48000 at x = 150, y = 150
Draw x+y=300, 3x+y=600 and x−y=−100 in the first quadrant; the feasible region is the bounded region on the origin side of the first two lines and below y=x+100.
In a school, the probability of holding a debate competition is 31 and that of a quiz competition is 32. In the two participating teams, A has 4 girls and 6 boys and B has 7 girls and 3 boys. If a debate competition is held, the students are selected from team A and for the quiz competition they are selected from team B. If only two students are to be chosen from the teams, then find the probability that one will be a girl and the other a boy.
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Answer:4522
Let E1: debate (team A), E2: quiz (team B); P(E1)=31, P(E2)=32. Let G: one girl and one boy chosen.
Find the equation of a line (in vector and cartesian form) that passes through the point of intersection of lines 2x−1=3y−2=4z−3 and 5x−4=2y−1=z and is parallel to the vector 3i^+2j^−8k^.
At a birthday party, children are being served orange juice in conical cups, as shown in the figure. Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0.1 cm3/s. On the basis of the above information, answer the following questions : (i) Establish a relation between the height h of the juice in the cup and radius r of the surface of the juice in the cup, if the semi-vertical angle of the cone is α. (1) (ii) At what rate is the juice level in the cup rising when the juice is 6 cm deep ? (1) (iii) When the juice is 6 cm deep, then find at what rate is the upper surface area of juice increasing ? (2) OR (iii) When the juice is 6 cm deep, then find the rate at which the wetted surface area of the cup is increasing. (2)
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Answer: (i) r=htanα, here tanα=155 so r=3h (ii) 40π1 cm/s (iii) 301 cm2/s; OR 3010 cm2/s
(i) In the right triangle formed by the axis, a radius of the juice surface and the slant side, tanα=hr, so r=htanα. For the cup tanα=155=31, so r=3h.
(ii) V=31πr2h=27πh3, so dtdV=9πh2dtdh.
At h=6: 0.1=4πdtdh⇒dtdh=40π1 cm/s.
(iii) Upper surface area A=πr2=9πh2; dtdA=92πhdtdh=912π⋅40π1=301 cm2/s.
OR (iii) Wetted surface S=πrl with l=h2+r2=310h, so S=910πh2.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length. On the basis of the above information, answer the following questions : (i) Write the equations representing the various dimensions and express them as the matrix equation AX = B. (1) (ii) Find if A−1 exists. Justify your answer. (1) (iii) Find A−1. (2) OR (iii) Find A2+7I. (2)
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Answer: (i) l+b−h=3, 2l+3b+h=10, 3l−b−7h=1; 12313−1−11−7lbh=3101 (ii) Yes, ∣A∣=8=0 (iii) A−1=81−2017−118−444−31; OR A2+7I=711−2051777−652
(i) Let length l, breadth b, height h (cm). l+b=h+3, 2l+3b+h=10, b+7h=3l−1.
So l+b−h=3, 2l+3b+h=10, 3l−b−7h=1: A=12313−1−11−7, X=lbh, B=3101.
(ii) ∣A∣=1(−21+1)−1(−14−3)−1(−2−9)=−20+17+11=8=0, so A is non-singular and A−1 exists.
An NGO organises a charity event in which they decide to distribute woollen caps to protect children from winter. The caps to be distributed are in three separate boxes, Box I has 30 red caps, Box II has 20 red and 10 green caps, and Box III has 30 green caps. The probability that a Box i is selected and a cap picked out is 6i, where i = 1, 2, 3. Based on the above information, answer the following questions : A person selects a cap. (i) What is the probability that he selects a red cap ? (2) (ii) If he selects a green cap, what is the probability that the cap has come from Box II ? (2)
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Answer: (i) 187 (ii) 112
Let Ei: Box i is selected; P(E1)=61, P(E2)=62, P(E3)=63.