CBSE Class 12 Maths 2023 Question Paper 65/5/1 with Solutions
All 45 questions from the CBSE Class 12 Mathematics board paper, Set 65/5/1 (2023),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
Assertion (A) : Two coins are tossed simultaneously. The probability of getting two heads, if it is known that at least one head comes up, is 31. Reason (R) : Let E and F be two events with a random experiment, then P(F/E)=P(E)P(E∩F).
(A)Both (A) and (R) are true and (R) is the correct explanation of (A).
(B)Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C)(A) is true and (R) is false.
(D)(A) is false, but (R) is true.
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Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Sample space {HH, HT, TH, TT}. Let E = at least one head = {HH, HT, TH}, F = two heads = {HH}.
P(F/E)=P(E)P(E∩F)=3/41/4=31. So (A) is true.
(R) is the definition of conditional probability (with P(E)=0), so it is true and it is exactly what gives (A).
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 53 be the probability that he knows the answer and 52 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 31. What is the probability that the student knows the answer, given that he answered it correctly ?
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Answer:119
Let E1: student knows the answer, E2: student guesses, A: answer is correct.
P(E1)=53, P(E2)=52, P(A∣E1)=1, P(A∣E2)=31.
By Bayes' theorem, P(E1∣A)=53×1+52×3153×1=151153.
Q35 (OR) (OR)5 marksLong AnswerProbabilityNot in current syllabus
A box contains 10 tickets, 2 of which carry a prize of ₹ 8 each, 5 of which carry a prize of ₹ 4 each, and remaining 3 carry a prize of ₹ 2 each. If one ticket is drawn at random, find the mean value of the prize.
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Answer: ₹ 4.20
Let X be the prize. X takes values 8, 4, 2 with probabilities 102, 105, 103.
An organization conducted bike race under two different categories – Boys and Girls. There were 28 participants in all. Among all of them, finally three from category 1 and two from category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his college project. Let B={b1,b2,b3} and G={g1,g2}, where B represents the set of Boys selected and G the set of Girls selected for the final race. Based on the above information, answer the following questions : (I) How many relations are possible from B to G ? (1) (II) Among all the possible relations from B to G, how many functions can be formed from B to G ? (1) (III) Let R : B → B be defined by R = {(x, y) : x and y are students of the same sex}. Check if R is an equivalence relation. (2) OR (III) A function f : B → G be defined by f={(b1,g1),(b2,g2),(b3,g1)}. Check if f is bijective. Justify your answer. (2)
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Answer: (I) 64 (II) 8 (III) Yes, R is an equivalence relation. OR (III) f is not bijective (onto but not one-one).
(I) n(B×G)=3×2=6, so the number of relations from B to G is 26=64.
(II) Each of the 3 elements of B can be mapped to either of the 2 elements of G: 23=8 functions.
(III) All elements of B are boys, so every pair of B is of the same sex and R=B×B.
R is reflexive ((x,x)∈R), symmetric and transitive (every pair is in R), so R is an equivalence relation.
OR (III) f(b1)=f(b3)=g1 with b1=b3, so f is not one-one.
Range of f = {g1,g2}=G, so f is onto. Since f is not one-one, f is not bijective.
Gautam buys 5 pens, 3 bags and 1 instrument box and pays a sum of ₹ 160. From the same shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a sum of ₹ 190. Also Ankur buys 1 pen, 2 bags and 4 instrument boxes and pays a sum of ₹ 250. Based on the above information, answer the following questions : (I) Convert the given above situation into a matrix equation of the form AX = B. (1) (II) Find | A |. (1) (III) Find A−1. (2) OR (III) Determine P=A2−5A. (2)
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Answer: (I) 521312134xyz=160190250 (II) ∣A∣=−22 (III) A−1=22125−310−197−8131 OR (III) P=7585831323
(I) Let the prices of a pen, a bag and an instrument box be ₹ x, ₹ y and ₹ z. Then 5x+3y+z=160, 2x+y+3z=190, x+2y+4z=250.
An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form dxdy=F(x,y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogenous function of degree n if F(λx,λy)=λnF(x,y). To solve a homogeneous differential equation of the type dxdy=F(x,y)=g(xy), we make the substitution y=vx and then separate the variables. Based on the above, answer the following questions : (I) Show that (x2−y2)dx+2xydy=0 is a differential equation of the type dxdy=g(xy). (2) (II) Solve the above equation to find its general solution. (2)
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Answer: (I) Proved. (II) x2+y2=Cx
(I) dxdy=2xyy2−x2. Dividing numerator and denominator by x2: dxdy=2(xy)(xy)2−1=g(xy).
(II) Put y=vx, dxdy=v+xdxdv: v+xdxdv=2vv2−1.
xdxdv=2vv2−1−2v2=−2v1+v2, so 1+v22vdv=−xdx.
Integrating: log(1+v2)=−log∣x∣+logC, so x(1+v2)=C.