CBSE Class 12 Maths 2023 Question Paper 65/4/2 with Solutions
All 45 questions from the CBSE Class 12 Mathematics board paper, Set 65/4/2 (2023),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
Assertion (A) : The lines r=a1+λb1 and r=a2+μb2 are perpendicular, when b1⋅b2=0. Reason (R) : The angle θ between the lines r=a1+λb1 and r=a2+μb2 is given by cosθ=∣b1∣∣b2∣b1⋅b2
(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 and Reason (R) is false.
(D)Assertion (A) is false and Reason (R) is true.
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Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
The angle between the lines is the angle between b1 and b2, with cosθ=∣b1∣∣b2∣b1⋅b2. R is true.
If b1⋅b2=0 then cosθ=0, θ=2π, so the lines are perpendicular. A is true and follows from R.
Solve the following linear programming problem graphically : Maximise z=−3x−5y subject to the constraints −2x+y≤4, x+y≥3, x−2y≤2, x≥0, y≥0.
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Answer: Maximum z=−329 at (38,31)
Boundary lines: −2x+y=4 through (0, 4), (− 2, 0); x+y=3 through (3, 0), (0, 3); x−2y=2 through (2, 0), (0, − 1).
The feasible region lies on or above x+y=3, between y=2x+4 and y=2x−2, with x≥0; it is unbounded, with corner points A(0, 3), B(0, 4) and C(38,31) (where x+y=3 meets x−2y=2).
z at corners: A: − 15; B: − 20; C: −8−35=−329.
As the region is unbounded, check −3x−5y>−329, i.e. 3x+5y<329: this open half plane has no point in common with the feasible region (since 3x+5y=311(x+y)−32(x−2y)≥11−34=329 there).
If N denotes the set of all natural numbers and R is the relation on N×N defined by (a, b) R (c, d), if ad(b+c)=bc(a+d). Show that R is an equivalence relation.
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Answer: Proved.
Since all numbers are natural, divide by abcd: ad(b+c)=bc(a+d)⟺c1+b1=d1+a1⟺b1−a1=d1−c1.
Reflexive: ab(b+a)=ba(a+b), so (a, b) R (a, b).
Symmetric: if (a, b) R (c, d) then b1−a1=d1−c1, which read the other way gives (c, d) R (a, b).
Transitive: if (a, b) R (c, d) and (c, d) R (e, f), then b1−a1=d1−c1=f1−e1, so (a, b) R (e, f).
A building contractor undertakes a job to construct 4 flats on a plot along with parking area. Due to strike the probability of many construction workers not being present for the job is 0.65. The probability that many are not present and still the work gets completed on time is 0.35. The probability that work will be completed on time when all workers are present is 0.80. Let : E1 : represent the event when many workers were not present for the job; E2 : represent the event when all workers were present; and E : represent completing the construction work on time. Based on the above information, answer the following questions : (i) What is the probability that all the workers are present for the job ? (1) (ii) What is the probability that construction will be completed on time ? (1) (iii) What is the probability that many workers are not present given that the construction work is completed on time ? (2) OR (iii) What is the probability that all workers were present given that the construction job was completed on time ? (2)
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Answer: (i) 0.35 (ii) 0.5075 (iii) 2913; OR 2916
Let f(x) be a real valued function. Then its Left Hand Derivative (L.H.D.) : Lf′(a)=limh→0−hf(a−h)−f(a) Right Hand Derivative (R.H.D.) : Rf′(a)=limh→0hf(a+h)−f(a) Also, a function f(x) is said to be differentiable at x = a if its L.H.D. and R.H.D. at x = a exist and both are equal. For the function f(x)={∣x−3∣,4x2−23x+413,x≥1x<1 answer the following questions : (i) What is R.H.D. of f(x) at x = 1 ? (1) (ii) What is L.H.D. of f(x) at x = 1 ? (1) (iii) Check if the function f(x) is differentiable at x = 1. (2) OR (iii) Find f′(2) and f′(−1). (2)
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Answer: (i) − 1 (ii) − 1 (iii) Yes, f is differentiable at x = 1 (f′(1)=−1); OR f′(2)=−1, f′(−1)=−2
f(1)=∣1−3∣=2. For x near 1 with x≥1, f(x)=3−x.
(i) R.H.D. =limh→0h(3−1−h)−2=limh→0h−h=−1.
(ii) f(1−h)=4(1−h)2−23(1−h)+413=2+h+4h2, so L.H.D. =limh→0−hh+4h2=−1.
(iii) L.H.D. = R.H.D. = − 1, so f is differentiable at x = 1 with f′(1)=−1.
OR (iii) For 1≤x<3, f(x)=3−x, so f′(2)=−1. For x<1, f′(x)=2x−23, so f′(−1)=−21−23=−2.
Sooraj’s father wants to construct a rectangular garden using a brick wall on one side of the garden and wire fencing for the other three sides as shown in the figure. He has 200 metres of fencing wire. Based on the above information, answer the following questions : (i) Let ‘x’ metres denote the length of the side of the garden perpendicular to the brick wall and ‘y’ metres denote the length of the side parallel to the brick wall. Determine the relation representing the total length of fencing wire and also write A(x), the area of the garden. (2) (ii) Determine the maximum value of A(x). (2)
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Answer: (i) 2x+y=200; A(x)=x(200−2x)=200x−2x2 (ii) 5000 m²
(i) Fencing covers the two sides perpendicular to the wall and the side opposite the wall: 2x+y=200.
Area A=xy=x(200−2x)=200x−2x2.
(ii) A′(x)=200−4x=0 gives x=50; A′′(x)=−4<0, so this is a maximum.