CBSE Class 12 Maths 2024 Question Paper 65/1/2 with Solutions
All 45 questions from the CBSE Class 12 Mathematics board paper, Set 65/1/2 (2024),
with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.
Assertion (A) : A line in space cannot be drawn perpendicular to x, y and z axes simultaneously. Reason (R) : For any line making angles, α, β, γ with the positive directions of x, y and z axes respectively, cos2α+cos2β+cos2γ=1.
(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: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Reason is the standard identity for direction cosines, so it is true.
If a line were perpendicular to all three axes, then α=β=γ=90∘ and cos2α+cos2β+cos2γ=0=1.
So no such line exists: Assertion is true, and the Reason explains it.
A function f is defined from R→R as f(x)=ax+b, such that f(1)=1 and f(2)=3. Find function f(x). Hence, check whether function f(x) is one-one and onto or not.
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Answer:f(x)=2x−1; f is both one-one and onto.
f(1)=a+b=1 and f(2)=2a+b=3; subtracting, a=2, so b=−1.
Equations of sides of a parallelogram ABCD are as follows : AB : 1x+1=−2y−2=2z−1 BC : 3x−1=−5y+2=3z−5 CD : 1x−4=−2y+7=2z−8 DA : 3x−2=−5y+3=3z−4 Find the equation of diagonal BD.
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Answer:1x−1=−1y+2=−1z−5
B is the intersection of AB and BC. A point of AB is (λ−1,−2λ+2,2λ+1); a point of BC is (3μ+1,−5μ−2,3μ+5).
Equating x and y: λ=3μ+2 and −2λ+2=−5μ−2, giving μ=0, λ=2; z: 5=5. So B = (1, – 2, 5).
D is the intersection of CD and DA. A point of CD is (s+4,−2s−7,2s+8); a point of DA is (3t+2,−5t−3,3t+4).
Equating x and y: s=3t−2 and −2s−7=−5t−3, giving t=0, s=−2; z: 4=4. So D = (2, – 3, 4).
Direction ratios of BD: 2−1,−3+2,4−5 = 1, – 1, – 1.
According to recent research, air turbulence has increased in various regions around the world due to climate change. Turbulence makes flights bumpy and often delays the flights. Assume that, an airplane observes severe turbulence, moderate turbulence or light turbulence with equal probabilities. Further, the chance of an airplane reaching late to the destination are 55%, 37% and 17% due to severe, moderate and light turbulence respectively. On the basis of the above information, answer the following questions : (i) Find the probability that an airplane reached its destination late. (2) (ii) If the airplane reached its destination late, find the probability that it was due to moderate turbulence. (2)
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Answer: (i) 300109 (ii) 10937
Let E1,E2,E3 be severe, moderate and light turbulence, P(Ei)=31; let A be reaching late.
P(A∣E1)=0.55, P(A∣E2)=0.37, P(A∣E3)=0.17.
(i) P(A)=31(0.55+0.37+0.17)=31.09=300109.
(ii) By Bayes' theorem, P(E2∣A)=31×1.0931×0.37=10937.
If a function f:X→Y defined as f(x)=y is one-one and onto, then we can define a unique function g:Y→X such that g(y)=x, where x∈X and y=f(x), y∈Y. Function g is called the inverse of function f. The domain of sine function is R and function sine : R→R is neither one-one nor onto. The following graph shows the sine function. Let sine function be defined from set A to [– 1, 1] such that inverse of sine function exists, i.e., sin−1x is defined from [– 1, 1] to A. On the basis of the above information, answer the following questions : (i) If A is the interval other than principal value branch, give an example of one such interval. (1) (ii) If sin−1(x) is defined from [– 1, 1] to its principal value branch, find the value of sin−1(−21)−sin−1(1). (1) (iii) Draw the graph of sin−1x from [– 1, 1] to its principal value branch. (2) OR (iii) Find the domain and range of f(x)=2sin−1(1−x). (2)
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Answer: (i) e.g. [2π,23π] (ii) −32π (iii) graph of y=sin−1x from (−1,−2π) through O to (1,2π) OR (iii) domain [0, 2], range [−π,π]
(i) Sine is one-one and onto [– 1, 1] on [2π,23π] (or [−23π,−2π]), so A can be such an interval.
(ii) sin−1(−21)−sin−1(1)=−6π−2π=−32π.
(iii) Reflect the part of y=sinx on [−2π,2π] in the line y=x: an increasing curve from (−1,−2π) through (0, 0) to (1,2π), with domain [– 1, 1] and range [−2π,2π].
OR (iii) Need −1≤1−x≤1, i.e. 0≤x≤2: domain [0, 2].
sin−1(1−x)∈[−2π,2π], so f(x)∈[−π,π]: range [−π,π].
The traffic police has installed Over Speed Violation Detection (OSVD) system at various locations in a city. These cameras can capture a speeding vehicle from a distance of 300 m and even function in the dark. A camera is installed on a pole at the height of 5 m. It detects a car travelling away from the pole at the speed of 20 m/s. At any point, x m away from the base of the pole, the angle of elevation of the speed camera from the car C is θ. On the basis of the above information, answer the following questions : (i) Express θ in terms of height of the camera installed on the pole and x. (1) (ii) Find dxdθ. (1) (iii) Find the rate of change of angle of elevation with respect to time at an instant when the car is 50 m away from the pole. (2) OR (iii) If the rate of change of angle of elevation with respect to time of another car at a distance of 50 m from the base of the pole is 1013 rad/s, then find the speed of the car. (2)
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Answer: (i) θ=tan−1(x5) (ii) dxdθ=−x2+255 (iii) −1014 rad/s (angle decreasing at 1014 rad/s) OR (iii) 15 m/s
(i) tanθ=x5, so θ=tan−1(x5).
(ii) dxdθ=1+x2251⋅(−x25)=−x2+255.
(iii) dtdθ=dxdθ⋅dtdx=−x2+255×20.
At x=50: dtdθ=−2525100=−1014 rad/s.
OR (iii) Magnitude of dtdθ at x=50 is 25255v=505v.