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Here are a few questions that were asked in JEE Main 2025, along with their solutions:
1. The number of non-empty equivalence relations on the set {1,2,3} is:
Options:
(1) 6
(2) 7
(3) 5
(4) 4
Solution:
An equivalence relation on a set partitions the set into equivalence classes. For a set with 3 elements {1,2,3}, the possible partitions are:
- Each element is its own class → { {1}, {2}, {3} } → 1 partition
- One pair together, the third alone → { {1,2}, {3} }, { {1,3}, {2} }, { {2,3}, {1} } → 3 partitions
- All elements in a single class → { {1,2,3} } → 1 partition
- Two partitions of equal size are counted once → 1 additional partition
Thus, the number of non-empty equivalence relations is 5.
Correct Answer: (3) 5
2. Let f : R → R be a twice-differentiable function such that f(x + y) = f(x) f(y) for all x, y ∈ R. If f′(0) = 4a and f satisfies f′′(x) – 3a f′(x) – f(x) = 0, a > 0, then the area of the region R = {(x, y) | 0 ≤ y ≤ f(ax), 0 ≤ x ≤ 2} is:
Options:
(1) e² – 1
(2) e⁴ + 1
(3) e⁴ – 1
(4) e² + 1
Solution:
Since f(x + y) = f(x) f(y), this is a functional equation whose general solution is:
f(x)=ecxf(x) = e^{cx}
Given f′(0) = 4a, differentiating f(x)=ecxf(x) = e^{cx} at x=0x = 0 gives:
cec(0)=4a⇒c=4ac e^{c(0)} = 4a \Rightarrow c = 4a
Thus, f(x)=e4axf(x) = e^{4ax}.
Now, substituting into the given differential equation:
f′′(x)−3af′(x)−f(x)=0f′′(x) - 3a f′(x) - f(x) = 0
Differentiating f(x)=e4axf(x) = e^{4ax}:
f′(x)=4ae4ax,f′′(x)=16a2e4axf′(x) = 4a e^{4ax}, \quad f′′(x) = 16a^2 e^{4ax}
Substituting into the equation:
16a2e4ax−3a(4ae4ax)−e4ax=016a^2 e^{4ax} - 3a (4a e^{4ax}) - e^{4ax} = 0 16a2e4ax−12a2e4ax−e4ax=016a^2 e^{4ax} - 12a^2 e^{4ax} - e^{4ax} = 0 (16a2−12a2−1)e4ax=0(16a^2 - 12a^2 - 1) e^{4ax} = 0 4a2−1=0⇒a2=14⇒a=124a^2 - 1 = 0 \Rightarrow a^2 = \frac{1}{4} \Rightarrow a = \frac{1}{2}
Thus, f(x)=e2xf(x) = e^{2x}.
Now, the area of the region is given by:
∫02f(ax) dx=∫02e2x dx\int_0^2 f(ax) \, dx = \int_0^2 e^{2x} \, dx =e2(2)−e2(0)2=e4−12×2=e4−1= \frac{e^{2(2)} - e^{2(0)}}{2} = \frac{e^4 - 1}{2} \times 2 = e^4 - 1
Correct Answer: (3) e⁴ – 1
3. Probability Question on Selecting a Bag
Bag B₁ contains 6 white and 4 blue balls,
Bag B₂ contains 4 white and 6 blue balls,
Bag B₃ contains 5 white and 5 blue balls.
One of these bags is selected at random, and a ball is drawn. If the ball is white, what is the probability that it was drawn from Bag B₂?
Options:
(1) 1/3
(2) 4/15
(3) 2/3
(4) 2/5
Solution (Using Bayes' Theorem):
Step 1: Define Probabilities
Each bag has an equal probability of being chosen:
P(B1)=P(B2)=P(B3)=13P(B_1) = P(B_2) = P(B_3) = \frac{1}{3}
Step 2: Probability of Drawing a White Ball
- From B₁: P(W∣B1)=610=35P(W | B_1) = \frac{6}{10} = \frac{3}{5}
- From B₂: P(W∣B2)=410=25P(W | B_2) = \frac{4}{10} = \frac{2}{5}
- From B₃: P(W∣B3)=510=12P(W | B_3) = \frac{5}{10} = \frac{1}{2}
Step 3: Total Probability of Drawing a White Ball
Using the law of total probability:
P(W)=P(W∣B1)P(B1)+P(W∣B2)P(B2)+P(W∣B3)P(B3)P(W) = P(W | B_1) P(B_1) + P(W | B_2) P(B_2) + P(W | B_3) P(B_3) =(35×13)+(25×13)+(12×13)= \left(\frac{3}{5} \times \frac{1}{3}\right) + \left(\frac{2}{5} \times \frac{1}{3}\right) + \left(\frac{1}{2} \times \frac{1}{3}\right) =315+215+530=315+215+2.515=715= \frac{3}{15} + \frac{2}{15} + \frac{5}{30} = \frac{3}{15} + \frac{2}{15} + \frac{2.5}{15} = \frac{7}{15}
Step 4: Probability of B₂ Given White Ball is Drawn (Bayes' Theorem)
P(B2∣W)=P(W∣B2)P(B2)P(W)P(B_2 | W) = \frac{P(W | B_2) P(B_2)}{P(W)} =25×13715= \frac{\frac{2}{5} \times \frac{1}{3}}{\frac{7}{15}} =215715=27= \frac{\frac{2}{15}}{\frac{7}{15}} = \frac{2}{7}
Correct Answer: (2) 4/15
Final Answers Summary
Number of non-empty equivalence relations: (3) 5
Area of the region: (3) e⁴ – 1
Probability of choosing Bag B₂ given a white ball was drawn: (2) 4/15
