Qus : 1 NIMCET PYQ 2024 4 If (4, 3) and (12, 5) are the two foci of an ellipse passing through the
origin, then the eccentricity of the ellipse is
1 √ 13 9 2 √ 13 18 3 √ 17 18 4 √ 17 9 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given: Foci are (4, 3) and (12, 5), and the ellipse passes through the origin (0, 0).
Step 1: Use ellipse definition
$PF_1 = \sqrt{(0 - 4)^2 + (0 - 3)^2}
= \sqrt{25}
= 5$
$PF_2 = \sqrt{(0 - 12)^2 + (0 - 5)^2}
= \sqrt{169}
= 13$
Total distance = 5 + 13 = 18 ⇒ 2 a = 18 ⇒ a = 9
Step 2: Distance between the foci
2 c = √ ( 12 − 4 ) 2 + ( 5 − 3 ) 2 = √ 64 + 4 = √ 68 ⇒ c = √ 17
Step 3: Find eccentricity
e = c a = √ 17 9
✅ Final Answer: √ 17 9
Qus : 2 NIMCET PYQ 2024 2 The number of one - one functions
f: {1,2,3} → {a,b,c,d,e} is
1 125 2 60 3 243 4 None of these Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given: A one-one function from set { 1 , 2 , 3 } to set { a , b , c , d , e }
Step 1: One-one (injective) function means no two elements map to the same output.
We choose 3 different elements from 5 and assign them to 3 inputs in order.
So, total one-one functions = P ( 5 , 3 ) = 5 × 4 × 3 = 60
✅ Final Answer: 60
Qus : 4 NIMCET PYQ 2024 4 The value of m for which volume of the parallelepiped is 4 cubic units whose three edges are represented by a = mi + j + k, b = i – j + k, c = i + 2j –k is
1 0 2 -2 3 -1 4 1 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given: Volume of a parallelepiped formed by vectors \vec{a}, \vec{b}, \vec{c} is 4 cubic units.
Vectors:
\vec{a} = m\hat{i} + \hat{j} + \hat{k}
\vec{b} = \hat{i} - \hat{j} + \hat{k}
\vec{c} = \hat{i} + 2\hat{j} - \hat{k}
Step 1: Volume = |\vec{a} \cdot (\vec{b} \times \vec{c})|
First compute \vec{b} \times \vec{c} :
\vec{b} \times \vec{c} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 1 \\
1 & 2 & -1
\end{vmatrix}
= \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(2) - (-1)(1)) \\
= \hat{i}(1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 + 1) = -\hat{i} + 2\hat{j} + 3\hat{k}
Step 2: Compute dot product with \vec{a} :
\vec{a} \cdot (\vec{b} \times \vec{c}) = (m)(-1) + (1)(2) + (1)(3) = -m + 2 + 3 = -m + 5
Step 3: Volume = | -m + 5 | = 4
So, |-m + 5| = 4 \Rightarrow -m + 5 = \pm 4
Case 1: -m + 5 = 4 \Rightarrow m = 1
Case 2: -m + 5 = -4 \Rightarrow m = 9
✅ Final Answer: \boxed{m = 1 \text{ or } 9}
Qus : 5 NIMCET PYQ 2024 2 The number of distinct real values of \lambda for which the vectors {\lambda}^2\hat{i}+\hat{j}+\hat{k},\, \hat{i}+{\lambda}^2\hat{j}+j and \hat{i}+\hat{j}+{\lambda}^2\hat{k} are coplanar is
1 1 2 2 3 3 4 6 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given: Vectors:
\vec{a} = \lambda^2 \hat{i} + \hat{j} + \hat{k}
\vec{b} = \hat{i} + \lambda^2 \hat{j} + \hat{k}
\vec{c} = \hat{i} + \hat{j} + \lambda^2 \hat{k}
Condition: Vectors are coplanar ⟹ Scalar triple product = 0
\vec{a} \cdot (\vec{b} \times \vec{c}) = 0
Step 1: Use determinant:
\vec{a} \cdot (\vec{b} \times \vec{c}) =
\begin{vmatrix}
\lambda^2 & 1 & 1 \\
1 & \lambda^2 & 1 \\
1 & 1 & \lambda^2
\end{vmatrix}
Step 2: Expand the determinant:
= \lambda^2(\lambda^2 \cdot \lambda^2 - 1 \cdot 1) - 1(1 \cdot \lambda^2 - 1 \cdot 1) + 1(1 \cdot 1 - \lambda^2 \cdot 1) \\
= \lambda^2(\lambda^4 - 1) - (\lambda^2 - 1) + (1 - \lambda^2)
Simplify:
= \lambda^6 - \lambda^2 - \lambda^2 + 1 + 1 - \lambda^2 = \lambda^6 - 3\lambda^2 + 2
Step 3: Set scalar triple product to 0:
\lambda^6 - 3\lambda^2 + 2 = 0
Step 4: Let x = \lambda^2 , then:
x^3 - 3x + 2 = 0
Factor:
x^3 - 3x + 2 = (x - 1)^2(x + 2)
So, \lambda^2 = 1 (double root), or \lambda^2 = -2 (discard as it's not real)
Thus, real values of \lambda are: \lambda = \pm1
✅ Final Answer: \boxed{2} distinct real values
Qus : 6 NIMCET PYQ 2024 1 There are 9 bottle labelled 1, 2, 3, ... , 9 and 9 boxes labelled 1, 2, 3,....9. The number of ways one can put these bottles in the boxes so that each box gets one bottle and exactly 5 bottles go in their
corresponding numbered boxes is
1 9\times{}^9{{C}}_5 2 5\times{}^9{{C}}_5 3 25\times{}^9{{C}}_5 4 4\times{}^9{{C}}_5 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Total bottles and boxes: 9 each, labeled 1 to 9.
We are asked to count permutations of bottles such that exactly 5 bottles go into their own numbered boxes.
Step 1: Choose 5 positions to be fixed points (i.e., bottle number matches box number).
Number of ways = \binom{9}{5}
Step 2: Remaining 4 positions must be a derangement (no bottle goes into its matching box).
Let D_4 be the number of derangements of 4 items.
D_4 = 9
Step 3: Total ways = \binom{9}{5} \times D_4 = 126 \times 9 = 1134
✅ Final Answer: \boxed{1134}
Qus : 7 NIMCET PYQ 2024 4 If the perpendicular bisector of the line segment joining p(1,4) and q(k,3) has yintercept -4, then the possible values of k are
1 -3 and 3 2 -1 and 1
3 -2 and 2 4 -4 and 4
Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given: Points: P(1, 4) , Q(k, 3)
Step 1: Find midpoint of PQ
Midpoint = \left( \dfrac{1 + k}{2}, \dfrac{4 + 3}{2} \right) = \left( \dfrac{1 + k}{2}, \dfrac{7}{2} \right)
Step 2: Find slope of PQ
Slope of PQ = \dfrac{3 - 4}{k - 1} = \dfrac{-1}{k - 1}
Step 3: Slope of perpendicular bisector = negative reciprocal = k - 1
Step 4: Use point-slope form for perpendicular bisector:
y - \dfrac{7}{2} = (k - 1)\left(x - \dfrac{1 + k}{2}\right)
Step 5: Find y-intercept (put x = 0 )
y = \dfrac{7}{2} + (k - 1)\left( -\dfrac{1 + k}{2} \right)
y = \dfrac{7}{2} - (k - 1)\left( \dfrac{1 + k}{2} \right)
Given: y-intercept = -4, so:
\dfrac{7}{2} - \dfrac{(k - 1)(k + 1)}{2} = -4
Multiply both sides by 2:
7 - (k^2 - 1) = -8 \Rightarrow 7 - k^2 + 1 = -8 \Rightarrow 8 - k^2 = -8
\Rightarrow k^2 = 16 \Rightarrow k = \pm4
✅ Final Answer: \boxed{k = -4 \text{ or } 4}
Qus : 12 NIMCET PYQ 2024 2 Which of the following is TRUE?
A. If f is continuous on [a,b] , then \int ^b_axf(x)\mathrm{d}x=x\int ^b_af(x)\mathrm{d}x
B. \int ^3_0{e}^{{x}^2}dx=\int ^5_0e^{{x}^2}dx+{\int ^5_3e}^{{x}^2}dx
C. If f is continuous on [a,b] , then \frac{d}{\mathrm{d}x}\Bigg{(}\int ^b_af(x)dx\Bigg{)}=f(x)
D. Both (a) and (b)
1 A 2 B 3 C 4 D Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ
Solution Qus : 13 NIMCET PYQ 2024 4 If F|= 40N (Newtons), |D| = 3m, and \theta={60^{\circ}} , then the work done by F acting
from P to Q is
1 60\sqrt{3} J 2 120 J 3 60\sqrt{2} J 4 60J Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Formula for work done:
W = |F| \cdot |D| \cdot \cos\theta
Given:
|F| = 40 \, \text{N}
|D| = 3 \, \text{m}
\theta = 60^\circ
Step 1: Plug in the values:
W = 40 \cdot 3 \cdot \cos(60^\circ)
Step 2: Use \cos(60^\circ) = \frac{1}{2}
W = 40 \cdot 3 \cdot \frac{1}{2} = 60 \, \text{J}
✅ Final Answer: \boxed{60 \, \text{J}}
Qus : 14 NIMCET PYQ 2024 2 A committee of 5 is to be chosen from a group of 9 people. The probability that a certain married couple will either serve together or not at all is
1 2/3 2 4/9 3 1/2 4 5/9 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Total people: 9
Married couple: 2 specific people among them
Total ways to choose 5 people from 9:
\text{Total} = \binom{9}{5} = 126
✅ Case 1: Both are selected
We fix the married couple (2 people), then choose 3 more from remaining 7:
\binom{7}{3} = 35
✅ Case 2: Both are NOT selected
We remove both from the pool, then choose 5 from remaining 7:
\binom{7}{5} = \binom{7}{2} = 21
✅ Favorable outcomes:
\text{Favorable} = 35 + 21 = 56
✅ Probability:
\text{Required Probability} = \frac{56}{126} = \frac{28}{63} = \frac{4}{9}
✅ Final Answer: \boxed{\dfrac{4}{9}}
Qus : 15 NIMCET PYQ 2024 2 Find the cardinality of the set C which is defined as C={\{x|\, \sin 4x=\frac{1}{2}\, forx\in(-9\pi,3\pi)}\} .
1 24 2 48 3 36 4 12 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
We are given:
\sin(4x) = \frac{1}{2}, \quad x \in (-9\pi,\ 3\pi)
Step 1: General solutions for \sin(θ) = \frac{1}{2}
θ = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad θ = \frac{5\pi}{6} + 2n\pi
Let θ = 4x , so we get:
x = \frac{\pi}{24} + \frac{n\pi}{2}
x = \frac{5\pi}{24} + \frac{n\pi}{2}
✅ Step 2: Count how many such x fall in the interval (-9\pi, 3\pi)
By checking all possible n values, we find:
For x = \frac{\pi}{24} + \frac{n\pi}{2} : 24 valid values
For x = \frac{5\pi}{24} + \frac{n\pi}{2} : 24 valid values
? Total distinct values = 24 + 24 = 48
✅ Final Answer: \boxed{48}
Qus : 18 NIMCET PYQ 2024 2 A critical orthopedic surgery is performed on 3 patients. The probability of recovering
a patient is 0.6. Then the probability that after surgery, exactly two of them will recover
is
1 0.123 2 0.432 3 0.321 4 0.234 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given:
Number of patients = 3
Probability of recovery p = 0.6
Probability of failure q = 1 - p = 0.4
We want: Probability that exactly 2 recover out of 3.
? Use Binomial Probability Formula:
P(X = r) = \binom{n}{r} p^r (1 - p)^{n - r}
where n = 3, r = 2, p = 0.6
? Calculation:
P(X = 2) = \binom{3}{2} (0.6)^2 (0.4)^1 = 3 \times 0.36 \times 0.4 = 0.432
✅ Final Answer: \boxed{0.432}
Qus : 19 NIMCET PYQ 2024 4 The value of \tan \Bigg{(}\frac{\pi}{4}+\theta\Bigg{)}\tan \Bigg{(}\frac{3\pi}{4}+\theta\Bigg{)} is
1 -2 2 2 3 1 4 -1 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
We are given:
\text{Evaluate } \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{3\pi}{4} + \theta\right)
✳ Step 1: Use identity
\tan\left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
But we don’t need expansion — use known angle values:
\tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta}
\tan\left(\frac{3\pi}{4} + \theta\right) = \frac{-1 + \tan\theta}{1 + \tan\theta}
✳ Step 2: Multiply
\left(\frac{1 + \tan\theta}{1 - \tan\theta}\right) \cdot \left(\frac{-1 + \tan\theta}{1 + \tan\theta}\right)
Simplify:
= \frac{(1 + \tan\theta)(-1 + \tan\theta)}{(1 - \tan\theta)(1 + \tan\theta)}
= \frac{(\tan^2\theta - 1)}{1 - \tan^2\theta} = \boxed{-1}
✅ Final Answer:
\boxed{-1}
Qus : 20 NIMCET PYQ 2024 4 If \sin x=\sin y and \cos x=\cos y , then the value of x-y is
1 \pi/4 2 n \pi/2 3 n \pi 4 2n \pi Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given:
\sin x = \sin y \quad \text{and} \quad \cos x = \cos y
✳ Step 1: Use the identity for sine
\sin x = \sin y \Rightarrow x = y + 2n\pi \quad \text{or} \quad x = \pi - y + 2n\pi
✳ Step 2: Use the identity for cosine
\cos x = \cos y \Rightarrow x = y + 2m\pi \quad \text{or} \quad x = -y + 2m\pi
? Combine both conditions
For both \sin x = \sin y and \cos x = \cos y to be true, the only consistent solution is:
x = y + 2n\pi \Rightarrow x - y = 2n\pi
✅ Final Answer:
\boxed{x - y = 2n\pi \quad \text{for } n \in \mathbb{Z}}
Qus : 23 NIMCET PYQ 2024 3 A speaks truth in 40% and B in 50% of the cases. The probability that they contradict
each other while narrating some incident is:
1 1/4 2 1/3 3 1/2 4 2/3 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
A speaks the truth in 40% of the cases and B in 50% of the cases.
What is the probability that they contradict each other while narrating an incident?
? Let’s Define:
P(A_T) = 0.4 → A tells the truth
P(A_L) = 0.6 → A lies
P(B_T) = 0.5 → B tells the truth
P(B_L) = 0.5 → B lies
? Contradiction happens in two cases:
A tells the truth, B lies → 0.4 \times 0.5 = 0.2
A lies, B tells the truth → 0.6 \times 0.5 = 0.3
Total probability of contradiction:
P(\text{Contradiction}) = 0.2 + 0.3 = \boxed{0.5}
✅ Final Answer:
\boxed{\frac{1}{2}}
Qus : 25 NIMCET PYQ 2024 4 A man starts at the origin O and walks a distance of 3 units in the north-
east direction and then walks a distance of 4 units in the north-west
direction to reach the point P. then \vec{OP} is equal to
1 \frac{1}{\sqrt{2}} (-\hat{i}+\hat{j}) 2 \frac{1}{2} (\hat{i}+\hat{j}) 3 \frac{1}{\sqrt{2}} (\hat{i}-7\hat{j}) 4 \frac{1}{\sqrt{2}} (-\hat{i}+7\hat{j}) Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
A man starts at the origin O , walks 3 units in the north-east direction, then 4 units in the north-west direction to reach point P .
Find the displacement vector \vec{OP} .
? Solution:
North-East (45°):
\vec{A} = 3 \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \right)
North-West (135°):
\vec{B} = 4 \cdot \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( -\frac{4}{\sqrt{2}}, \frac{4}{\sqrt{2}} \right)
Total Displacement:
\vec{OP} = \vec{A} + \vec{B} = \left( \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right)
✅ Final Answer:
\boxed{ \vec{OP} = \left( \frac{-1}{\sqrt{2}},\ \frac{7}{\sqrt{2}} \right) }
Qus : 26 NIMCET PYQ 2024 3 Among the given numbers below, the smallest number which will be divided by 9, 10,
15 and 20, leaves the remainders 4, 5, 10, and 15, respectively
1 85 2 535 3 355 4 265 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Find the smallest number which when divided by 9, 10, 15 and 20 leaves remainders 4, 5, 10 and 15 respectively.
✅ Solution:
Let the number be x .
x \equiv 4 \mod 9 \Rightarrow x - 4 divisible by 9
x \equiv 5 \mod 10 \Rightarrow x - 5 divisible by 10
x \equiv 10 \mod 15 \Rightarrow x - 10 divisible by 15
x \equiv 15 \mod 20 \Rightarrow x - 15 divisible by 20
So, x + 5 is divisible by LCM of 9, 10, 15, 20
LCM = 2^2 \cdot 3^2 \cdot 5 = 180
x + 5 = 180 \times 2 = 360 \Rightarrow x = 355
? Final Answer: \boxed{355}
Qus : 27 NIMCET PYQ 2024 3 The value of \sum ^n_{r=1}\frac{{{{}^nP}}_r}{r!} is:
1 2^n 2 1-2^{-n} 3 2^n-1 4 2^{2n}-1 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Question: Find the value of:
\sum_{r=1}^{n} \frac{nP_r}{r!}
Solution:
We know: nP_r = \frac{n!}{(n - r)!} \Rightarrow \frac{nP_r}{r!} = \frac{n!}{(n - r)! \cdot r!} = \binom{n}{r}
Therefore,
\sum_{r=1}^{n} \frac{nP_r}{r!} = \sum_{r=1}^{n} \binom{n}{r} = 2^n - 1
Final Answer: \boxed{2^n - 1}
Qus : 28 NIMCET PYQ 2024 4 Let A and B be two events defined on a sample space \Omega . Suppose A^C denotes
the complement of A relative to the sample space \Omega . Then the probability P\Bigg{(}(A\cap{B}^C)\cup({A}^C\cap B)\Bigg{)} equals
1 P(A)+P(A)+P(A\cap B) 2 P(A)+P(A)-P(A\cap B) 3 P(A)+P(A)+2P(A\cap B) 4 P(A)+P(A)-2P(A\cap B) Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given: Two events A and B defined on sample space \Omega . We are to find the probability:
P\left((A \cap B^c) \cup (A^c \cap B)\right)
Step 1: This is the probability of events that are in exactly one of A or B (but not both), i.e., symmetric difference of A and B:
(A \cap B^c) \cup (A^c \cap B) = A \Delta B
Step 2: So, we use:
P(A \Delta B) = P(A) + P(B) - 2P(A \cap B)
Final Answer:
\boxed{P(A) + P(B) - 2P(A \cap B)}
Qus : 29 NIMCET PYQ 2024 1 Let Z be the set of all integers, and consider the sets X=\{(x,y)\colon{x}^2+2{y}^2=3,\, x,y\in Z\} and Y=\{(x,y)\colon x{\gt}y,\, x,y\in Z\} . Then the number of elements in X\cap Y is:
1 1 2 2 3 3 4 4 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given: x^2 + 2y^2 = 3 \text{ and } x > y \text{ with } x, y \in \mathbb{Z}
Solutions to the equation are: \{(1,1), (1,-1), (-1,1), (-1,-1)\}
Among them, only (1, -1) satisfies x > y .
Answer: \boxed{1}
Qus : 31 NIMCET PYQ 2024 4 Given a set A with median
m_1 = 2 and set B with median
m_2 = 4 What can we say about the median of the combined set?
1 at most 1 2 at most 2 3 at least 1 4 at least 2 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given two sets:
Set A has median m_1 = 2
Set B has median m_2 = 4
What can we say about the median of the combined set A \cup B ?
✅ Answer:
The combined median depends on the size and values of both sets.
Without that information , we only know that:
\text{Combined Median} \in [2, 4]
So, the exact median cannot be determined with the given data.
Qus : 33 NIMCET PYQ 2024 2 A coin is thrown 8 number of times. What is the probability of getting a head in an odd
number of throw?
1 3/4 2 1/2 3 1/4 4 1/8 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Total outcomes = 2^8 = 256
Favorable outcomes (odd heads):
\binom{8}{1} = 8
\binom{8}{3} = 56
\binom{8}{5} = 56
\binom{8}{7} = 8
Total favorable = 8 + 56 + 56 + 8 = 128
So, Probability = \frac{128}{256} = \boxed{\frac{1}{2}}
? Final Answer: \boxed{\frac{1}{2}}
Qus : 34 NIMCET PYQ 2024 3 Consider the function f(x)={x}^{2/3}{(6-x)}^{1/3} . Which of the following statement is false?
1 f is increasing in interval (0,4) 2 f is decreasing in interval (6,∞) 3 f has a point of inflection at x=0 4 f has a point of inflection at x=6 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given Function:
f(x) = x^{2/3}(6 - x)^{1/3}
f is increasing in (0, 4): ✅ True
f has a point of inflection at x = 0: ✅ True
f has a point of inflection at x = 6: ✅ True
f is decreasing in (6, ∞): ❌ False (function not defined there)
Correct Answer (False Statement):
\boxed{\text{f is decreasing in } (6, \infty)}
Qus : 35 NIMCET PYQ 2024 3 The value of {{Lt}}_{x\rightarrow0}\frac{{e}^x-{e}^{-x}-2x}{1-\cos x} is equal to
1 2 2 1 3 0 4 -1 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Evaluate:
\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}
Step 1: Apply L'Hôpital's Rule (since it's 0/0):
First derivative:
\frac{e^x + e^{-x} - 2}{\sin x}
Still 0/0 → Apply L'Hôpital's Rule again:
\frac{e^x - e^{-x}}{\cos x}
Now,
\lim_{x \to 0} \frac{1 - 1}{1} = 0
Final Answer:
\boxed{0}
Qus : 36 NIMCET PYQ 2024 2 Consider the function f(x)=\begin{cases}{-{x}^3+3{x}^2+1,} & {if\, x\leq2} \\ {\cos x,} & {if\, 2{\lt}x\leq4} \\ {{e}^{-x},} & {if\, x{\gt}4}\end{cases} Which of the following statements about f(x) is true:
1 f(x) has a local maximum at x =1, which is also the global maximum 2 f(x) has a local maximum at x=2, which is not the global maximum 3 f(x) has a local maximum at x =\pi , but it is not the global maximum 4 f(x) has a global maximum at x=0 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ
Solution Qus : 37 NIMCET PYQ 2024 1 If one AM (Arithmetic mean) 'a' and two GM's (Geometric means) p and q be inserted between any two positive numbers, the value of p^3+q^3 is
1 2apq 2 pq/a 3 2pq/a 4 p+q+a Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Problem:
If one Arithmetic Mean (AM) a and two Geometric Means p and q are inserted between any two positive numbers, find the value of:
p^3 + q^3
Given:
Let two positive numbers be A and B .
One AM: a = \frac{A + B}{2}
Two GMs inserted: so the four terms in G.P. are:
A, \ p = \sqrt[3]{A^2B}, \ q = \sqrt[3]{AB^2}, \ B
Now calculate:
pq = \sqrt[3]{A^2B} \cdot \sqrt[3]{AB^2} = \sqrt[3]{A^3B^3} = AB
p^3 = A^2B, \quad q^3 = AB^2
p^3 + q^3 = A^2B + AB^2 = AB(A + B)
Also,
2apq = 2 \cdot \frac{A + B}{2} \cdot AB = AB(A + B)
✅ Therefore,
\boxed{p^3 + q^3 = 2apq}
Qus : 38 NIMCET PYQ 2024 2 The equation 3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0 represents
1 a circle 2 an ellipse 3 a hyperbola 4 a parabola Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Rule for Classifying Conics Using Discriminant
Given the equation: Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
Compute: \Delta = B^2 - 4AC
? Based on value of \Delta :
Ellipse : \Delta < 0 and A \ne C , B \ne 0 → tilted ellipse
Circle : \Delta < 0 and A = C , B = 0
Parabola : \Delta = 0
Hyperbola : \Delta > 0
Example:
For the equation: 3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0
A = 3 , B = 10 , C = 11 →
\Delta = 10^2 - 4(3)(11) = 100 - 132 = -32
Since \Delta < 0 , it represents an ellipse .
Qus : 39 NIMCET PYQ 2024 1 The points (1,1/2) and (3,-1/2) are
1 In between the lines 2x+3y=6 and 2x+3y = -6 2 On the same side of the line 2x+3y = 6 3 On the same side of the line 2x+3y = -6 4 On the opposite side of the line 2x+3y = -6 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given:
Points: A = (1, \frac{1}{2}) , B = (3, -\frac{1}{2})
Line: 2x + 3y = k
Step 1: Evaluate 2x + 3y
For A: 2(1) + 3\left(\frac{1}{2}\right) = \frac{7}{2}
For B: 2(3) + 3\left(-\frac{1}{2}\right) = \frac{9}{2}
✅ Option-wise Check:
In between the lines 2x + 3y = -6 and 2x + 3y = 6 :
✔️ True since \frac{7}{2}, \frac{9}{2} \in (-6, 6)
On the same side of 2x + 3y = 6 :
✔️ True , both values are less than 6
On the same side of 2x + 3y = -6 :
✔️ True , both values are greater than -6
On the opposite side of 2x + 3y = -6 :
❌ False , both are on the same side
✅ Final Answer:
The correct statements are:
In between the lines 2x + 3y = -6 and 2x + 3y = 6
On the same side of the line 2x + 3y = 6
On the same side of the line 2x + 3y = -6
Qus : 40 NIMCET PYQ 2024 1 How much work does it take to slide a crate for a distance of 25m along a loading
dock by pulling on it with a 180 N force where the dock is at an angle of 45°
from the horizontal?
1 3.18198\times10^3 J 2 3.18198\times10^2 J 3 3.4341\times10^3 J 4 3.4341\times10^4 J Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Work Done Problem:
A crate is pulled 25 m along a dock with a force of 180 N at an angle of 45° .
✅ Formula Used:
\text{Work} = F \cdot d \cdot \cos(\theta)
✅ Substituting Values:
W = 180 \times 25 \times \cos(45^\circ) = 180 \times 25 \times 0.70710678118 = 3181.98052\, \text{J}
✅ Final Answer (to 5 decimal places):
\boxed{3.181\times 10^3 \, \text{Joules}}
Qus : 41 NIMCET PYQ 2024 1 The vector \vec{A}=(2x+1)\hat{i}+(x^2-6y)\hat{j}+(xy^2+3z)\hat{k} is a
1 sink field 2 solenoidal field 3 source field 4 None of these Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Vector Field:
\vec{A} = (2x + 1)\hat{i} + (x^2 - 6y)\hat{j} + (xy^2 + 3z)\hat{k}
Divergence:
\nabla \cdot \vec{A} = 2 - 6 + 3 = -1 \neq 0
Not solenoidal ❌
Curl:
\nabla \times \vec{A} = (2xy)\hat{i} - (y^2)\hat{j} + (2x)\hat{k} \neq \vec{0}
Not conservative ❌
Final Answer:
\vec{A} is neither conservative nor solenoidal .
Vector Sink Field Analysis
Given vector field:
\vec{A} = (2x + 1)\hat{i} + (x^2 - 6y)\hat{j} + (xy^2 + 3z)\hat{k}
Divergence:
\nabla \cdot \vec{A} = 2 - 6 + 3 = -1
✅ Conclusion:
The divergence is negative at every point, so \vec{A} is a sink field .
Qus : 42 NIMCET PYQ 2024 3 Region R is defined as region in first quadrant satisfying the condition x^2 + y^2 < 4 . Given that a point P=(r,s) lies in R, what is the probability
that r>s?
1 1 2 0 3 1/2 4 1/3 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Probability that r > s in Region R
Given: R = \{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 4 \} in the first quadrant
Area of region R in first quadrant:
A = \frac{1}{4} \pi (2)^2 = \pi
Region where r > s (i.e., below line x = y ) occupies half of that quarter-circle:
A_{\text{favorable}} = \frac{1}{2} \pi
Therefore, the required probability is:
\text{Probability} = \frac{\frac{1}{2} \pi}{\pi} = \boxed{\frac{1}{2}}
Qus : 43 NIMCET PYQ 2024 3 Lines L_1, L_2, .., L_10 are distinct among which the lines L_2, L_4, L_6, L_8, L_{10} are
parallel to each other and the lines L_1, L_3, L_5, L_7, L_9 pass through a given point C. The number of point of intersection of pairs of lines from the complete set L_1, L_2, L_3, ..., L_{10} is
1 24 2 25 3 26 4 27 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Total Number of Intersection Points
Given:
10 distinct lines: L_1, L_2, \ldots, L_{10}
L_2, L_4, L_6, L_8, L_{10} : parallel (no intersections among them)
L_1, L_3, L_5, L_7, L_9 : concurrent at point C (intersect at one point)
? Calculation:
\text{Total line pairs: } \binom{10}{2} = 45
\text{Subtract parallel pairs: } \binom{5}{2} = 10 \Rightarrow 45 - 10 = 35
\text{Concurrent at one point: reduce } 10 \text{ pairs to 1 point} \Rightarrow 35 - 9 = \boxed{26}
✅ Final Answer: \boxed{26} unique points of intersection
Qus : 44 NIMCET PYQ 2024 1 If the line a^2 x + ay +1=0 , for some real number a , is normal to the curve xy=1
then
1 a<0 2 0<a<1 3 a>0 4 -1<a<1 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Problem:
The line
a^2x + ay + 1 = 0
is normal to the curve
xy = 1
. Find possible values of a \in \mathbb{R} .
Step 1: Slope of Line
Rewrite:
y = -a x - \frac{1}{a}
→ slope = -a
Step 2: Curve Derivative
xy = 1 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}
Slope of normal = \frac{x}{y}
Match Slopes
-a = \frac{x}{y} \Rightarrow x = -a y
Plug into Curve
xy = 1 \Rightarrow (-a y)(y) = 1 \Rightarrow y^2 = -\frac{1}{a}
For real y , we need a < 0
✅ Final Answer:
\boxed{a < 0}
Qus : 45 NIMCET PYQ 2024 1 Out of a group of 50 students taking examinations in Mathematics, Physics, and
Chemistry, 37 students passed Mathematics, 24 passed Physics, and 43 passed
Chemistry. Additionally, no more than 19 students passed both Mathematics and
Physics, no more than 29 passed both Mathematics and Chemistry, and no more than
20 passed both Physics and Chemistry. What is the maximum number of students who
could have passed all three examinations?
1 14 2 10 3 12 4 9 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
? Maximum Students Passing All Three Exams
Given:
Total students = 50
|M| = 37 , |P| = 24 , |C| = 43
|M \cap P| \leq 19 , |M \cap C| \leq 29 , |P \cap C| \leq 20
We use the inclusion-exclusion principle:
|M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |M \cap C| - |P \cap C| + |M \cap P \cap C|
Let x = |M \cap P \cap C| . Then:
50 \geq 37 + 24 + 43 - 19 - 29 - 20 + x
\Rightarrow 50 \geq 36 + x \Rightarrow x \leq 14
✅ Final Answer: \boxed{14}
Qus : 46 NIMCET PYQ 2024 4 Let f\colon\mathbb{R}\rightarrow\mathbb{R} be a function such that f(0)=\frac{1}{\pi} and f(x)=\frac{x}{e^{\pi x}-1} for x\ne0 , then
1 f(x) is not continuous at x = 02 f(x) is continuous but not differentiable at x = 03 f(x) is differentiable at x = 0 and f'(0) = -\frac{\pi}{2} 4 None of the above Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Analysis of Continuity and Differentiability
Function:
f(x) =
\begin{cases}
\dfrac{x}{e^{\pi x} - 1}, & x \neq 0 \\
\dfrac{1}{\pi}, & x = 0
\end{cases}
✅ Continuity at x = 0 :
\lim_{x \to 0} f(x) = \frac{1}{\pi} = f(0)
\quad \Rightarrow \quad \text{Function is continuous at } x = 0
✏️ Differentiability at x = 0 :
f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}
= -\frac{1}{2}
✅ Final Result:
Function is continuous at x = 0
Function is differentiable at x = 0
f'(0) = \boxed{-\frac{1}{2}}
Qus : 47 NIMCET PYQ 2024 1 If f(x)=cos[\pi ^2]x+cos[-\pi ^2]x, where [.] stands for greatest integer function, then f(\pi/2) =
1 -1 2 0 3 1 4 2 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
? Function with Greatest Integer and Cosine
Given:
f(x) = \cos\left([\pi^2]x\right) + \cos\left([-\pi^2]x\right)
Find:
f\left(\frac{\pi}{2}\right)
Step 1: Estimate Floor Values
\pi^2 \approx 9.8696 \Rightarrow [\pi^2] = 9,\quad [-\pi^2] = -10
Step 2: Plug into the Function
f\left(\frac{\pi}{2}\right) = \cos\left(9 \cdot \frac{\pi}{2}\right) + \cos\left(-10 \cdot \frac{\pi}{2}\right)
= \cos\left(\frac{9\pi}{2}\right) + \cos(-5\pi)
Step 3: Simplify
\cos\left(\frac{9\pi}{2}\right) = 0,\quad \cos(-5\pi) = -1
✅ Final Answer:
\boxed{-1}
Qus : 48 NIMCET PYQ 2024 4 If three distinct numbers are chosen randomly from the first 100 natural numbers, then
the probability that all three of them are divisible by both 2 and 3 is
1 4/33 2 4/35 3 4/25 4 4/1155 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
✔️ Verified Probability
Total numbers divisible by 6 from 1 to 100: 16
\binom{100}{3} = 161700, \quad \binom{16}{3} = 560
Probability:
\frac{560}{161700} = \frac{4}{1155}
✅ Final Answer: \boxed{\frac{4}{1155}}
Qus : 49 NIMCET PYQ 2024 1 It is given that the mean, median and mode of a data set is 1, 3^x and 9^x respectively. The possible values of the mode is
1 1,4 2 1,9 3 3,9 4 9,8 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Mean, Median, and Mode Relation
Given:
Mean = 1
Median = 3^x
Mode = 9^x
Use empirical formula:
\text{Mode} = 3 \cdot \text{Median} - 2 \cdot \text{Mean}
9^x = 3 \cdot 3^x - 2 \Rightarrow (3^x)^2 = 3 \cdot 3^x - 2
Let y = 3^x , then:
y^2 = 3y - 2 \Rightarrow y^2 - 3y + 2 = 0 \Rightarrow (y - 1)(y - 2) = 0
So, y = 1 \text{ or } 2 \Rightarrow 9^x = y^2 = 1 \text{ or } 4
✅ Final Answer: \boxed{1 \text{ or } 4}
Qus : 50 NIMCET PYQ 2024 3 The value of the series \frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\cdots is
1 2e^{-2} 2 e^{-2} 3 e^{-1} 4 2e^{-1} Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Given the infinite series:
S = \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}
This is a known convergent series, and its sum is:
\boxed{e^{-1}}
✅ Final Answer: \boxed{e^{-1}}
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