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NIMCET Previous Year Questions (PYQs)

NIMCET 2024 PYQ


NIMCET PYQ 2024
If (4, 3) and (12, 5) are the two foci of an ellipse passing through the origin, then the eccentricity of the ellipse is





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NIMCET Previous Year PYQNIMCET 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=182a=18a=9

Step 2: Distance between the foci

2c=(124)2+(53)2=64+4=68c=17

Step 3: Find eccentricity

e=ca=179

✅ Final Answer: 179


NIMCET PYQ 2024
The number of one - one functions f: {1,2,3} → {a,b,c,d,e} is





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NIMCET Previous Year PYQNIMCET 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


NIMCET PYQ 2024
The value of the limit limx0(1x+2x+3x+4x4)1/x is





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Solution


NIMCET PYQ 2024
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





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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ

Solution

Given: Volume of a parallelepiped formed by vectors a,b,c is 4 cubic units.

Vectors:

  • a=mˆi+ˆj+ˆk
  • b=ˆiˆj+ˆk
  • c=ˆi+2ˆjˆk

Step 1: Volume = |a(b×c)|

First compute b×c:

b×c=|ˆiˆjˆk111121|=ˆi((1)(1)(1)(2))ˆj((1)(1)(1)(1))+ˆk((1)(2)(1)(1))=ˆi(12)ˆj(11)+ˆk(2+1)=ˆi+2ˆj+3ˆk

Step 2: Compute dot product with a:

a(b×c)=(m)(1)+(1)(2)+(1)(3)=m+2+3=m+5

Step 3: Volume = |m+5|=4

So, |m+5|=4m+5=±4

  • Case 1: m+5=4m=1
  • Case 2: m+5=4m=9

✅ Final Answer: m=1 or 9


NIMCET PYQ 2024
The number of distinct real values of λ for which the vectors λ2ˆi+ˆj+ˆk,ˆi+λ2ˆj+j and ˆi+ˆj+λ2ˆk are coplanar is





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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ

Solution

Given: Vectors:

  • a=λ2ˆi+ˆj+ˆk
  • b=ˆi+λ2ˆj+ˆk
  • c=ˆi+ˆj+λ2ˆk

Condition: Vectors are coplanar ⟹ Scalar triple product = 0

a(b×c)=0

Step 1: Use determinant:

a(b×c)=|λ2111λ2111λ2|

Step 2: Expand the determinant:

=λ2(λ2λ211)1(1λ211)+1(11λ21)=λ2(λ41)(λ21)+(1λ2)

Simplify:

=λ6λ2λ2+1+1λ2=λ63λ2+2

Step 3: Set scalar triple product to 0:

λ63λ2+2=0

Step 4: Let x=λ2, then:

x33x+2=0

Factor:

x33x+2=(x1)2(x+2)

So, λ2=1 (double root), or λ2=2 (discard as it's not real)

Thus, real values of λ are: λ=±1

✅ Final Answer: 2 distinct real values


NIMCET PYQ 2024
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 





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NIMCET Previous Year PYQNIMCET 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}


NIMCET PYQ 2024
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





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NIMCET Previous Year PYQNIMCET 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}


NIMCET PYQ 2024
Let C denote the set of all tuples (x,y) which satisfy x^2 -2^y=0 where x and y are natural numbers. What is the cardinality of C?





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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ

Solution


NIMCET PYQ 2024
If x=1+\sqrt[{6}]{2}+\sqrt[{6}]{4}+\sqrt[{6}]{8}+\sqrt[{6}]{16}+\sqrt[{6}]{32} then {\Bigg{(}1+\frac{1}{x}\Bigg{)}}^{24} =





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Solution


NIMCET PYQ 2024
The number of solutions of {5}^{1+|\sin x|+|\sin x{|}^2+\ldots}=25 for x\in(-\mathrm{\pi},\mathrm{\pi}) is





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Solution


NIMCET PYQ 2024
The system of equations x+2y+2z=5, x+2y+3z=6, x+2y+\lambda z=\mu has infinitely many solutions if 





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Solution


NIMCET PYQ 2024
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)





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Solution


NIMCET PYQ 2024
If F|= 40N (Newtons), |D| = 3m, and \theta={60^{\circ}}, then the work done by F acting
from P to Q is





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NIMCET Previous Year PYQNIMCET 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}}


NIMCET PYQ 2024
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





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NIMCET Previous Year PYQNIMCET 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}}


NIMCET PYQ 2024
Find the cardinality of the set C which is defined as C={\{x|\, \sin 4x=\frac{1}{2}\, forx\in(-9\pi,3\pi)}\}.





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NIMCET Previous Year PYQNIMCET 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}


NIMCET PYQ 2024
At how many points the following curves intersect \frac{{y}^2}{9}-\frac{{x}^2}{16}=1 and \frac{{x}^2}{4}+\frac{{(y-4)}^2}{16}=1





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Solution


NIMCET PYQ 2024
If for non-zero x, cf(x)+df\Bigg{(}\frac{1}{x}\Bigg{)}=|\log |x||+3, where c\ne 0, then \int ^e_1f(x)dx=





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Solution


NIMCET PYQ 2024
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





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NIMCET Previous Year PYQNIMCET 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}


NIMCET PYQ 2024
The value of \tan \Bigg{(}\frac{\pi}{4}+\theta\Bigg{)}\tan \Bigg{(}\frac{3\pi}{4}+\theta\Bigg{)} is





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NIMCET Previous Year PYQNIMCET 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}


NIMCET PYQ 2024
If \sin x=\sin y and \cos x=\cos y, then the value of x-y is





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NIMCET Previous Year PYQNIMCET 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}}


NIMCET PYQ 2024
For an invertible matrix A, which of the following is not always true:





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Solution


NIMCET PYQ 2024
For what values of \lambda does the equation 6x^2 - xy + \lambda y^2 = 0 represents two perpendicular lines and two lines inclined at an angle of \pi/4.





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Solution


NIMCET PYQ 2024
A speaks truth in 40% and B in 50% of the cases. The probability that they contradict each other while narrating some incident is:





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NIMCET Previous Year PYQNIMCET 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}}


NIMCET PYQ 2024
The two parabolas y^2 = 4a(x + c) and y^2 = 4bx, a > b > 0 cannot have a common normal unless





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Solution


NIMCET PYQ 2024
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





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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) }


NIMCET PYQ 2024
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





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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}


NIMCET PYQ 2024
The value of \sum ^n_{r=1}\frac{{{{}^nP}}_r}{r!} is:





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NIMCET Previous Year PYQNIMCET 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}


NIMCET PYQ 2024
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





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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)}


NIMCET PYQ 2024
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:





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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}


NIMCET PYQ 2024
The value of f(1) for f\Bigg{(}\frac{1-x}{1+x}\Bigg{)}=x+2 is





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Solution

Given:
f\left(\frac{1 - x}{1 + x}\right) = x + 2

To Find: f(1)

Let \frac{1 - x}{1 + x} = 1 \Rightarrow x = 0

Then, f(1) = f\left(\frac{1 - 0}{1 + 0}\right) = 0 + 2 = 2

Answer: \boxed{2}


NIMCET PYQ 2024
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?





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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.


NIMCET PYQ 2024
Let f(x)=\begin{cases}{{x}^2\sin \frac{1}{x}} & {,\, x\ne0} \\ {0} & {,x=0}\end{cases}
Then which of the follwoing is true





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Solution


NIMCET PYQ 2024
A coin is thrown 8 number of times. What is the probability of getting a head in an odd number of throw?





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NIMCET Previous Year PYQNIMCET 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}}


NIMCET PYQ 2024
Consider the function f(x)={x}^{2/3}{(6-x)}^{1/3}. Which of the following statement is false?





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NIMCET Previous Year PYQNIMCET 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)}


NIMCET PYQ 2024
The value of {{Lt}}_{x\rightarrow0}\frac{{e}^x-{e}^{-x}-2x}{1-\cos x} is equal to 





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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}


NIMCET PYQ 2024
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:





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Solution


NIMCET PYQ 2024
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





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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}


NIMCET PYQ 2024
The equation 3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0 represents





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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.


NIMCET PYQ 2024
The points (1,1/2) and (3,-1/2) are





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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:

  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

NIMCET PYQ 2024
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?





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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}}


NIMCET PYQ 2024
The vector \vec{A}=(2x+1)\hat{i}+(x^2-6y)\hat{j}+(xy^2+3z)\hat{k} is a





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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.


NIMCET PYQ 2024
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?





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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}}


NIMCET PYQ 2024
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 





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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


NIMCET PYQ 2024
If the line a^2 x + ay +1=0, for some real number a, is normal to the curve xy=1 then





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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}


NIMCET PYQ 2024
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?





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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}


NIMCET PYQ 2024
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





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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}}

NIMCET PYQ 2024
If f(x)=cos[\pi^2]x+cos[-\pi^2]x, where [.] stands for greatest integer function, then f(\pi/2)=





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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}


NIMCET PYQ 2024
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





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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}}


NIMCET PYQ 2024
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





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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}


NIMCET PYQ 2024
The value of the series \frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\cdots is





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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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