Qus : 2 Phrases PYQ 2023 1
If f(x) is a polynomial of degree 4, f(n) = n + 1 & f(0) = 25, then find f(5) = ?
1 30 2 20 3 25 4 None of these Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2023 PYQ Solution
Correct Shortcut Method — Find f ( 5 )
Step 1: Define a helper polynomial:
g ( x ) = f ( x ) − ( x + 1 )
Given: f ( 1 ) = 2 , f ( 2 ) = 3 , f ( 3 ) = 4 , f ( 4 ) = 5 ⇒ g ( 1 ) = g ( 2 ) = g ( 3 ) = g ( 4 ) = 0
So,
g ( x ) = A ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ⇒ f ( x ) = A ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) + ( x + 1 )
Step 2: Use f ( 0 ) = 25 to find A:
f ( 0 ) = A ( − 1 ) ( − 2 ) ( − 3 ) ( − 4 ) + ( 0 + 1 ) = 24 A + 1 = 25 ⇒ A = 1
Step 3: Compute f ( 5 ) :
f ( 5 ) = ( 5 − 1 ) ( 5 − 2 ) ( 5 − 3 ) ( 5 − 4 ) + ( 5 + 1 ) = 4 ⋅ 3 ⋅ 2 ⋅ 1 + 6 = 24 + 6 = 30
✅ Final Answer: f ( 5 ) = 30
Qus : 3 Phrases PYQ 2023 2
The maximum value of f ( x ) = ( x – 1 ) 2 ( x + 1 ) 3 is equal to 2 p 3 q 3125
then the ordered pair of (p, q) will be
1 (3,7) 2 (7,3) 3 (5,5) 4 (4,4) Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2023 PYQ Solution
Maximum Value of f ( x ) = ( x − 1 ) 2 ( x + 1 ) 3
Step 1: Let’s define the function:
f ( x ) = ( x − 1 ) 2 ( x + 1 ) 3
Step 2: Take derivative to find critical points
Use product rule:
Let u = ( x − 1 ) 2 , v = ( x + 1 ) 3
f ′ ( x ) = u ′ v + u v ′ = 2 ( x − 1 ) ( x + 1 ) 3 + ( x − 1 ) 2 ⋅ 3 ( x + 1 ) 2
f ′ ( x ) = ( x − 1 ) ( x + 1 ) 2 [ 2 ( x + 1 ) + 3 ( x − 1 ) ]
f ′ ( x ) = ( x − 1 ) ( x + 1 ) 2 ( 5 x − 1 )
Step 3: Find critical points
Set f ′ ( x ) = 0 :
( x − 1 ) ( x + 1 ) 2 ( 5 x − 1 ) = 0 ⇒ x = 1 , − 1 , 1 5
Step 4: Evaluate f ( x ) at these points
f ( 1 ) = 0
f ( − 1 ) = 0
f ( 1 5 ) = ( 1 5 − 1 ) 2 ( 1 5 + 1 ) 3 = ( − 4 5 ) 2 ( 6 5 ) 3
f ( 1 5 ) = 16 25 ⋅ 216 125 = 3456 3125
Step 5: Compare with given form:
It is given that maximum value is 3456 3125 = 2 p ⋅ 3 q / 3125
Factor 3456:
3456 = 2 7 ⋅ 3 3 ⇒ So p = 7 , q = 3
✅ Final Answer:
( p , q ) = ( 7 , 3 )
Qus : 7 Phrases PYQ 2019 1 Let X i , i = 1 , 2 , . . , n be n observations and w i = p x i + k , i = 1 , 2 , , n where p and k are constants. If the mean of x ′ i s is 48 and the standard deviation is 12, whereas the mean of w ′ i s is 55 and the standard deviation is 15, then the value of p and k should be
1 p = 1.25, k = -5 2 p=-1.25, k = 5 3 p = 2.5, k = -5 4 p = 25, k = 5 Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2019 PYQ Qus : 9 Phrases 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 Phrases Previous Year PYQ Phrases 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 : 13 Phrases PYQ 2024 1 If f(x)=cos[π ^2]x+cos[-π ^2]x, where [.] stands for greatest integer function, then f ( π / 2 ) =
1 -1 2 0 3 1 4 2 Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2024 PYQ Solution
? Function with Greatest Integer and Cosine
Given:
f ( x ) = cos ( [ π 2 ] x ) + cos ( [ − π 2 ] x )
Find: f ( π 2 )
Step 1: Estimate Floor Values
π 2 ≈ 9.8696 ⇒ [ π 2 ] = 9 , [ − π 2 ] = − 10
Step 2: Plug into the Function
f ( π 2 ) = cos ( 9 ⋅ π 2 ) + cos ( − 10 ⋅ π 2 ) = cos ( 9 π 2 ) + cos ( − 5 π )
Step 3: Simplify
cos ( 9 π 2 ) = 0 , cos ( − 5 π ) = − 1
✅ Final Answer:
− 1
Qus : 16 Phrases PYQ 2017 4 If the graph of y = (x – 2)2 – 3 is shifted by 5 units up along y-axis and 2 units to the right along
the x-axis, then the equation of the resultant graph is
1 y=x2 +2 2 y=(x-2)2 +5 3 y=(x+2)2 +2 4 y = (x - 4)2 + 2 Go to Discussion Phrases Previous Year PYQ Phrases NIMCET 2017 PYQ Solution When y= f (x) is shifted by k units to the right along x
– axis, it become y= f (x - k )
Hence, new equation of
graph is y = (x - 4)2 + 2
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