Verified Solution • 2025–26 Session moderate

QJEEM21Maccf5: JEE Main 2021 Mathematics - Class 11 JEEM21Maccf5 Solution: Let f(x) be a polynomial of degree 3 such that f ( k ) = − 2 k…

Let f(x) be a polynomial of degree 3 such that
f(k)=−2k for k = 2, 3, 4, 5. Then the value of 52 − 10f(10) is equal to :

Step-by-Step Verified Solution

kf(k)+2=λ(x−2)(x−3)(x−4)(x−5) .... (1)

put x = 0

we get λ=160

Now, put λ in equation (1)

⇒kf(k)+2=160(x−2)(x−3)(x−4)(x−5)

Put x = 10

⇒10f(10)+2=160(8)(7)(6)(5)

⇒52−10f(10)=52−26=26

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