Verified Solution • 2025–26 Session moderate

QJEEM22Ma6782: JEE Main 2022 Mathematics - Class 11 JEEM22Ma6782 Solution: . If , then is equal to :

 Let f(x)=ax2+bx+c be such that f(1)=3,f(−2)=λ and \text { Let } f(x)=a x^{2}+b x+c \text { be such that } f(1)=3, f(-2)=\lambda \text { and } f(3)=4f(3)=4 . If f(0)+f(1)+f(−2)+f(3)=14f(0)+f(1)+f(-2)+f(3)=14 , then λ\lambda is equal to :

Multiple Choice Options

  • A −- 4
  • B 132\frac{13}{2}
  • C 232\frac{23}{2}
  • D 4

Step-by-Step Verified Solution

f(1)=a+b+c=3f(1) = a + b + c = 3 ..... (i)

f(3)=9a+3b+c=4f(3) = 9a + 3b + c = 4 .... (ii)

f(0)+f(1)+f(−2)+f(3)=14f(0) + f(1) + f( - 2) + f(3) = 14

OR c+3+(4a−2b+c)+4=14c + 3 + (4a - 2b + c) + 4 = 14

OR 4a−2b+2c=74a - 2b + 2c = 7 ..... (iii)

From (i) and (ii) 8a+2b=18a + 2b = 1 ..... (iv)

From (iii) −(2)×- (2) \times (i)

⇒2a−4b=1\Rightarrow 2a - 4b = 1 ..... (v)

From (iv) and (v) a=16, b=−16a = {1 \over 6},\,b = {{ - 1} \over 6} and c=3c = 3

f(−2)=4a−2b+cf( - 2) = 4a - 2b + c

=46+26+3=4= {4 \over 6} + {2 \over 6} + 3 = 4

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