Q54 / 160 · AP EAMCET 2021 (20 Aug Shift 2) · Mathematics

If f(10-x)=3x2+4x-5 and f(x)=px2+qx+r then p+q+r=

  • A
    272
  • Correct answer
    B
    274
  • C
    275
  • D
    273

Solution

Given: f10-x=3x2+4x-5

fx=px2+qx+r

p+q+r=f1

p+q+r=f10-9

⇒ p+q+r=3×92+4×9-5

⇒ p+q+r=243+36-5

⇒ p+q+r=274

Hence, the correct answer is option: 'B'.

Relevant Questions