JEE Main 2021 · 24 Feb · Shift 1 · Mathematics Q62

MathematicsSingle correct+4 / -1Question 62 of 90

Let p and q be two positive numbers such that p + q = 2 and $p^4+q^4=272$ . Then p and q are roots of the equation :

x^2-2x+136=0

x^2-2x+16=0

x^2-2x+8=0

x^2-2x+2=0

Answer and worked solution

Official answer: B

Product. With $t=pq$ : $p^2+q^2=4-2t$ and $(4-2t)^2-2t^2=272$ , so $t^2-8t-128=0$ , $t=16$ (or −8). Then p, q are roots of $x^2-2x+16=0$ .

Note. That quadratic has complex roots, so no positive p and q actually exist; the key takes the equation the algebra gives.

(B)

Topics:Roots and coefficients

JEE Main 2021 · 24 Feb · Shift 1, Mathematics question 62