Verified Solution • 2025–26 Session moderate

QJEEM18Ma725a: JEE Main 2018 Mathematics - Class 11 JEEM18Ma725a Solution: Let p, q and r be real numbers (p q, r 0), such that the roots…

Let p, q and r be real numbers (p ≠\ne q, r ≠\ne 0), such that the roots of the equation 1x+p+1x+q=1r{1 \over {x + p}} + {1 \over {x + q}} = {1 \over r} are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

Multiple Choice Options

  • A p2+q22{{{p^2} + {q^2}} \over 2}
  • B p2 + q2
  • C 2(p2 + q2)
  • D p2 + q2 + r2

Step-by-Step Verified Solution

Given,

1x+p+1x+q=1r{1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}

⇒\Rightarrow    \,\,\, x+p+x+q(x+p)(x+q)=1r{{x + p + x + q} \over {\left( {x + p} \right)\left( {x + q} \right)}} = {1 \over r}

⇒\Rightarrow    \,\,\, (2x + p + q) r = x2 + px + qx + pq

⇒\Rightarrow    \,\,\, x2 + (p + q −- 2r) x + pq −- pr −- qr = 0

Let α\alpha and β\beta are the roots,

∴   \therefore\,\,\, α\alpha + β\beta = −- (p + q −- 2r)

and α\alpha β\beta = pq −- pr −- qr

Given that, α\alpha = −- β\beta ⇒\Rightarrow α\alpha + β\beta = 0

∴   \therefore\,\,\, −- (p + q −- 2r) = 0

Now, α\alpha 2 + β\beta 2

= ( α\alpha + β\beta )2 −- 2 α\alpha β\beta

= ( −- (p + q −- 2r))2 −- 2 (pq −- pr −- qr)

= p2 +q2 + 4r2 + 2pq −- 4pr −- 4qr −- 2pq + 2pr + 2qr

= p2 + q2 + 4r2 −- 2pr −- 2qr

= p2 + q2 −- 2r (p + q −- 2r)

= p2 + q2 −- 2r (0)

= p2 + q2

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