Verified Solution • 2025–26 Session moderate

QJEEM22Mafdb0: JEE Main 2022 Mathematics - Class 11 JEEM22Mafdb0 Solution: Let , be the roots of the equation and be the roots of the…

Let α\alpha , β\beta be the roots of the equation x2−2x+6=0x^{2}-\sqrt{2} x+\sqrt{6}=0 and 1α2+1,1β2+1\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1 be the roots of the equation x2+ax+b=0x^{2}+a x+b=0 . Then the roots of the equation x2−(a+b−2)x+(a+b+2)=0x^{2}-(a+b-2) x+(a+b+2)=0 are :

Multiple Choice Options

  • A non-real complex numbers
  • B real and both negative
  • C real and both positive
  • D real and exactly one of them is positive

Step-by-Step Verified Solution

α+β=2\alpha + \beta = \sqrt 2 , αβ=6\alpha \beta = \sqrt 6

1α2+1+1β2+1=2+α2+β26{1 \over {{\alpha ^2}}} + 1 + {1 \over {{\beta ^2}}} + 1 = 2 + {{{\alpha ^2} + {\beta ^2}} \over 6}

=2+2−266=−a= 2 + {{2 - 2\sqrt 6 } \over 6} = - a

(1α2+1)(1β2+1)=1+1α2+1β2+1α2β2\left( {{1 \over {{\alpha ^2}}} + 1} \right)\left( {{1 \over {{\beta ^2}}} + 1} \right) = 1 + {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} + {1 \over {{\alpha ^2}{\beta ^2}}}

=76+2−266=b= {7 \over 6} + {{2 - 2\sqrt 6 } \over 6} = b

⇒a+b=−56\Rightarrow a + b = {{ - 5} \over 6}

So, equation is x2+17x6+76=0{x^2} + {{17x} \over 6} + {7 \over 6} = 0

OR 6x2+17x+7=06{x^2} + 17x + 7 = 0

Both roots of equation are −- ve and distinct

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