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Previous Years AIEEE/JEE Mains Questions
(1) Greater than –4ab (2) Less than –4ab (3) Greater than 4ab (4) Less than 4ab [AIEEE-2010]
since the roots of $\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{a}=0$ are imaginary $\because c^{2}-4 a b<0 \Rightarrow c^{2}<4 a b$ for exp. $3 b^{2} x^{2}+6 b c x+2 c^{2}$ Min. value $=\frac{-\mathrm{D}}{4 \mathrm{a}}=\frac{\left(36 \mathrm{b}^{2} \mathrm{c}^{2}-24 \mathrm{b}^{2} \mathrm{c}^{2}\right)}{12 \mathrm{b}^{2}}=\frac{-12 \mathrm{b}^{2} \mathrm{c}^{2}}{12 \mathrm{b}^{2}}=-\mathrm{c}^{2}$ $\therefore-\mathrm{c}^{2}>-4 \mathrm{ab}$ So exp. is greater than (– 4ab)
(1) –2 (2) –1 (3) 1 (4) 2 [AIEEE-2010]
Roots of equation $\mathrm{x}^{2}-\mathrm{x}+1=0 \mathrm{ar}$ $\alpha=-\omega, \quad \beta=-\omega^{2}$ $\alpha^{2009}+\beta^{2009}=(-\omega)^{2009}+\left(-\omega^{2}\right)^{2009}$c $=-\left(\omega^{2}+\omega\right)=1$
If $\mathrm{p}(\mathrm{x})=0$ only for $\mathrm{x}=-1$ and $\mathrm{p}(-2)=2,$ then the value of $\mathrm{p}(2)$ is: (1) 18 (2) 3 (3) 9 (4) 6 [AIEEE-2011]
$\mathrm{P}(\mathrm{x})=\mathrm{k}(\mathrm{x}+1)^{2}$ $\mathrm{P}(-2)=2=\mathrm{k}(-1)^{2}$ $\Rightarrow \mathrm{k}=2$ $\therefore \mathrm{P}(\mathrm{x})=2(\mathrm{x}+1)^{2}$ $\Rightarrow \mathrm{P}(2)=18$ Aliter : $\mathrm{P}(\mathrm{x})=\left(\mathrm{a}-\mathrm{a}_{1}\right) \mathrm{x}^{2}+\left(\mathrm{b}-\mathrm{b}_{1}\right) \mathrm{x}+\left(\mathrm{c}-\mathrm{c}_{1}\right)=0$ only x = –1 P(x) = 0 So roots are equal. it means D = 0 $\left(b-b_{1}\right)^{2}=4\left(a-a_{1}\right)\left(c-c_{1}\right) \Rightarrow q^{2}=4 p r$ Let $\mathrm{a}-\mathrm{a}_{1}=\mathrm{p}$ $\mathrm{b}-\mathrm{b}_{1}=\mathrm{q}$ $\mathrm{c}-\mathrm{c}_{1}=\mathrm{r}$ P(–1) = 0 p – q + r = 0 …… (1) 4p – 2q + r = 2 ….. (2) 4p + 2q + r = ? 4p + 2q + r = ? $(p+r)^{2}-4 p r=0$ $(p-r)^{2}=0$ from eq. (1) q = 2r So from eq. (2) 4r – 4r + r = 2 r = 2 So 4p + 2q + r = 4r + 4r + r = 9r = 18
(1) –4, –3 (2) 6, 1 (3) 4, 3 (4) –6, –1 [AIEEE-2011]
(1) exactly four real roots. (2) infinite number of real roots. (3) no real roots. (4) exactly one real root. [AIEEE-2012]
(1) 1 : 2 : 3 (2) 3 : 2 : 1 (3) 1 : 3 : 2 (4) 3 : 1 : 2 [JEE-MAIN-2013]
$\mathrm{x}^{2}+2 \mathrm{x}+3=0$ $\mathrm{D}<0$ $\therefore \quad \mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0 \mathrm{has}$ both roots common $\therefore \quad a: b: c=1: 2: 3$
(1) 3 (2) – 3 (3) 6 (4) – 6 [JEE-MAIN-2015]
(1) 5 (2) 3 (3) –4 (4) 6 [JEE-MAIN-2016]
$\mathrm{x}^{2}-5 \mathrm{x}+5=1 \Rightarrow \mathrm{x}=1,4$ $x^{2}-5 x+5=-1 \Rightarrow x=2,3$ but 3 is rejected $\mathrm{x}^{2}+4 \mathrm{x}-60=0 \Rightarrow \mathrm{x}=-10,6$ Sum = 3
If $\alpha, \beta \in \mathrm{C}$ are the distinct roots of the equation $\mathrm{x}^{2}-\mathrm{x}+1=0,$ then $\alpha^{101}+\beta^{107}$ is equal to- [JEE-MAIN-2018]
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