Quadratic Equation – JEE Main Previous Year Question with Solutions

Class 9-10, JEE & NEET

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Q. If the roots of the equation $b x^{2}+c x+a=0$ be imaginary, then for all real values of x, the expression $3 b^{2} x^{2}+6 b c x+2 c^{2}$ is :- (1) Greater than –4ab (2) Less than –4ab (3) Greater than 4ab (4) Less than 4ab [AIEEE-2010]

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Sol. (1) 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)

Q. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-x+1=0,$ then $\alpha^{2009}+\beta^{2009}=$ (1) –2 (2) –1 (3) 1 (4) 2 [AIEEE-2010]

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Sol. (3) 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 $=-\left(\omega^{2}+\omega\right)=1$

Q. Let for $a \neq a_{1} \neq 0, f(x)=a x^{2}+b x+c, g(x)=a_{1} x^{2}+b_{1} x+c_{1}$ and $p(x)=f(x)-g(x)$ 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]

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Sol. (1) $\mathrm{P}(\mathrm{x})=\mathrm{k}(\mathrm{x}+1)^{2}$ $\mathrm{P}(-2)=2=\mathrm{k}(-1)^{2}$ $\Rightarrow \mathrm{k}=2$ $\therefore \mathrm{P}(\mathrm{x})=2(\mathrm{x}+1)^{2}$ $\Rightarrow \mathrm{P}(2)=18$ 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

Q. Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4, 3). Rahul made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of equation are: (1) –4, –3 (2) 6, 1 (3) 4, 3 (4) –6, –1 [AIEEE-2011]

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Sol. (2)

Q. The equation $\mathrm{e}^{\mathrm{sinx}}-\mathrm{e}^{-\mathrm{sinx}}-4=0$ has : (1) exactly four real roots. (2) infinite number of real roots. (3) no real roots. (4) exactly one real root. [AIEEE-2012]

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Sol. (3)

Q. If the equations $\mathrm{x}^{2}+2 \mathrm{x}+3=0$ and $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0, \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}$, have a common root, then a : b : c is : (1) 1 : 2 : 3 (2) 3 : 2 : 1 (3) 1 : 3 : 2 (4) 3 : 1 : 2 [JEE-MAIN-2013]

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Sol. (1) $\mathrm{x}^{2}+2 \mathrm{x}+3=0$ $\mathrm{D}<0$ $\therefore \quad \mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0 \mathrm{has}$ both roots common $\therefore \quad a: b: c=1: 2: 3$

Q. Let  and $\beta$ be the roots of equation $\mathrm{x}^{2}-6 \mathrm{x}-2=0 .$ If $\mathrm{a}_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}},$ for $\mathrm{n} \geq$ 1, then the value of $\frac{\mathrm{a}_{10}-2 \mathrm{a}_{8}}{2 \mathrm{a}_{9}}$ is equal to : (1) 3 (2) – 3 (3) 6 (4) – 6 [JEE-MAIN-2015]

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Sol. (1)

Q. The sum of all real values of x satisfying the equation $\left(x^{2}-5 x+5\right)^{x^{2}+4 x-60}=1$ is : (1) 5 (2) 3 (3) –4 (4) 6 [JEE-MAIN-2016]

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Sol. (2) $\mathrm{x}^{2}-5 \mathrm{x}+5=1 \Rightarrow \mathrm{x}=1,4$ $x^{2}-5 x+5=-1 \Rightarrow x=2,3$ but 3 is rejected $\mathrm{x}^{2}+4 \mathrm{x}-60=0 \Rightarrow \mathrm{x}=-10,6$ Sum = 3

Q. 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- 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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Sol. (2)

• February 3, 2021 at 3:10 pm

I don’t want to leave anything 😡🤬

0
• November 11, 2020 at 6:48 pm