Quadratic Equation MCQ Questions & Answers in Algebra | Maths

It is an identity in $$x$$ if $${a^2} - 3a + 2 = 0,{a^2} - 5a + 6 = 0,{a^2} - 4 = 0$$         hold at the same time.

$$\eqalign{ & {\text{Let }}{3^{\tan x}} = y.\,{\text{Then }}ay + \frac{a}{y} - 2 = 0\,\,{\text{or, }}a{y^2} - 2y + a = 0. \cr & D \geqslant 0 \cr} $$
$$ \Rightarrow \,\,4 - 4{a^2} \geqslant 0.$$    Also roots are positive as $$y = {3^{\tan x}} > 0.$$
∴ sum of the root $$ = \frac{2}{a} > 0$$
$$ \Rightarrow \,\,a > 0.$$

$$\eqalign{ & f\left( x \right) = {x^2} - 3x + a + \frac{1}{a};f\left( 3 \right) = 9 - 9 + a + \frac{1}{a} > 0 \cr & \Rightarrow a + \frac{1}{a} > 0 \cr & \Rightarrow a > 0 \cr & f\left( 2 \right) = 4 - 6 + a + \frac{1}{a} \leqslant 0 \cr & \Rightarrow \frac{{{a^2} - 2a + 1}}{a} \leqslant 0 \cr & \Rightarrow \frac{{{{\left( {a - 1} \right)}^2}}}{a} \leqslant 0 \cr & \Rightarrow a = 1 \cr} $$
Therefore, $$f\left( x \right) = {x^2} - 3x + 2 = 0$$     has root 1 and 2.
$$\therefore {\alpha ^2} + {\beta ^2} = 5$$

$$\eqalign{ & x + \frac{1}{x} = 1\,\,\,\,{\text{or }}{x^2} - x + 1 = 0 \cr & \therefore x = \frac{1}{2} \pm i\frac{{\sqrt 3 }}{2}\,\,\,\,\,{\text{or }}x = {e^{\frac{{i\pi }}{3}}} \cr & \therefore {x^a} + {x^{ - a}} = {e^{\frac{{ia\pi }}{3}}} + {e^{\frac{{ - ia}}{3}}} = 2\cos \frac{{ar}}{3} \cr & {\text{Hence, }}\cos \frac{{a\pi }}{3} + \cos \frac{{b\pi }}{3} + \cos \frac{{c\pi }}{3} = 0 \cr & a,b,c \in I \cr & \therefore {\left. {a + b + c} \right|_{\min }} = \left( {1 + 3 + 5} \right) = 9 \cr} $$

Here, $$xy \geqslant 100.$$  Now, $${\left( {x + y} \right)^2} = {\left( {x - y} \right)^2} + 4xy \geqslant 400 + {\left( {x - y} \right)^2} \geqslant 400$$
∴ the smallest possible value of $$x + y$$  is $$20$$ $$\left( {\because \,\,x > 0,y > 0} \right).$$

$$\eqalign{ & {\alpha ^2} + {\beta ^2} = \frac{1}{{{a^2}}}\,\,{\text{and }}{\alpha ^2}{\beta ^2} = \frac{{1 - {a^2}}}{{{a^2}}} \cr & \Rightarrow {\alpha ^2} + {\beta ^2} - 1 = {\alpha ^2}{\beta ^2} \cr & \Rightarrow \left( {{\alpha ^2} - 1} \right)\left( {{\beta ^2} - 1} \right) = 0 \cr & \because {\beta ^2} \ne 1, \cr & \Rightarrow {\alpha ^2} = 1,\,\,{\text{so,}}\,\,{\beta ^2} = \frac{{1 - {a^2}}}{{{a^2}}} \cr} $$

KEY CONCEPT : If both roots of a quadratic equation
$$a{x^2} + bx + c = 0$$    are less than $$k$$
then $$af\left( k \right) > 0,D \geqslant 0,\alpha + \beta < 2k.$$

$$\eqalign{ & f\left( x \right) = {x^2} - 2ax + {a^2} + a - 3 = 0, \cr & f\left( 3 \right) > 0,\alpha + \beta < 6,D \geqslant 0. \cr & \Rightarrow \,\,{a^2} - 5a + 6 > 0,a < 3, - 4a + 12 \geqslant 0 \cr & \Rightarrow \,\,a < 2\,\,{\text{or }}a > 3,a < 3,a < 3 \cr & \Rightarrow \,\,a < 2. \cr} $$

$$\eqalign{ & {x^2} - 2x\sec \theta + 1 = 0 \cr & \Rightarrow x = \sec \,\,\theta \pm \,\tan \theta \cr & {\text{and }}{x^2} + 2x\tan \theta - 1 = 0 \cr & \Rightarrow \,\,x = - \tan \theta \pm \sec \theta \cr & \because - \frac{\pi }{6} < \theta < \, - \frac{\pi }{{12}} \cr & \Rightarrow \sec \frac{\pi }{6} > \sec \theta > \sec \frac{\pi }{{12}} \cr & {\text{and}}\,\, - \tan \frac{\pi }{6} < \tan \theta < - \frac{{\tan \pi }}{{12}} \cr & {\text{also}}\,\,\tan \frac{\pi }{{12}} < - \tan \theta < \tan \frac{\pi }{6} \cr & {\alpha _1},{\beta _1}\,{\text{are}}\,{\text{roots}}\,{\text{of }}{x^2} - 2x\,\sec \theta \, + 1 = 0 \cr & {\text{and}}\,{\alpha _1} > {\beta _1} \cr & \therefore {\alpha _1} = \sec \theta - \tan \theta \,\,{\text{and }}{\beta _1} = \sec \theta + \tan \theta \cr & {\alpha _2},{\beta _2}\,{\text{are}}\,{\text{roots of }}{x^2} + 2x\,\tan {\text{ }}\theta - 1 = 0\,\,{\text{and}}\,{\alpha _2} > {\beta _2} \cr & \therefore {\alpha _2} = - \tan \,\theta + \sec \theta ,{\beta _2} = - \tan \,\theta - \sec \theta \cr & \therefore {\alpha _1} + {\beta _2} = \sec \theta \, - \tan \theta - \tan \theta - \sec \theta = - 2\tan \theta \cr} $$

Given equation is $$x - \frac{2}{{x - 1}} = 1 - \frac{2}{{x - 1}}$$
Clearly $$x \ne 1$$  for the given eqn. to be defined. If $$x - 1 \ne 0,$$   we can cancel the common term $$\frac{{ - 2}}{{x - 1}}$$ on both sides to get $$x = 1,$$  but it is not possible. So given eqn. has no roots.
∴ $${\text{(A)}}$$ is the correct answer.

$$\eqalign{ & {\text{Let }}y = 2{\log _{10}}x - {\log _x}0.01 \cr & = 2{\log _{10}}x - \frac{{{{\log }_{10}}0.01}}{{{{\log }_{10}}x}} = 2{\log _{10}}x + \frac{2}{{{{\log }_{10}}x}} \cr & = 2\left[ {{{\log }_{10}}x + \frac{2}{{{{\log }_{10}}x}}} \right] \cr & \left[ {{\text{Here }}x > 1\,\, \Rightarrow \,{{\log }_{10}}x > 0} \right] \cr} $$
Now since sum ofa real $$+ ve$$  number and its reciprocal is always greater than or equal to 2.
$$\eqalign{ & \therefore \,\,\,y \geqslant 2 \times 2 \cr & \Rightarrow \,y \geqslant 4 \cr} $$
∴ Least value of $$y$$ is $$4.$$