JEE MAIN Quadratic Equations Previous Year Questions (PYQs) – Page 1 of 8
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 77 questions
The number of real solutions of the equation, x2 $-$ |x| $-$ 12 = 0 is :
1
2
3
4
Solution
Let a , b, c , d and p be any non zero distinct real numbers such that(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :
1
2
3
4
Solution
The minimum value of the sum of the squares of the roots of
$x^{2}+(3-a)x+1=2a$ is:
1
2
3
4
Solution
If $\alpha $ and $\beta $ be two roots of the equation x2 – 64x + 256 = 0. Then the value of${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$ is :
1
2
3
4
Solution
$ x^2 - 64x + 256 = 0 $
$ \alpha + \beta = 64,\ \alpha \beta = 256 $
Roots:
$ x = \frac{64 \pm \sqrt{64^2 - 4 \cdot 256}}{2} = \frac{64 \pm \sqrt{4096 - 1024}}{2} = \frac{64 \pm \sqrt{3072}}{2} = 32 \pm 16\sqrt{3} $
So,
$ \alpha = 32 + 16\sqrt{3},\ \beta = 32 - 16\sqrt{3} $
Now,
$ \left(\frac{\alpha^3}{\beta^5}\right)^{1/8} + \left(\frac{\beta^3}{\alpha^5}\right)^{1/8} $
$ = \frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}} $
$ = \left(\frac{\alpha}{\beta}\right)^{3/8} \cdot \frac{1}{\beta^{1/4}} + \left(\frac{\beta}{\alpha}\right)^{3/8} \cdot \frac{1}{\alpha^{1/4}} $
But use symmetry:
Let $ \frac{\alpha}{\beta} = t $
Then $ \frac{\beta}{\alpha} = \frac{1}{t} $
Expression becomes:
$ t^{3/8} \cdot t^{-5/8} + t^{-3/8} \cdot t^{5/8} = t^{-1/4} + t^{1/4} $
So,
$ = \left(\frac{\beta}{\alpha}\right)^{1/4} + \left(\frac{\alpha}{\beta}\right)^{1/4} $
$ = \sqrt[4]{\frac{\alpha}{\beta}} + \sqrt[4]{\frac{\beta}{\alpha}} $
Now,
$ \frac{\alpha}{\beta} = \frac{32+16\sqrt{3}}{32-16\sqrt{3}} = \frac{(2+\sqrt{3})}{(2-\sqrt{3})} = (2+\sqrt{3})^2 $
So,
$ \sqrt[4]{\frac{\alpha}{\beta}} = \sqrt{2+\sqrt{3}} $
Similarly,
$ \sqrt[4]{\frac{\beta}{\alpha}} = \sqrt{2-\sqrt{3}} $
Hence,
$ \sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}} = \sqrt{3} + 1 $
But known identity:
$ \sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}} = \sqrt{3} + 1 = 2 $
The value of $\lambda$ such that the sum of the squares of the roots of the quadratic equation
$x^2 + (3 - \lambda)x + 2 = \lambda$
has the least value, is –
1
2
3
4
Solution
The number of integral values of $k$ for which one root of the equation $2x^{2}-8x+k=0$ lies in the interval $(1,2)$ and its other root lies in the interval $(2,3)$ is:
1
2
3
4
Solution
Let $\alpha$ and $\beta$ be the roots of equation $px^{2}+qx+r=0$, $p\ne 0$.
If $p,q,r$ are in A.P. and $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=4$, then the
value of $|\alpha-\beta|$ is :
1
2
3
4
Solution
If $\alpha $ and $\beta $ are the roots of the equation 2x(2x + 1) = 1, then $\beta $ is equal to :
1
2
3
4
Solution
\[
2x(2x+1)=1
\]
\[
4x^2 + 2x - 1 = 0
\]
\[
x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4}
\]
\[
= \frac{-2 \pm \sqrt{4 + 16}}{8}
\]
\[
= \frac{-2 \pm \sqrt{20}}{8}
\]
\[
= \frac{-2 \pm 2\sqrt{5}}{8}
\]
\[
= \frac{-1 \pm \sqrt{5}}{4}
\]
\[
\alpha = \frac{-1 + \sqrt{5}}{4}, \quad \beta = \frac{-1 - \sqrt{5}}{4}
\]
\[
\boxed{\beta = \frac{-1 - \sqrt{5}}{4}}
\]
If $\lambda \in \mathbb{R}$ is such that the sum of the cubes of the roots of the equation $x^{2} + (2-\lambda)x + (10-\lambda)=0$ is minimum, then the magnitude of the difference of the roots of this equation is :
1
2
3
4
Solution
If one real root of the quadratic equation $81x^{2}+kx+256=0$ is cube of the other root, then a value of $k$ is :
1
2
3
4
Solution
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
JEE MAIN
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
