1
JEE Advanced 2026 Paper 2 Online
MCQ (More than One Correct Answer)
+4
-1

Let a, b, c be positive integers in arithmetic progression such that the equation

$$ax^2 + bx + c = 0$$

has only integer solutions.

Then which of the following statements is (are) TRUE?

A

c - b is an integer multiple of a

B

Both the roots of the equation $ax^2 + bx + c = 0$ are odd integers

C

If $c = 15$, then $ab = 8$

D

If $b = 8$, then $x = 3$ is a root of the equation $ax^2 + bx + c = 0$

2
JEE Advanced 2019 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-1
Change Language
Let $$\alpha $$ and $$\beta $$ be the roots of$${x^2} - x - 1 = 0$$, with $$\alpha $$ > $$\beta $$. For all positive integers n, define

$${a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1$$

$${b_1} = 1\,and\,{b_n} = {a_{n - 1}} + {a_{n + 1}},\,n \ge 2$$

Then which of the following options is/are correct?
A
$$\sum\limits_{n = 1}^\infty {{{{b_n}} \over {{{10}^n}}}} = {8 \over {89}}$$
B
bn = $$\alpha $$n + $$\beta $$n for all n $$ \ge $$ 1
C
a1 + a2 + a3 + ... + an = an+2 $$ - $$ 1 for all n $$ \ge $$ 1
D
$$\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{{10}^n}}}} = {10 \over {89}}$$
3
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
Let $$S$$ be the set of all non-zero real numbers $$\alpha $$ such that the quadratic equation $$\alpha {x^2} - x + \alpha = 0$$ has two distinct real roots $${x_1}$$ and $${x_2}$$ satisfying the inequality $$\left| {{x_1} - {x_2}} \right| < 1.$$ Which of the following intervals is (are) $$a$$ subset(s) os $$S$$?
A
$$\left( { - {1 \over 2} - {1 \over {\sqrt 5 }}} \right)$$
B
$$\left( { - {1 \over {\sqrt 5 }},0} \right)$$
C
$$\left( {0,{1 \over {\sqrt 5 }}} \right)$$
D
$$\left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$$
4
JEE Advanced 2013 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
If $${3^x}\, = \,{4^{x - 1}},$$ then $$x\, = $$
A
$${{2{{\log }_3}\,2} \over {2{{\log }_3}\,2 - 1}}$$
B
$${2 \over {2 - {{\log }_2}\,3}}$$
C
$${1 \over {1 - {{\log }_4}\,3}}$$
D
$${{2{{\log }_2}\,3} \over {2{{\log }_2}\,3 - 1}}$$

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