1. Let $f\left(x\right) = ax^{2}+bx+c,a,b,c \epsilon R$       . If f (x) takes real values for real values of x and non-real values for non-real values of x, then.
a) a=0
b) b=0
c) c=0
d) nothing can be said about a, b, c.

Explanation: Suppose a $\neq$ 0. We rewrite f (x) as follows:

2. The condition that the equation
$\frac{1}{x}+\frac{1}{x+b}=\frac{1}{m}+\frac{1}{m+b}$
has real roots that are equal in magnitude but opposite in sign is
a) $b^{2}=m^{2}$
b) $b^{2}=2m^{2}$
c) $2b^{2}=m^{2}$
d) none of these

Explanation: Clearly x = m is a root of the equation. Therefore, the other root must be – m. That is,

3. If x is real, and $k=\frac{x^{2}-x+1}{x^{2}+x+1}$
then
a) $1/3\leq k\leq 3$
b) $k\geq 5$
c) $k\leq 0$
d) $2/3\leq k\leq 1$

Explanation:

4. Let a > 0, b > 0 and c > 0. Then both the roots of the equation $ax^{2}+bx+c=0$
a) are real and negative
b) have negative real parts
c) are rational numbers
d) none of these

Explanation:

5. Let a, b, c be non-zero real numbers such that
$\int_{0}^{1} \left(e^{-x}+e^{x}\right)\left(ax^{2}+bx+c\right)dx=\int_{0}^{2} \left(e^{-x}+e^{x}\right)\left(ax^{2}+bx+c\right)dx$
Then the quadratic equation ax2 + bx + c = 0 has
a) no root in (0, 1)
b) at least one root in (1, 2)
c) a double root in (0, 1)
d) none of these

Explanation:

6.The equation$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$
has
a) no solution
b) only one solution
c) only two solution
d) more than two solutions

Explanation:

7. The equation $3^{x-1}+5^{x-1}=34$    has
a) no solution
b) one solution
c) two solution
d) more than two solutions

Explanation: It is quite clear that x = 3 satisfies the given

solution.

8. If the harmonic mean between roots of
$\left(5+\sqrt{2}\right)x^{2}-bx+8+2\sqrt{5}=0$              (1)
is 4 then b equals
a) 2
b) $4-\sqrt{5}$
c) 3
d) $4+\sqrt{5}$

Explanation: Let $\alpha$ , $\beta$ be the roots of (1), then

9. If $a \leq0$   , then number of real roots of
$x^{2}+2a\mid x-a\mid +3a^{2}=0$              (1)
is
a) 0
b) 1
c) 2
d) infinite

10. Let $\left( a_{1},a_{2},a_{3},a_{4},a_{5}\right)$     denote a rearrangement of (3, – 5, 7, 4, – 9), then the equation $a_{1}x^{4}+a_{2}x^{3}+a_{3}x^{2}+a_{4}x+a_{5}=0$