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 ax^{2} + 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

Explanation: Put x – a = t and rewrite (1) as

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
\[ a_{1}x^{4}+a_{2}x^{3}+a_{3}x^{2}+a_{4}x+a_{5}=0\]

has

a) at least two real roots

b) all four real roots

c) only imaginary roots

d) none of these

Explanation: x = 1 is always a root of the equation