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1

### WB JEE 2010

MCQ (Single Correct Answer)

The quadratic equation $${x^2} + 15|x| + 14 = 0$$ has

A
only positive solutions
B
only negative solutions
C
no solution
D
both positive and negative solution

## Explanation

$${x^2} + 15|x| + 14 = 0$$

$$\Rightarrow |x{|^2} + 15|x| + 14 = 0$$

$$\Rightarrow (|x| + 14)(|x| + 1) = 0$$

$$\Rightarrow |x| = - 14, - 1$$ (Both are impossible as $$|x| \ge 0$$)

$$\Rightarrow$$ No solution.

2

### WB JEE 2010

MCQ (Single Correct Answer)

The roots of the quadratic equation $${x^2} - 2\sqrt 3 x - 22 = 0$$ are

A
imaginary
B
real, rational and equal
C
real, irrational and unequal
D
real, rational and unequal

## Explanation

$$D = {( - 2\sqrt 3 )^2} - 4\,.\,1( - 22)$$

$$= 12 + 88 = 100 > 0$$

$$\therefore$$ Roots are real & unequal

Moreover, coefficient of x is irrational.

Hence, roots are real, irrational and unequal.

3

### WB JEE 2010

MCQ (Single Correct Answer)

If $$\alpha$$, $$\beta$$ be the roots of the quadratic equation x2 + x + 1 = 0 then the equation whose roots are $$\alpha$$19, $$\beta$$7 is

A
$${x^2} - x + 1 = 0$$
B
$${x^2} - x - 1 = 0$$
C
$${x^2} + x - 1 = 0$$
D
$${x^2} + x + 1 = 0$$

## Explanation

$$\because$$ $$\alpha$$, $$\beta$$ are roots of $${x^2} + x + 1 = 0$$ (given)

But $$\omega$$, $$\omega$$2 are roots of $${x^2} + x + 1 = 0$$

$$\therefore$$ $$\alpha$$ = $$\omega$$, $$\beta$$ = $$\omega$$2 (say)

$$\therefore$$ $${\alpha ^{19}} + {\beta ^7} = {\omega ^{19}} + {\omega ^{14}} = \omega + {\omega ^2} = - 1$$ [$$\because$$ $${\omega ^3} = 1$$ & $$1 + \omega + {\omega ^2} = 0$$]

$${\alpha ^{19}}\,.\,{\beta ^7} = {\omega ^{19}}\,.\,{\omega ^{14}} = {\omega ^{33}} = 1$$

$$\therefore$$ Required equation is $${x^2} - ( - 1)x + 1 = 0$$

i.e. $${x^2} + x + 1 = 0$$

4

### WB JEE 2009

MCQ (Single Correct Answer)

If a, b, c are real, then both the roots of the equation $$(x - b)(x - c) + (x - c)(x - a) + (x - a)(x - b) = 0$$ are always

A
positive
B
negative
C
real
D
imaginary

## Explanation

Given equation is

$$(x - b)(x - c) + (x - c)(x - a) + (x - a)(x - b) = 0$$

or, $$3{x^2} - 2(a + b + c)x + ab + bc + ca = 0$$

$$D = {b^2} - 4ac = 4{(a + b + c)^2} - 4\,.\,3(ab + bc + ca)$$

$$= 4[{a^2} + {b^2} + {c^2} - ab - bc - ca]$$

$$= 2[{(a - b)^2} + {(b - c)^2} + {(c - a)^2}] \ge 0$$

So roots are real.

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