*Year-wise Solution of IIT-JEE paper on Theory of Equations*

**Q.[2004] For all 'x' , x² + 2ax + (10 − 3a) > 0, then the interval in which 'a' lies is:**

(a) a < -5 (b) -5<a<2

(a) a < -5 (b) -5<a<2

**(c) a>5 (d) 2<a<5**

Sol:- (b) We know that

ax² + bx + c > 0 for all x, so b² − 4ac < 0 and a>0.

So, here 4a² − 4(10−3a) < 0

⇒ (a+5)(a−2) < 0

So, −5 < a < 2.

**Q.[2003] If x² + (a−b)x + (1−a−b) = 0 where a,b ∈ ℝ, then find the values of a for which equation has unequal real roots for all values of b.**

Sol: We know that

ax² + bx + c > 0 for all x, so D = b² − 4ac < 0 and a>0.

Given equation has real and distinct roots if D>0

⇒ (a−b)² − 4(1−a−b) > 0

⇒ b² − (2a−4)b + (a² + 4a − 4) > 0, for all b∈ ℝ.

So, here D < 0

⇒ (2a − 4)² − 4(a² + 4a − 4) < 0

⇒ a > 1.

**Q.[2002] The number of values of k for which the system of equations**

(k+1)x + 8y = 4k

kx + (k+3)y = 3k − 1

has infinitely many solutions, is:

(a) 0 (b) 1 (c) 2 (d) infinite

(k+1)x + 8y = 4k

kx + (k+3)y = 3k − 1

has infinitely many solutions, is:

(a) 0 (b) 1 (c) 2 (d) infinite

Sol:- (b) The equations has infinitely many solutions if

(k+1)/k = 8/(k+3) = 4k/(3k−1)

Solving first two members you get k=1,3.

From last two members k=1,2.

Thus k=1, we have only one value of k.

**Q.[2000] For the equation 3x² + px + 3 = 0, p>0, if one of the roots is square of the other, then p is equal to**

(a) 1/3 (b) 1 (c) 3 (d) 2/3Sol:- (c) We know that the roots of x² + x + 1 = 0 are 1,ω,ω².

(a) 1/3 (b) 1 (c) 3 (d) 2/3

So, comparing with given equation p=3.

**Q.[2000] If b > a, then the equation (x − a)(x − b) − 1 = 0 has**

(a) both roots in (a,b)

(b) both roots in (−∞,a)

(c) both roots in (b,+∞)

(d) one root in (−∞,a) and other in (b,+∞)

(a) both roots in (a,b)

(b) both roots in (−∞,a)

(c) both roots in (b,+∞)

(d) one root in (−∞,a) and other in (b,+∞)

Sol:- (d). Let f(x)=(x − a)(x − b) − 1

So, f(a)=f(b)=−1

Given equation represents an upward parabola. So,the roots of f(x)=0 are m,n.

Draw the figure & check that m is lying in (−∞,a) and n is lying in (b,+∞).

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