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IIT-JEE Previous Year Solved Paper On Theory of Equations

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    

(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

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

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,+∞).

Note:- Each set on different topics  will contain 5 distinct Solved problems from IIT-JEE papers. Keep visiting.