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For real x, the function will assume all realvalues provided (1984  3 Marks)
Let
⇒ (x  c) y = x^{2 } (a + b)x+ ab
⇒ x^{2 } (a +b + y) x +ab +cy=0
Here, Δ = (a + b + y)^{2}  4(ab+ cy)
= y^{2} + 2 y(a +b  2c) + (ab)^{2}
Since x is real and y assumes all real values.
∴ Δ≥ 0 for all real values of y
⇒ y^{2} + 2 y(a + b  2c) + (a b)^{2}≥0
Now we know that the sign of a quad is same as of coeff of y^{2} provided its descriminant B^{2} – 4AC < 0
This will be so if, 4(a +b  2c)^{2}  4(a b)^{2}< 0
or 4 (a +b  2c+ a  b)(a +b  2c a +b) <0
⇒ 16 (a – c) (b – c) < 0
⇒ 16 (c – a) (c – b) < 0 ....(1)
Now,
If a < b then from inequation (1), we get c ∈ (a, b)
⇒ a < c < b
or If a > b then from inequation (1) we get, c ∈ (b,a)
⇒ b < c < a or a > c > b
Thus, we observe that both (c) and (d) are the correct answer.
If S is the set of all real x such that is positive,then S contains (1986  2 Marks)
KEY CONCEPT : Wavy curve method :
Let f (x) = (x α_{1})(x α_{2}) ... (x α_{n}) To find sign of f(x), plot α_{1}, α_{2}, ... α_{n }on number line in ascending order of magnitude. Starting from right extreme put + ve, –ve signs alternately. f (x) is positive in the intervals having + ve sign and negative in the intervals having –ve sign.
We have,
NOTE THIS STEP : Critical points are x = 1/2, 0, –1/2, –1 On number line by wavy method, we have
For f (x) > 0, when
Clearly S contains (∞,3 / 2) and (1/2, 3)
If a, b and c are distinct positive numbers, then the expression (b + c – a)(c + a – b)(a + b – c) – abc is (1986  2 Marks)
Given that a, b, c are distinct +ve numbers. The expression whose sign is to be checked is ( b + c – a) (c + a – b)(a + b – c) – abc.
As this expression is symmetric in a, b, c, without loss of generality, we can assume that a < b < c.
Then c – a = + ve and c – b = + ve
∴ b + c – a = + ve and c + a – b = + ve
But a + b – c may be + ve or – ve.
Case I : If a + b – c = + ve then we can say that a, b, c, are such that sum of any two is greater than the 3rd.
Consider x = a + b – c, y = b + c – a, z = c + a – b
then x, y, z all are + ve.
and then
Now we know that A.M. > G.M. for distinct real numbers
⇒ abc > (a +b  c)(b + c  a)(c + ab)
⇒ (b +c a)(c +a b)(a +b  c)  abc< 0
Case II : If a + b c= ve then
(b + c  a)(c + a  b)(a + b  c) abc
= (+ ve)(+ ve)(ve)  (+ve)
= ( ve)  (+ ve) = (ve)
⇒ (b + c  a)(c + a  b)(a + b  c)< abc
Hence in either case given expression is –ve.
If a, b, c, d and p are distinct real numbers such that (a^{2} + b^{2 }+ c^{2})p^{2} – 2 (ab + bc + cd)p + (b^{2} + c^{2 }+ d^{2}) ≤ 0 then a, b, c, d (1987  2 Marks)
Given that a, b, c, d, p are real and distinct numbers such that
(a^{2} + b^{2} + c^{2}) p^{2 } 2(ab + bc + cd) p + (b^{2 }+ c^{2} +d^{2})≤0
⇒ (a^{2} p^{2} + b^{2} p^{2 }+ c^{2} p^{2} )  (2abp + 2bcp+ 2cdp) +(b^{2 }+ c^{2} +d^{2} )≤0
⇒ (a^{2} p^{2}  2abp + b^{2}) + (b^{2} p^{2}  2bcp+ c^{2}) +(c^{2} p^{2}  2cdp +d^{2} )≤0
⇒ (ap  b)^{2} + (bp  c)^{2} + (cp  d)^{2 }≤ 0
Being sum of perfect squares, LHS can never be –ve, therefore the only possibility is
(ap  b)^{2} + (bp  c)^{2 }+ (cp d )^{2}=0
Which is possible only when each term is zero individually i.e.
ap b = 0; bp c = 0; cp  d=0
⇒ a,b,c,d are in G.P.
The equation has
The given equation is For x > 0, taking log on both sides to the base x, we get
Let log_{2} x = y, then we get,
⇒ 3y^{3} +4y^{2} 5y2=0
⇒ ( y  1)( y + 2)(3y + 1)= 0 ⇒ y = 1, 2,1/3
⇒ log_{2 }x = 1, 2, 1 / 3 ⇒x= 2, 2^{2} , 2^{1 /3}
(All accepted as > 0)
∴ There are three real solution in which one is irrational.
The product of n positive numbers is unity Then their sum is (1991  2 Marks)
Let x_{1},x_{2},....,x_{n} be the n +ve numbers
According to the question,
x_{1} x_{2} x_{3}....x_{n} = 1 ....(1)
We know for +ve no.’s A.M. ≥ G.M.
[Using eq. (1)]
Number of divisor of the form 4n + 2 (n ≥ 0) of the integer 240 is (1998  2 Marks)
We have 240 = 2^{4}.3.5.
Divisors of 240 are
Out of these divisors just 4 divisors viz., 2, 6, 10, 30 are of the form 4n +2.
If 3^{x} = 4^{x–1}, then x = (JEE Adv. 2013)
Also
Let S be the set of all nonzero real numbers α such that the quadratic equation αx^{2} – x + α = 0 has two distinct real roots x_{1} and x_{2} satisfying the inequality x_{1 }– x_{2} < 1. Which of the following intervals is(are) a subset(s) of S? (JEE Adv. 2015)
αx^{2} – x + α = 0 has distinct real roots.
∴ D > 0 ⇒ 1 – 4α^{2} > 0
...(i)
...(ii)
Combining (i) and (ii)
∴ Subsets of S can be
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