IIT-JEE 2010 - Career Point

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Expert’s Solution for Question asked by IIT-JEE Aspirants
1. Suppose f(x) = x 3 + ax 2 + bx + c, where a, b, c are
chosen respectively by throwing a die three times.
Find the probability that f(x) is an increasing
function.
Sol. f´(x) = 3x 2 + 2ax + b
y = f(x) is strictly increasing
⇒ f´(x) > 0 ∀ x
⇒ (2a) 2 – 4.3.b < 0
This is true for exactly 15 ordered pairs (a, b); 1 ≤ a,
15 5
b ≤ 6, so probability = = 36 12
3. Let g be a real valued function satisfying g(x) + g(x +
4) = g(x + 2) + g(x + 6), then prove that
∫ x +8
x
is a constant function.
g(t)
dt
Sol. given that g(x) + g(x + 4) = g(x + 2) + g(x + 6) ...(1)
putting x = x + 2 in (1) ........
g(x + 2) + g(x + 6) = g(x + 4) + g(x + 8) ...(2)
from (1) & (2)
g(x) = g(x + B)
Now, f(x) =
∫ x +8
x
g(t)
dt
2. If (a, b, c) is a point on the plane 3x + 2y + z = 7,
then find the least value of a 2 + b 2 + c 2 , using vector
methods.
Sol. Let → A = a î + b ĵ + c kˆ
⇒ → B = 3î + 2 ĵ + kˆ
f´(x) = g(x + 8) – g(x) = 0
⇒ g is constant function
4. If exactly three distinct chords from (h, 0) point to the
circle x 2 + y 2 = a 2 are bisected by the parabola
y 2 = 4ax, a > 0, then find the range of 'h' parameter.
Sol. Let M(at 2 , 2at) is mid-point of chord AB, then chord
⇒
→ →
(A.B)
2
≤ | → A | 2 | → B| 2
AB = T = S 1
3a + 2b + c ≤
(7) 2 ≤ (a 2 + b 2 + c 2 ) (14)
2
2
a + b + c 14
2
B
M
A
{Q 3a + 2b + c = 7, point lies on the plane}
a 2 + b 2 + c 2 49 7
≥ = 14 2
AB : x.at 2 + y.2at = a 2 t 4 + 4a 2 t 2
since AB chord passes through (h, 0)
XtraEdge for IIT-JEE 49 DECEMBER 2009