KEAM Paper 2 Mock Test - 1 - JEE MCQ

(a, b)
g(x) = f(x) + f(–x)
g(–x) = f(–x) + f(x) = g(x)
therefore g(x) is even

Tangent at (1, − 2) to x² + y² = 5 is 
x − 2y − 5 = 0 ... (i). 
Centre and radius of 
x² + y² − 8x + 6y + 20 = 0 are 
C (4, − 3) and radius r = √5.  
Perpendicular distance from 
C (4, − 3) to (i) is radius. 
∴ (i) is also a tangent to the second circle. 
Let P (h, k) be the foot of the drawn circle from C (4, − 3) on (i)
h-4/1 = k=3/-2 = - [1.4 -2. (-3)-5]  
∴ (h, k) = (3, −1)

f(x) = x³ – 3x + [a]
Let [a] = t (where t will be an integer)
f(x) = x³ – 3x + t ……….(i)
⇒ f ’(x) = 3x² – 3
⇒ f ‘(x) = 0 has two real and distinct solution which are x = 1 and x = -1
so  f(x) = 0 will have three distinct and real solution when  f (1). f(-1) < 0 ……………. (ii)
Now,
f(1) = (1)³ -3(1) + t = t – 2
f(–1) = (–1)³ – 3 (–1) + t = t + 2
From equation (ii)
(t –2)  (t + 2) < 0
⇒ t ∈ (-2, 2)
Now t = [a]
Hence [a] ∈ (-2, 2)
 ⇒ a ∈ [-1, 2)

f(0) = cos 0 = 1 and f(2π) = cos (2π) = 1
So, f is not one to one and cos(x) lie between - 1 and 1.
Therefore, range is not equal to its codomain.
Hence, it is not onto.