Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

Maths 35 Years JEE Mains & Advance Past year Papers Class 12

JEE : Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

The document Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev is a part of the JEE Course Maths 35 Years JEE Mains & Advance Past year Papers Class 12.
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Q. 1. The values of  f(x)  Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev lie in the interval

Ans.  Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

Solution. For the given function to be defined

Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

and sine function increases on [0, π/4]

Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

Q. 2. For the function 

Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

the derivative from the right, f '(0+) = ......................., and the derivative from the left, f '(0–) = ...............

Ans. 0, 1

Solution. 

Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev
Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

Thus  f ' (0+) = 0 and f ' (0) = 1

Q. 3. The domain  of  the function  Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev given by .................

Ans. [–2, –1] ∪ [1, 2]

Solution. To find domain of function  Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

For f (x) to be defined we should have  Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

NOTE THIS STEP :

Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev


Q. 4. Let A be a set of n distinct elements. Then the total number of distinct functions  from A to A is ................. and out of these ................. are onto functions.

Ans. nn, n!

Solution. Set A has n distinct elements. Then to define a function from A to A, we need to associate each element of set A to any one the n elements of set A. So total number of functions from set A to set A is equal to the number of ways of doing n jobs where each job can be done in n ways. The total number such ways is n × n × n × .... × n (n - times).
Hence the total number of functions from A to A is nn.
Now for an onto function from A to A, we need  to associate each element of A to one and only one element of A. So total number of onto functions from set A to A is equal to number of ways of arranging n elements in range (set A) keeping n elements fixed in domain (set A). n elements can be arranged in n! ways.
Hence, the total number of functions from A to A is n!.

Q. 5. If  Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev, then domain of f(x) is .... and its range is .................

Ans. (-2, 1), [-1, 1]

Solution. The given function is,

  Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev  

Combining these two inequalities, we get x ∈ (– 2, 1)
∴ Domain of f is (– 2, 1)
Also sin q always lies in [– 1, 1].
∴ Range of f is [– 1, 1]


Q. 6. There are exactly two distinct linear functions, ................., and ........... which map [– 1, 1] onto [0, 2].

Ans. x + 1 and - x+ 1

Solution. KEY CONCEPT : Every linear function is either strictly increasing or strictly decreasing. If  f (x) = ax + b,  Df  = [p, q], Rf  = [m, n]
Then f (p) = m and  f (q) = n,  if f (x) is strictly increasing and f (p) = n, f (q) = m, If f (x) is strictly decreasing function.
Let f (x) = ax + b be the linear function which maps [–1, 1] onto [0, 2]. then f (–1) = 0 and f (1) = 2 or f (–1) = 2 and f (1) = 0
Depending upon f (x) is increasing or decreasing respectively.
⇒ – a + b = 0 and a + b = 2 ....(1)
or – a + b = 2 and a + b = 0 ....(2)
Solving (1), we get a = 1, b = 1.
Solving (2), we get a = – 1, b = 1
Thus there are only two functions i.e., x + 1 and – x + 1.

Q. 7. If f is an even function defined on the interval (-5, 5), then four real values of x satisfying the equation Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

are ................., ................., ................., and ...........

Ans. Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

Solution. Given that   Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev and f is an even function
Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev
Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev
Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

Q. 8. If f(x) = sin2 x + Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev then (gof) (x) = .................

Ans. 1

Solution. 

Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRevFill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

Fill Ups of Functions, Past year Questions JEE Advance, Class 12, Maths JEE Notes | EduRev

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