हिंदी

If F ( X ) = Sin 2 X and the Composite Function G ( F ( X ) ) = | Sin X | Then G(X) is Equal to (A) √ X − 1 (B) √ X (C) √ X + 1 (D) − √ X - Mathematics

Advertisements
Advertisements

प्रश्न

If  \[f\left( x \right) = \sin^2 x\] and the composite function   \[g\left( f\left( x \right) \right) = \left| \sin x \right|\] then g(x) is equal to

विकल्प

  • \[\sqrt{x - 1}\]

  • \[\sqrt{x}\]

  • \[\sqrt{x + 1}\]

  • \[- \sqrt{x}\]

MCQ
Advertisements

उत्तर

(b) \[\text{If we takeg}\left( x \right) = \sqrt{x}, \text{then}\] 
\[g\left( f\left( x \right) \right) = g\left( \sin^2 x \right) = \sqrt{\sin^2 x} = \pm \sin x = \left| \sin x \right|\]

So, the answer is (b).

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Functions - Exercise 2.6 [पृष्ठ ७९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 2 Functions
Exercise 2.6 | Q 46 | पृष्ठ ७९

वीडियो ट्यूटोरियलVIEW ALL [5]

संबंधित प्रश्न

Show that the function f in `A=R-{2/3} ` defined as `f(x)=(4x+3)/(6x-4)` is one-one and onto hence find f-1


Check the injectivity and surjectivity of the following function:

f : N → N given by f(x) = x2


Check the injectivity and surjectivity of the following function:

f : N → N given by f(x) = x3


Show that the modulus function f : R → R given by f(x) = |x| is neither one-one nor onto, where |x| is x if x is positive or 0 and |x| is − x if x is negative.


Show that the function f : R → R given by f(x) = x3 is injective.


Give examples of two functions fN → Z and gZ → Z such that g o f is injective but gis not injective.

(Hint: Consider f(x) = x and g(x) =|x|)


Let A = {−1, 0, 1} and f = {(xx2) : x ∈ A}. Show that f : A → A is neither one-one nor onto.


Classify the following function as injection, surjection or bijection :

f : Q − {3} → Q, defined by `f (x) = (2x +3)/(x-3)`


Show that f : R→ R, given by f(x) = x — [x], is neither one-one nor onto.


Let R+ be the set of all non-negative real numbers. If f : R+ → R+ and g : R+ → R+ are defined as `f(x)=x^2` and `g(x)=+sqrtx` , find fog and gof. Are they equal functions ?


Let f : R → R and g : R → R be defined by f(x) = + 1 and (x) = x − 1. Show that fog = gof = IR.


Give examples of two functions f : N → N and g : N → N, such that gof is onto but f is not onto.


If f : A → B and g : B → C are one-one functions, show that gof is a one-one function.


If f : A → B and g : B → C are onto functions, show that gof is a onto function.


Find fog and gof  if : f(x) = sin−1 x, g(x) = x2


Find fog and gof  if : f(x)= x + 1, g (x) = 2x + 3 .


Find fog and gof  if : f(x) = c, c ∈ R, g(x) = sin `x^2`


Let  f  be any real function and let g be a function given by g(x) = 2x. Prove that gof = f + f.


   if `f (x) = sqrt(1-x)` and g(x) = `log_e` x are two real functions, then describe functions fog and gof.


Let

f (x) =`{ (1 + x, 0≤ x ≤ 2) , (3 -x , 2 < x ≤ 3):}`

Find fof.


State with reason whether the following functions have inverse :
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}


Let A and B be two sets, each with a finite number of elements. Assume that there is an injective map from A to B and that there is an injective map from B to A. Prove that there is a bijection from A to B.


Let `f : R - {- 3/5}` → R be a function defined as `f  (x) = (2x)/(5x +3).` 

f-1 : Range of f → `R -{-3/5}`.


Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. State whether f is one-one or not.


Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}                                                                                                        [NCERT EXEMPLAR]


The function f : R → R defined by

`f (x) = 2^x + 2^(|x|)` is 

 


Let 

\[A = \left\{ x \in R : - 1 \leq x \leq 1 \right\} = B\] Then, the mapping\[f : A \to \text{B given by} f\left( x \right) = x\left| x \right|\] is 

 


Let M be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then, the function f : M→ R defined by f(A) = |A| for every A ∈ M, is

 


If \[f : R \to R is given by f\left( x \right) = 3x - 5, then f^{- 1} \left( x \right)\] 

 


If \[g \left( f \left( x \right) \right) = \left| \sin x \right| \text{and} f \left( g \left( x \right) \right) = \left( \sin \sqrt{x} \right)^2 , \text{then}\]

 


Mark the correct alternative in the following question:
Let A = {1, 2, ... , n} and B = {a, b}. Then the number of subjections from A into B is


Let f: R → R be given by f(x) = tan x. Then f–1(1) is ______.


Let f : R → R be a function defined by f(x) `= ("e"^abs"x" - "e"^-"x")/("e"^"x" + "e"^-"x")` then f(x) is


The function f: R → R defined as f(x) = x3 is:


A general election of Lok Sabha is a gigantic exercise. About 911 million people were eligible to vote and voter turnout was about 67%, the highest ever


Let I be the set of all citizens of India who were eligible to exercise their voting right in the general election held in 2019. A relation ‘R’ is defined on I as follows:

R = {(V1, V2) ∶ V1, V2 ∈ I and both use their voting right in the general election - 2019}

  • Three friends F1, F2, and F3 exercised their voting right in general election-2019, then which of the following is true?

Raji visited the Exhibition along with her family. The Exhibition had a huge swing, which attracted many children. Raji found that the swing traced the path of a Parabola as given by y = x2.

Answer the following questions using the above information.

  • Let f: {1,2,3,....} → {1,4,9,....} be defined by f(x) = x2 is ____________.

Function f: R → R, defined by f(x) = `x/(x^2 + 1)` ∀ x ∈ R is not


Let [x] denote the greatest integer ≤ x, where x ∈ R. If the domain of the real valued function f(x) = `sqrt((|[x]| - 2)/(|[x]| - 3)` is (–∞, a) ∪ [b, c) ∪ [4, ∞), a < b < c, then the value of a + b + c is ______.


Let f(x) = ax (a > 0) be written as f(x) = f1(x) + f2(x), where f1(x) is an even function and f2(x) is an odd function. Then f1(x + y) + f1(x – y) equals ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×