English

The trigonometric equation tan–1x = 3tan–1a has solution for ______.

Advertisements
Advertisements

Question

The trigonometric equation tan–1x = 3tan–1 a has solution for ______.

Options

  • `|a| ≤ 1/sqrt(3)`

  • `|a| > 1/sqrt(3)`

  • `|a| < 1/sqrt(3)`

  • all real value of a.

MCQ
Fill in the Blanks
Advertisements

Solution

The trigonometric equation tan–1x = 3tan–1a has solution for `underlinebb(|a| < 1/sqrt(3))`.

Explanation:

To solve the equation tan–1(x) = 3tan–1(a), we use the tangent function properties and transformations.

Let θ = tan–1(a).

Then:

x = tan(3θ)

Using the triple-angle formula for tangent:

tan(3θ) = `(3tan(θ) - tan^3(θ))/(1 - 3tan^3(θ))`

Since tan(θ) = a, substituting a in gives us:

x = `(3a - a^3)/(1 - 3a^2)`

For the function tan⁡−1(x) = 3tan−1(a) to have a solution, the argument of tan (which is 3θ) must be within the range of the tan function, which is `(-π/2, π/2)`.

Therefore, 3θ must also be `(-π/2, π/2)`.

Given that θ = tan−1(a) is within `(-π/2, π/2)`, the condition for 3θ to remain in this interval is:

`-π/6 < θ < π/6`

This translates to:

`-π/6 < tan^-1(a) < π/6`

Taking the tangent of the bounds:

`-1/sqrt(3) < a < 1/sqrt(3)`

Thus, the condition for a is:

`|a| < 1/sqrt(3)`

shaalaa.com
  Is there an error in this question or solution?
2024-2025 (March) Specimen Paper

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

Let f : R → R be defined as f(x) = 3x. Choose the correct answer.


Show that the function f : R → {x ∈ R : –1 < x < 1} defined by f(x) = `x/(1 + |x|)`, x ∈ R is one-one and onto function.


Let fR → R be the Signum Function defined as

f(x) = `{(1,x>0), (0, x =0),(-1, x< 0):}`

and gR → be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1]?


Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.


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


Find gof and fog when f : R → R and g : R → R is defined by  f(x) = x and g(x) = |x| .


Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.


Find fog and gof  if : f(x) = `x^2` + 2 , g (x) = 1 − `1/ (1-x)`.


Let f(x) = x2 + x + 1 and g(x) = sin x. Show that fog ≠ gof.


If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?


If f : R → R is given by f(x) = x3, write f−1 (1).


Which of the following functions form Z to itself are bijections?

 

 

 
 

Let

\[A = \left\{ x : - 1 \leq x \leq 1 \right\} \text{and} f : A \to \text{A such that f}\left( x \right) = x|x|\]

 


Mark the correct alternative in the following question:
If the set A contains 7 elements and the set B contains 10 elements, then the number one-one functions from A to B is


Mark the correct alternative in the following question:
Let f :  \[-\] \[\left\{ \frac{3}{5} \right\}\] \[\to\]  R be defined by f(x) = \[\frac{3x + 2}{5x - 3}\] Then,

 


Let the function f: R → R be defined by f(x) = 4x – 1, ∀ x ∈ R. Then, show that f is one-one.


Let f: R → R be defined by f(x) = 3x – 4. Then f–1(x) is given by ______.


The domain of the function `"f"("x") = 1/(sqrt ({"sin x"} + {"sin" ( pi + "x")}))` where {.} denotes fractional part, is


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. Based on the given information, f is best defined as:


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}

  • Mr. ’X’ and his wife ‘W’ both exercised their voting right in the general election-2019, Which of the following is true?

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 ____________.

A function f: x → y is said to be one – one (or injective) if:


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


Consider a set containing function A= {cos–1cosx, sin(sin–1x), sinx((sinx)2 – 1), etan{x}, `e^(|cosx| + |sinx|)`, sin(tan(cosx)), sin(tanx)}. B, C, D, are subsets of A, such that B contains periodic functions, C contains even functions, D contains odd functions then the value of n(B ∩ C) + n(B ∩ D) is ______ where {.} denotes the fractional part of functions)


The graph of the function y = f(x) is symmetrical about the line x = 2, then ______.


Let S = {1, 2, 3, 4, 5, 6, 7}. Then the number of possible functions f: S `rightarrow` S such that f(m.n) = f(m).f(n) for every m, n ∈ S and m.n ∈ S is equal to ______.


For x ∈ R, x ≠ 0, let f0(x) = `1/(1 - x)` and fn+1 (x) = f0(fn(x)), n = 0, 1, 2, .... Then the value of `f_100(3) + f_1(2/3) + f_2(3/2)` is equal to ______.


A function f : [– 4, 4] `rightarrow` [0, 4] is given by f(x) = `sqrt(16 - x^2)`. Show that f is an onto function but not a one-one function. Further, find all possible values of 'a' for which f(a) = `sqrt(7)`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×