हिंदी

The number of real solutions of the equation 1+cos2x=2cos-1(cosx) in [π2,π] is ______. - Mathematics

Advertisements
Advertisements

प्रश्न

The number of real solutions of the equation `sqrt(1 + cos 2x) = sqrt(2) cos^-1 (cos x)` in `[pi/2, pi]` is ______.

विकल्प

  • 0

  • 1

  • 2

  • Infinite

MCQ
रिक्त स्थान भरें
Advertisements

उत्तर

The number of real solutions of the equation `sqrt(1 + cos 2x) = sqrt(2) cos^-1 (cos x)` in `[pi/2, pi]` is 0.

Explanation:

We have `sqrt(1 + cos 2x) = sqrt(2) cos^-1 (cos x)` 

⇒ `sqrt(2 cos^2x) = sqrt(2)x`        ...`[because cos^-1 (cos x) = x]`

⇒ `sqrt(2) cos x = sqrt(2)x`

⇒ cos x = x

∴ There are no solution for given equation.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Inverse Trigonometric Functions - Exercise [पृष्ठ ३९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 2 Inverse Trigonometric Functions
Exercise | Q 36 | पृष्ठ ३९

संबंधित प्रश्न

Prove `tan^(-1)  2/11 + tan^(-1)  7/24 = tan^(-1)  1/2`


Find the value of the following:

`tan^-1 [2 cos (2  sin^-1  1/2)]`


Find the value of the following:

`tan  1/2 [sin^(-1)  (2x)/(1+ x^2) + cos^(-1)  (1-y^2)/(1+y^2)]`, |x| < 1, y > 0 and xy < 1


Find the value of the given expression.

`sin^(-1) (sin  (2pi)/3)`


Find the value of the given expression.

`tan^(-1) (tan  (3pi)/4)`


Prove that:

`sin^(-1)  8/17 + sin^(-1)  3/5 = tan^(-1)  77/36`


Prove that:

`tan^(-1) sqrtx = 1/2 cos^(-1)  (1-x)/(1+x)`, x ∈ [0, 1]


Solve the following equation:

2 tan−1 (cos x) = tan−1 (2 cosec x)


sin–1 (1 – x) – 2 sin–1 x = `pi/2`, then x is equal to ______.


Solve  `tan^(-1) -  tan^(-1)  (x - y)/(x+y)` is equal to

(A) `pi/2`

(B). `pi/3` 

(C) `pi/4` 

(D) `(-3pi)/4`


Solve for x : \[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\] .


Find: ∫ sin x · log cos x dx


Find the value of the expression in terms of x, with the help of a reference triangle

sin (cos–1(1 – x))


Find the value of the expression in terms of x, with the help of a reference triangle

cos (tan–1 (3x – 1))


Find the value of `cot[sin^-1  3/5 + sin^-1  4/5]`


Find the value of  `tan(sin^-1  3/5 + cot^-1  3/2)`


Prove that `tan^-1x + tan^-1y + tan^-1z = tan^-1[(x + y + z - xyz)/(1 - xy - yz - zx)]`


Choose the correct alternative:

If `cot^-1(sqrt(sin alpha)) + tan^-1(sqrt(sin alpha))` = u, then cos 2u is equal to


Evaluate: `sin^-1 [cos(sin^-1 sqrt(3)/2)]`


Evaluate `cos[sin^-1  1/4 + sec^-1  4/3]`


Prove that cot–17 + cot–18 + cot–118 = cot–13


If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression.

`tan[tan^-1("d"/(1 + "a"_1 "a"_2)) + tan^-1("d"/(21 + "a"_2 "a"_3)) + tan^-1("d"/(1 + "a"_3 "a"_4)) + ... + tan^-1("d"/(1 + "a"_("n" - 1) "a""n"))]`


The value of cot–1(–x) for all x ∈ R in terms of cot–1x is ______.


The value of `"tan"^-1 (1/2) + "tan"^-1 (1/3) + "tan"^-1 (7/8)` is ____________.


If `"tan"^-1 ("cot"  theta) = 2theta, "then"  theta` is equal to ____________.


`"cot" (pi/4 - 2  "cot"^-1  3) =` ____________.


If `"cot"^-1 (sqrt"cos" alpha) - "tan"^-1 (sqrt"cos" alpha) = "x",` the sinx is equal to ____________.


The domain of the function defind by f(x) `= "sin"^-1 sqrt("x" - 1)` is ____________.


`"cot" ("cosec"^-1  5/3 + "tan"^-1  2/3) =` ____________.


`"cos"^-1["cos"(2"cot"^-1(sqrt2 - 1))]` = ____________.


The set of all values of k for which (tan–1 x)3 + (cot–1 x)3 = kπ3, x ∈ R, is the internal ______.


If `tan^-1 ((x - 1)/(x + 1)) + tan^-1 ((2x - 1)/(2x + 1)) = tan^-1 (23/36)` = then prove that 24x2 – 23x – 12 = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×