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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

If tan^ - 1 (1)/(2x + 1) + tan^ - 1 (1)/(4x + 1) = cot^ - 1 ( x^ 2...

If tan−112x+1+tan−114x+1=cot−1(x22),thenthenumberofallpossiblevaluesoftan^{- 1}\dfrac{1}{2x + 1} + tan^{- 1}\dfrac{1}{4x + 1} = cot^{- 1} \left(\dfrac{x^{2}}{2}\right), then the number of all possible values of xis/areis/are

Options

  1. A11
  2. B22
  3. C33
  4. D00

Correct answer

B. 22

Step-by-step solution

tan−1({1(2x+1)}+{1(4x+1)}1−{1(2x+1)}⋅{1(4x+1)})=tan−12x2tan^{- 1}\left(\dfrac{\left\{\dfrac{1}{\left(2x + 1\right)}\right\} + \left\{\dfrac{1}{\left(4x + 1\right)}\right\}}{1 - \left\{\dfrac{1}{\left(2x + 1\right)}\right\} ⋅ \left\{\dfrac{1}{\left(4x + 1\right)}\right\}}\right) = tan^{- 1}\dfrac{2}{x^{2}}
tan−1(6x+28x2+6x)=tan−12x2tan^{- 1}\left(\dfrac{6x + 2}{8x^{2} + 6x}\right) = tan^{- 1}\dfrac{2}{x^{2}}
Therefore, 3x+14x2+3x=2x2\dfrac{3x + 1}{4x^{2} + 3x} = \dfrac{2}{x^{2}}
x2(3x+1)=2x^{2} \left(3x + 1\right) = 2\left(4x^{2} + 3x\right)<br/><br />3x^{3} - 7x^{2} - 6x = 0<br/>x(3x2−7x−6)<br />x \left(3x^{2} - 7x - 6\right) = 0<br/>x(x−3)(3x+2)=0<br />x\left( x - 3\right)\left(3x + 2\right) = 0
We take x=3,0x = 3,0 because when x=−23,LHSx = - \dfrac{2}{3},LHS of the given equation will be negative whereas tan−1(2x2)tan^{- 1} \left(\dfrac{2}{x^{2}}\right)ispositive.<br/>Hence,thenumberofvaluespositive.<br />Hence, the number of values= 2
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