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Inverse Trigonometric Functions Test - 53

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Inverse Trigonometric Functions Test - 53
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  • Question 1
    1 / -0
    If $$\displaystyle \sin ^{ -1 }{ x } +\sin ^{ -1 }{ y } +\sin ^{ -1 }{ z } =\frac { 3\pi  }{ 2 } $$ and $$f\left( 1 \right) =1,f\left( p+q \right) =f\left( p \right) .f\left( q \right) \quad \forall p,q\in R$$ then $$\displaystyle { x }^{ f\left( 1 \right)  }+{ y }^{ f\left( 2 \right)  }+{ z }^{ f\left( 3 \right)  }-\frac { x+y+z }{ { x }^{ f\left( 1 \right)  }+{ y }^{ f\left( 2 \right)  }+{ z }^{ f\left( 3 \right)  } } =$$
    Solution
    Since $$\displaystyle -\frac { \pi  }{ 2 } \le \sin ^{ -1 }{ x } \le \frac { \pi  }{ 2 } $$
    $$\therefore \sin ^{ -1 }{ x } +\sin ^{ -1 }{ y } +\sin ^{ -1 }{ z } =\dfrac { 3\pi  }{ 2 } $$
    $$\displaystyle \sin ^{ -1 }{ x } =\sin ^{ -1 }{ y } =\sin ^{ -1 }{ z } =\frac { \pi  }{ 2 } $$
    $$\Rightarrow x=y=z=1$$
    Also $$f\left( p+q \right) =f\left( p \right) .f\left( q \right) \quad \forall p,q\in R$$   ....(1)
    Given $$f\left( 1 \right) =1$$
    From (1)
    $$f\left( 1+1 \right) =f\left( 1 \right) .f\left( 1 \right) \Rightarrow f\left( 2 \right) ={ 1 }^{ 2 }=1$$    ....(2)
    From (2) 
    $$f\left( 2+1 \right) =f\left( 2 \right) f\left( 1 \right) \Rightarrow f\left( 2 \right) ={ 1 }^{ 2 }.1=1$$ 
    Now $$\displaystyle { x }^{ f\left( 1 \right)  }+{ y }^{ f\left( 2 \right)  }+{ z }^{ f\left( 3 \right)  }-\frac { x+y+z }{ { x }^{ f\left( 1 \right)  }+{ y }^{ f\left( 2 \right)  }+{ z }^{ f\left( 3 \right)  } } =3-\frac { 3 }{ 3 } =2$$
  • Question 2
    1 / -0
    Exhaustive set of values of parameter $$a$$ so that $$\sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } =a\quad $$ has a solution is
    Solution
    Let $$f(x)=\sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } $$
    $$sin^{-1}x \space and \space \tan^{-1}x $$ are increasong functions between x=[-1,1].
    Minimum value of $$f(x) is -\pi/4$$ and  maximum value is $$\pi/4$$.
    Hence the correct option is answer B.
  • Question 3
    1 / -0
    If $$\displaystyle \sum _{ i=1 }^{ 2n }{ \sin ^{ -1 }{ { x }_{ i } }  } =n\pi $$, then $$\displaystyle \sum _{ i=1 }^{ 2n }{ { x }_{ i } } $$ is equal to
    Solution
    We have $$\sum _{ i=1 }^{ 2n }{ \sin ^{ -1 }{ { x }_{ i } }  } =n\pi $$
    $$\Rightarrow \sin ^{ -1 }{ { x }_{ 1 } } +\sin ^{ -1 }{ { x }_{ 2 } } +...+\sin ^{ -1 }{ { x }_{ 2n } } =n\pi $$
    Let $$\sin ^{ -1 }{ { x }_{ 1 } } ={ \alpha  }_{ 1 },\sin ^{ -1 }{ { x }_{ 2 } } ={ \alpha  }_{ 2 },...,\sin ^{ -1 }{ { x }_{ 2n } } ={ \alpha  }_{ n }$$
    $$\therefore { \alpha  }_{ 1 }+{ \alpha  }_{ 2 }+...+{ \alpha  }_{ 2n }=n\pi $$    ....(1)
    $$\Rightarrow { x }_{ 1 }=\sin { { \alpha  }_{ 1 } } ,{ x }_{ 2 }=\sin { { \alpha  }_{ 2 } } ,...,{ x }_{ 2n }=\sin { { \alpha  }_{ 2n } } $$
    $$\displaystyle \therefore \sum _{ i=1 }^{ 2n }{ { x }_{ i } } ={ x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 2n }=\sin { { \alpha  }_{ 1 } } +\sin { { \alpha  }_{ 2 } } +...+\sin { { \alpha  }_{ 2n } } $$
    Clearly, from (1) $$\displaystyle { \alpha  }_{ i }=\frac { \pi  }{ 2 } ,\quad \forall i=1,2,...2n$$
    $$\therefore \sum _{ i=1 }^{ 2n }{ { x }_{ i } } =1+1+...+1=2n$$
  • Question 4
    1 / -0
    The set of values of parameter $$a$$ so that the equation $$\displaystyle (\sin^{-1}x)^{3}+(\cos^{-1}x)^{3}=a\pi^{3}$$ has a solution. 
    Solution
    Let  $$y = (\sin^{-1}x)^{3}+(\cos^{-1}x)^{3} = (\sin^{-1}x+\cos^{-1}x)[(\sin^{-1}x)^2+(\cos^{-1}x)^2-\sin^{-1}x.\cos^{-1}x]$$
    $$\Rightarrow y = \cfrac{\pi}{2}[(\sin^{1-}x+\cos^{-1}x)^2-3\sin^{-1}x.\cos^{-1}x]$$
    $$\Rightarrow y = \cfrac{\pi}{2}\left[\cfrac{\pi^2}{4}-3\sin^{-1}x\left(\cfrac{\pi}{2}-\sin^{-1}x\right)\right ]$$
    $$\Rightarrow y = \cfrac{\pi}{2}\left[\cfrac{\pi^2}{16}+3\left(\cfrac{\pi}{4}-\sin^{-1}x\right)^2\right ]$$
    Now we know $$-\cfrac{\pi}{2} \leq \sin^{-1}x \leq \cfrac{\pi}{2}$$
    Thus Range of $$y$$ is $$\left[\cfrac{\pi^3}{32}, \cfrac{7\pi^3}{8}\right]$$
    Thus given equation to have solution $$a \in \left[\cfrac{1}{32}, \cfrac{7}{8}\right]$$
  • Question 5
    1 / -0
    The number of real solutions of $$(x, y)$$, where is $$\displaystyle \:\left | y \right |= \sin x,y= \cos ^{-1}\left ( \cos x  \right),-2\pi \leq x\leq 2\pi ,$$ is
    Solution

    $$\left| y \right| =\sin  x,\quad y=\cos ^{ -1 } \left( \cos  x \right) ,\quad -2\pi \leq x\leq 2\pi$$
    $$\left| y \right| =\sin  x\quad ----\boxed { 1 } \\ y=\cos ^{ -1 } \left( \cos  x \right) ,\quad -2\pi \leq x\leq 2\pi \quad ----\boxed { 2 } \\ \rightarrow y=x\quad \quad \quad \quad when\quad 0\leq x\leq \pi \\ \rightarrow y=2\pi -x\quad \quad \quad \quad \quad \quad \pi \leq x\leq 2\pi \\ \rightarrow y=-2\pi -x\quad \quad \quad \quad \quad -2\pi \leq x\leq -\pi \\ \rightarrow y=-x\quad \quad \quad \quad \quad \quad \quad \quad -\pi\leq x\leq 0$$
    from (1) and (2)
    three different cases are $$|x| = \sin{x}$$, $$|2\pi-x| = \sin{x}$$ and $$|2\pi+x| = \sin{x}$$

    For the above simultaneous equations, we get exactly 3 solutions.
    For $$x=0$$ we get $$y=0$$ for both the equations,
    For $$x=k(2\pi)$$ where $$k$$ is an integer, we get $$y=0$$ in both the cases.
    Since $$x\epsilon[-2\pi,2\pi]$$ we get
    $$x=-2\pi$$ and $$x=2\pi$$.
    Hence 3 solutions,
    at $$x=0$$
    $$x=-2\pi$$ and
    $$x=2\pi$$

    Hence, no of real solutions is 3

  • Question 6
    1 / -0
    $$\displaystyle \:\cos ^{-1}\left \{ \dfrac{1}{2}x^{2}+\sqrt{1-x^{2}.}\sqrt{1-\dfrac{x^{2}}{4}} \right \}= \cos ^{-1}\dfrac{x}{2}-\cos ^{-1}x$$ holds for 
    Solution
    For the domain of the following function,
    $$1-x^{2}\geq 0$$ and $$1-\frac{x^{2}}{4}\geq0$$
    Therefore
    $$1\geq x^2$$ and $$4\geq x^2$$
    $$|x|\leq 1$$ and $$|x|\leq 2$$
    Hence $$x\epsilon[-1,1]$$
    $$\displaystyle \:\cos ^{-1}\left \{ \frac{1}{2}x^{2}+\sqrt{1-x^{2}.}\sqrt{1-\frac{x^{2}}{4}} \right \}= \cos ^{-1}\frac{x}{2}-\cos ^{-1}x$$
    L.H.S is always positive.
    $$\Rightarrow \cos ^{-1}\frac{x}{2}-\cos ^{-1}x >0$$
    which is true for all $$0\leq x\leq 1$$
    $$\therefore$$ given equation holds for $$0\leq x\leq 1$$

  • Question 7
    1 / -0
    If $$\displaystyle \sin ^{ -1 }{ x } +\sin ^{ -1 }{ y } +\sin ^{ -1 }{ z } =\frac { 3\pi  }{ 2 } $$, then $$\displaystyle \frac { \sum _{ k=1 }^{ 2 }{ \left( { x }^{ 100k }+{ y }^{ 106k } \right)  }  }{ \sum { { x }^{ 207 }.{ y }^{ 207 } }  } $$ is
    Solution
    We have $$\displaystyle \sin ^{ -1 }{ x } +\sin ^{ -1 }{ y } +\sin ^{ -1 }{ z } =\frac { 3\pi  }{ 2 } $$

    It is possible only when 

    $$\displaystyle \sin ^{ -1 }{ x } =\frac { \pi  }{ 2 } ,\sin ^{ -1 }{ y } =\frac { \pi  }{ 2 } $$ and $$\displaystyle \sin ^{ -1 }{ z } =\frac { \pi  }{ 2 } $$

    [$$\because\dfrac{-\pi}{2}\leq\sin^{-1}x\leq\dfrac{\pi}{2}$$]

    $$\Rightarrow x=1,y=1$$ and $$z=1$$

    $$\displaystyle \quad \therefore \frac { \sum _{ k=1 }^{ 2 }{ \left( { x }^{ 100k }+{ y }^{ 106k } \right)  }  }{ \sum { { x }^{ 207 }.{ y }^{ 207 } }  } =\frac { \left( { x }^{ 100 }+{ y }^{ 106 } \right) +\left( { x }^{ 200 }+{ y }^{ 212 } \right)  }{ { x }^{ 207 }.{ y }^{ 207 }+{ y }^{ 207 }.{ z }^{ 207 }+{ z }^{ 207 }.{ x }^{ 207 } } $$

    $$\displaystyle =\frac { 1+1+1+1 }{ 1+1+1 } =\frac { 4 }{ 3 } $$
  • Question 8
    1 / -0
    The product of all real values of x satisfying the equation
    $$\sin^{-1}\cos \left ( \dfrac{2x^{2}+10\left | x \right |+4}{x^{2}+5\left | x \right |+3} \right )=\cot \left ( \cot ^{-1}\left ( \dfrac{2-18\left | x \right |}{9\left | x \right |} \right ) \right )+\dfrac{\pi }{2}$$ is
    Solution
    $$sin^{ -1 }cos\left( \cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 }  \right) =cot\left( cot^{ -1 }\left( \cfrac { 2-18x }{ 9x }  \right)  \right) +\cfrac { \pi  }{ 2 } \\ \Rightarrow sin^{ -1 }sin\left( \cfrac { \pi  }{ 2 } -\left( \cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 }  \right)  \right) =\cfrac { 2-18x }{ 9x } +\cfrac { \pi  }{ 2 } \\ \Rightarrow \cfrac { \pi  }{ 2 } -\cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 } =\cfrac { 2-18x }{ 9x } +\cfrac { \pi  }{ 2 } \\ \Rightarrow -18x^{ 3 }-90x^{ 2 }-36x=2x^{ 2 }+10x+6-18x^{ 3 }-90x-54x\\ \Rightarrow 2x^{ 2 }-8x+6=0\Rightarrow x^{ 2 }-4+3=0\Rightarrow \left( x-1 \right) \left( x-3 \right) =0$$
    Therefore product of roots is $$1\times 3=3$$
  • Question 9
    1 / -0
    The value of $$a$$ for which $$\displaystyle ax^{2}+sin^{-1}(x^{2}-2x+2)+cos^{-1}(x^{2}-2x+2)=0$$ has a real solution is 
    Solution
    $$ax^{2}+sin^{-1}((x-1)^{2}+1)+cos^{-1}((x-1)^{2}+1)=0$$
    $$ax^{2}+\frac{\pi}{2}=0$$
    $$x^{2}=\frac{-\frac{\pi}{2}}{a}$$
    $$x^{2}=\frac{-\pi}{2a}$$ ...(i)
    And
    $$-1\leq(x-1)^{2}+1\leq1$$
    $$-2\leq(x-1)^{2}\leq0$$
    $$(x-1)^{2}\leq0$$
    There is only one possibility
    $$(x-1)^{2}=0$$
    $$x=1$$
    Substituting in 1, we get
    $$a=-\frac{\pi}{2}$$
  • Question 10
    1 / -0
    If $$\displaystyle sin^{-1}x+sin^{-1}y+sin^{-1}z=\pi$$, then $$x^{4}+y^{4}+z^{4}+4x^{2}y^{2}z^{2}=k(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})$$, where $$k$$ is equal to
    Solution
    $$sin^{-1}(x)+sin^{-1}(y)=\pi-sin^{-1}(z)$$
    $$sin^{-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})=\pi-sin^{-1}(z)$$
    $$x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}=z$$
    $$x^{2}(1-y^{2})+y^{2}(1-x^{2})+2xy\sqrt{(1-x^{2})(1-y^{2})}=z^2$$
    $$x^{2}+y^{2}-2x^{2}y^{2}+2xy\sqrt{(1-x^{2})(1-y^{2})}=z^{2}$$
    $$x^{2}+y^{2}-z^{2}=2x^{2}y^{2}-2xy\sqrt{(1-x^{2})(1-y^{2})}$$
    Squaring and simplifying, we get
    $$x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+z^2y^2+x^2z^2)$$
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