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Inverse Trigonometric Functions Test - 54

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Inverse Trigonometric Functions Test - 54
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  • Question 1
    1 / -0
    If $$\displaystyle { \left( \tan ^{ -1 }{ x }  \right)  }^{ 2 }+{ \left( \cot ^{ -1 }{ x }  \right)  }^{ 2 }=\frac { 5{ \pi  }^{ 2 } }{ 8 } $$, then $$x$$ equals
    Solution
    We have $$\displaystyle { \left( \tan ^{ -1 }{ x }  \right)  }^{ 2 }+{ \left( \cot ^{ -1 }{ x }  \right)  }^{ 2 }=\frac { 5{ \pi  }^{ 2 } }{ 8 } $$

    $$\displaystyle \Rightarrow { \left( \tan ^{ -1 }{ x } +\cot ^{ -1 }{ x }  \right)  }^{ 2 }-2\tan ^{ -1 }{ x } \left( \frac { \pi  }{ 2 } -\tan ^{ -1 }{ x }  \right) =\frac { 5{ \pi  }^{ 2 } }{ 8 } $$

    $$\displaystyle \Rightarrow \frac { \pi ^2 }{ 4 } -2.\frac { \pi  }{ 2 } \tan ^{ -1 }{ x } +2{ \left( \tan ^{ -1 }{ x }  \right)  }^{ 2 }=\frac { \pi^2  }{ 8 } $$

    $$\displaystyle \Rightarrow 2{ \left( \tan ^{ -1 }{ x }  \right)  }^{ 2 }-\pi \tan ^{ -1 }{ x } -\frac { 3{ \pi  }^{ 2 } }{ 8 } =0$$

    $$\displaystyle \Rightarrow \tan ^{ -1 }{ x } =-\frac { \pi  }{ 4 } ,\frac { 3\pi  }{ 4 } $$

    $$\displaystyle \Rightarrow \tan ^{ -1 }{ x } =-\frac { \pi  }{ 4 } \Rightarrow x=-1$$
  • Question 2
    1 / -0

    Directions For Questions

    $$\displaystyle \cos^{-1}x+(\sin^{-1}y)^{2}= \frac{p\pi ^{2}}{4}$$ and $$\displaystyle(\cos^{-1}x)(\sin^{-1}y)^{2} = \frac{\pi ^{4}}{16},p\epsilon Z $$

    ...view full instructions

    The value of p for which system has a solution is
    Solution
    $$cos^{ -1 }x+\left( sin^{ -1 }y \right) ^{ 2 }=\cfrac { p\pi ^{ 2 } }{ 4 } $$   ...(1)
    $$\left( cos^{ -1 }x \right) \left( sin^{ -1 }y \right) ^{ 2 }=\cfrac { \pi ^{ 4 } }{ 16 } $$
    Therefore equation with roots $$cos^{ -1 }x$$ and $$\left( sin^{ -1 }y \right) ^{ 2 }$$ is
    $$x^{ 2 }-\cfrac { p\pi ^{ 2 } }{ 4 } x+\cfrac { \pi ^{ 4 } }{ 16 } =0$$ 
    And for real values of x
    $$\cfrac { p^{ 2 }\pi ^{ 4 } }{ 16 } -\cfrac { 4\pi ^{ 4 } }{ 16 } \ge 0\Rightarrow p^{ 2 }-4\ge 0$$
    $$\Rightarrow p^{ 2 }\ge 4\Rightarrow p\ge \pm 2$$

  • Question 3
    1 / -0
    $$\displaystyle ax+b(\sec(\tan^{-1}x))= c $$ and $$\displaystyle ay+b(\sec(\tan^{-1}y))= c $$, then the value of xy is,
    Solution
    $$ax+b\left( sec\left( tan^{ -1 }x \right)  \right) =c$$

    $$\\ \Rightarrow ax+b\left( sec\left( sec^{ -1 }\sqrt { 1+x^{ 2 } }  \right)  \right) =c$$

    $$\\ \Rightarrow ax+b\sqrt { 1+x^{ 2 } } =c\Rightarrow b\sqrt { 1+x^{ 2 } } =c-ax$$

    $$\\ \Rightarrow b^{ 2 }\left( 1+x^{ 2 } \right) =c^{ 2 }+a^{ 2 }x^{ 2 }-2acx$$

    $$\\ \Rightarrow \left( b^{ 2 }-a^{ 2 } \right) x^{ 2 }+\left( b^{ 2 }-c^{ 2 } \right) +2acx=0$$

    $$\\ \Rightarrow x=\cfrac { -2ac\pm \sqrt { 4a^{ 2 }c^{ 2 }-4\left( b^{ 2 }-a^{ 2 } \right) \left( b^{ 2 }-c^{ 2 } \right)  }  }{ 2\left( b^{ 2 }-a^{ 2 } \right)  } $$

    Similarly

    $$ay+bsec\left( tan^{ -1 }y \right) =c$$

    $$\\ \Rightarrow y=\cfrac { -2ac\pm \sqrt { 4a^{ 2 }c^{ 2 }-4\left( b^{ 2 }-a^{ 2 } \right) \left( b^{ 2 }-c^{ 2 } \right)  }  }{ 2\left( b^{ 2 }-a^{ 2 } \right)  } $$

    Therefore,
    $$xy=\cfrac { -2ac+\sqrt { 4a^{ 2 }c^{ 2 }-4\left( b^{ 2 }-a^{ 2 } \right) \left( b^{ 2 }-c^{ 2 } \right)  }  }{ 2\left( b^{ 2 }-a^{ 2 } \right)  } \times \cfrac { -2ac-\sqrt { 4a^{ 2 }c^{ 2 }-4\left( b^{ 2 }-a^{ 2 } \right) \left( b^{ 2 }-c^{ 2 } \right)  }  }{ 2\left( b^{ 2 }-a^{ 2 } \right)  } $$

    $$\\ =\cfrac { 4a^{ 2 }c^{ 2 }-4a^{ 2 }c^{ 2 }+4\left( b^{ 2 }-a^{ 2 } \right) \left( b^{ 2 }-c^{ 2 } \right)  }{ 4\left( b^{ 2 }-a^{ 2 } \right)^2  } =\cfrac { c^{ 2 }-b^{ 2 } }{ a^{ 2 }-b^{ 2 } } $$
  • Question 4
    1 / -0
    The number of solution of the equation $$ 1+x^{2}+2x\:\sin \left ( \cos^{-1}y \right )= 0 $$ is :
    Solution
    Given, $$1+x^2+2x(\sin(\cos^{-1}(y)))=0$$
    $$1+x^2+2x(\sin(\sin^{-1}(\sqrt{1-y^2})))=0$$
    $$1+x^2+2x(\sqrt{1-y^2})=0$$
    $$2x(\sqrt{1-y^2})=-(1+x^2)$$
    $$4x^2(1-y^2)=1+x^4+2x^2$$
    $$4x^2-4x^2y^2=1+x^4+2x^2$$
    $$-4x^{2}(y^2)=(1-x^2)^{2}$$
    Hence solution will be $$x=1$$ and $$y=\dfrac{1}{2}$$
  • Question 5
    1 / -0
    If $$\displaystyle [cot^{-1}x]+[cos^{-1}x]=0$$, where $$[\cdot]$$ denotes the greatest integer function, then the complete set of values of $$x$$ is 
    Solution
    $$0 < \cot^{-1}(x) < \pi$$ and $$0 < \cos^{-1}(x) < \pi$$.
    Hence,
    $$\left[\cot^{-1}(x)\right]+\left[\cos^{-1}(x)\right] = 0$$ when both $$\left[\cot^{-1}(x)\right] = 0$$ and $$\left[\cos^{-1}(x)\right] = 0$$.
    i.e.
    $$0 \le \cot^{-1} x < 1$$ and $$0 \le \cos^{-1} x < 1$$.
    Hence,
    $$x \in \left(\cos 1,1\right]$$ and $$x \in \left(\cot 1,\infty\right)$$. There is not such x.
  • Question 6
    1 / -0
    If $$ \sin^{-1}a +\sin^{-1}b+\sin^{-1}c= \displaystyle \frac{3\pi }{2} $$  and  $$ f\left ( 2 \right )=2,{f\left ( x+y \right )}= f\left ( x \right )\:f\left ( y \right )\:\:\forall \:\:x,\:y\:\epsilon \:R $$ then $$ a^{f\left ( 2 \right )}+\:b^{f\left ( 4 \right )}+\:c^{f\left ( 6 \right )}-\:\displaystyle \frac{3\left ( a^{f\left ( 2 \right )}. \:b^{f\left ( 4 \right )}.\:c^{f\left ( 6 \right )}\right )}{a^{f\left ( 2 \right )} +\:b^{f\left ( 4 \right )}+\:c^{f\left ( 6 \right )}} $$  equals
    Solution
    Since 
    $$sin^{-1}(a)+sin^{-1}(b)+sin^{-1}(c)=\dfrac{3\pi}{2}$$
    Hence
    $$a=b=c=1$$.
    Now 
    $$f(x+y)=f(x).f(y)$$ implies 
    $$f(x)=a^{x}$$
    Now 
    $$f(2)=2$$
    Or 
    $$a^{2}=2$$
    Or 
    $$a=\sqrt{2}$$.
    Hence
    $$f(x)=2^{\frac{x}{2}}$$.
    Substituting the values of f(x) and a,b,c in the above expression, we get 
    $$1^{2}+1^{2^{2}}+1^{2^{3}}-\dfrac{3(1^{2}.1^{2^{2}}.1^{2^{3}})}{1^{2}+1^{2^{2}}+1^{2^{3}}}$$

    $$=3-\dfrac{3.(1)}{3}$$
    $$=2$$.
  • Question 7
    1 / -0
    Find number of values of x which  satisfying $$\left [ \tan^{-1}x \right ]+\left [ \cot^{-1}x \right ]=2$$, where [.] represents the greatest integer function.

    Solution
    $$0< \cot^{-1}x< \pi $$ and $$-\pi /2< \tan^{-1}x< \pi $$

    $$\Rightarrow

    $$   $$\left [ \cot^{-1}x \right ]\in \left \{ 0, 1, 2, 3 \right

    \}$$ and $$\left [ \tan^{-1}x \right ]\in \left \{ -2, -1, 0, 1

    \right \}$$

    For $$\left [ \tan^{-1}x \right ]+\left [ \cot^{-1}x \right ]=2$$, following cases are possible

    Case (i): $$\left [ \cot^{-1}x \right ]=\left [ \tan^{-1}x \right ]=1$$

    $$\Rightarrow $$   $$1\leq \cot^{-1}x< 2$$ and $$1\leq \tan^{-1}x< \pi /2$$

    $$\Rightarrow $$   $$x\in \left ( \cot 2, \cot 1 \right ]$$ and $$x\in \left [ \tan 1, \infty  \right )$$

    $$\therefore $$   $$x\in \phi $$ as $$\cot 1< \tan 1$$

    Case (ii): $$\left [ \cot^{-1}x \right ]=2$$, $$\left [ \tan^{-1}x \right ]=0$$

    $$\Rightarrow $$   $$2\leq \cot^{-1}x< 3$$ and $$0\leq \tan^{-1}x< 1$$

    $$\Rightarrow $$   $$x\in \left ( \cot 3, \cot 2 \right ]$$ and $$x\in \left [ 0, \tan 1 \right )$$

    $$\therefore $$   $$x\in \phi $$ as $$\cot 2< 0$$

    So no solution.

    Case (iii): $$\left [ \cot^{-1}x \right ]=3$$, $$\left [ \tan^{-1}x \right ]=-1$$

    $$\Rightarrow $$   $$3\leq \cot^{-1}x< \pi $$ and $$-1\leq \tan^{-1}x< 0$$

    $$\Rightarrow $$   $$x\in \left ( -\infty , \cot 3 \right ]$$ and $$x\in \left [ -\tan 1, 0 \right )$$

    $$\therefore $$   $$x\in \phi $$ as $$\cot 3< -\tan 1$$

    Therefore, no such values of $$x$$ exist.
  • Question 8
    1 / -0
    $$\displaystyle sin^{-1}(3x-2-x^{2})+cos^{-1}(x^{2}-4x+3)=\frac{\pi}{4}$$ can have a solution for $$x\:\epsilon$$
    Solution
    For $$sin^{ -1 }\left( 3x-2-x^{ 2 } \right) \Rightarrow -1<3x-2-x^{ 2 }<1$$
    $$3x-2-x^{ 2 }<1$$ imaginary solution 
    $$3x-2-x^{ 2 }>-1\Rightarrow x\epsilon \left\{ \cfrac { 3-\sqrt { 5 }  }{ 2 } ,\cfrac { 3+\sqrt { 5 }  }{ 2 }  \right\} $$   ...(1)
    And for $$cos^{ -1 }\left( x^{ 2 }-4x+3 \right) \Rightarrow -1<\left( x^{ 2 }-4x+3 \right) <1$$
    $$\left( x^{ 2 }-4x+3 \right) <1\Rightarrow x=2$$   ...(2)
    $$x^{ 2 }-4x+3>-1\Rightarrow x\epsilon \left\{ 2-\sqrt { 2 } ,2+\sqrt { 2 }  \right\} $$   ...(3)
    From (1),(2) and (3)
    $$x\epsilon \left( 2-\sqrt { 2 } ,\cfrac { 3-\sqrt { 5 }  }{ 2 }  \right) \cup \left( \cfrac { 3-\sqrt { 5 }  }{ 2 } ,2+\sqrt { 2 }  \right) \cup \left\{ 2 \right\} $$
  • Question 9
    1 / -0
    The number of solutions of $$\sin^{-1}\left ( 1+b+b^{2}+\cdots \infty \right )+\cos^{-1}\left ( a-\displaystyle\frac{a^{2}}{3}+\frac{a^{2}}{9}\cdots \infty  \right )= \displaystyle\frac{\pi }{2}$$ is
    Solution
    Given : $$\sin^{-1}\left ( 1+b+b^{2}+\cdots \infty \right )+\cos^{-1}\left ( a-\displaystyle\frac{a^{2}}{3}+\frac{a^{2}}{9}\cdots \infty  \right )= \displaystyle\frac{\pi }{2}$$

    The given equation will be valid if $$-1\le 1+b+b^{2}+\cdots \infty \leq 1$$
    and $$a-\displaystyle \frac{a^{2}}{3}+\displaystyle \frac{a^{3}}{9}-\cdots\infty \epsilon \left [ -1,1 \right ]$$

    $$\sin ^{ -1 } \left( 1+b+b^{ 2 }+\cdots \infty  \right) =\sin ^{ -1 } \left( a-\dfrac { a^{ 2 } }{ 3 } +\dfrac { a^{ 2 } }{ 9 } \cdots \infty  \right) $$

    Also $$1+b+b^{2}+\cdots \infty=a-\displaystyle \frac{a^{2}}{3}+\displaystyle \frac{a^{3}}{9}\cdots\infty$$
    (Infinite GP. series)
    $$\Rightarrow \displaystyle\frac{1}{1-b}=\frac{a}{1+\displaystyle \frac{a}{3}}$$
    $$\Rightarrow \displaystyle\frac{1}{1-b}=\frac{3a}{a+3}$$
    which is true for infinite number of values of $$a$$ and $$b$$
    So, $$\exists$$ infinite solutions of the equation.
  • Question 10
    1 / -0
    The number of integral values of $$k$$ for which the equation $$\displaystyle sin^{-1}x+tan^{-1}x=2k+1$$ has a solution is 
    Solution
    $$-1\le x\le 1$$
    Here $$sin^{-1}(x)$$ c values can lie between -1 to 1 so $$tan^{-1}(x)$$ domain is also restricted to -1 to 1.
    Hence the range of the given expression is $$\frac { -3\pi  }{ 4 } \le 2k+1\le \frac { 3\pi  }{ 4 }$$
    Therefore possible values of k are -1,0. 
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