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Continuity and Differentiability Test - 32

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Continuity and Differentiability Test - 32
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  • Question 1
    1 / -0
    If $$y = Tan^{ -1 }\left(secx + tanx \right) then \dfrac { dy }{ dx }=$$
    Solution

  • Question 2
    1 / -0
    If $$x=\sin^ {-1}\left( \dfrac { 2\theta  }{ 1+{ \theta  }^{ 2 } }  \right),y=\sec^ {-1}\sqrt {1+\theta^ {2}}$$, then $$\dfrac {dy}{dx}=$$
    Solution

  • Question 3
    1 / -0
    If $$y={ tan }^{ -1 }\left( \frac { ax-b }{ bx+a }  \right)$$, the value of $$\frac { dy }{ dx } $$ is
    Solution

  • Question 4
    1 / -0
    Let $$f:[0,2]\rightarrow R$$ be a twice differentiable function such that $$f"(x)>0$$, for all $$x\in (0,2)$$ If $$\phi (x)=f(x)+f(2-x)$$, then $$\phi$$is:
    Solution
    $$\phi(x)=f(x)+f(2-x)$$
    $$\phi '(x)=f'(x)-f'(2-x)$$...(1)
    Since $$f"(x)>0$$
    $$\Rightarrow f'(x) is increasing \forall x\in (0,2)$$
    case I when $$x>2-x \Rightarrow  x>1$$
    $$\Rightarrow \phi'(x) >0\forall x \in (1,2)$$
    $$\therefore \phi (x) is increasing on (1,2)$$
    case II: when $$x<2-x \Rightarrow x<1$$
    $$\Rightarrow \phi '(x) <0\forall x\in (0,1)$$
    $$\therefore \phi (x) $$ is decreasing on $$(0,1)$$
  • Question 5
    1 / -0
    Let $$f(x)=15-|x-10|; x\in R$$. Then the set of all values of x, at which the function, $$g(x)=f(f(x))$$ is not differentiable, is?
    Solution
    $$f(x)=15-|x-10|, x\in R$$
    $$f(f(x))=15-|f(x)-10|$$
    $$=15-|15-|x-10|-10|$$
    $$=15-|5-|x-10||$$
    $$x=5, 10, 15$$ are points of non differentiability
    Aliter:
    At $$x=10$$ $$f(x)$$ is no differentiable
    also, when $$15-|x-10|=10$$
    $$\Rightarrow x=5, 15$$
    $$\therefore$$ non differentiability points are $$\{5, 10, 15\}$$.
  • Question 6
    1 / -0
    The set of points where the function $$f(x)=x|x|$$ is differentiable is?
    Solution
    Consider the given function.
    $$f(x)=x|x|$$

    $$f(x)=\left\{\begin{matrix} x^2, &  x>0\\ -x^2, & x<0 \\ 0 & x =0\end{matrix}\right.$$

    Since, it is polynomial function.
    Hence, it is differentiable at $$(-\infty, \infty)$$.
  • Question 7
    1 / -0
    If $$x=3\sin {t} ,\ y=3\cos {t} ,$$ find $$\dfrac {dy}{dx}$$ at $$t=\dfrac { \pi  }{ 3 } $$
    Solution
    $$x=3\sin t$$
    $$\dfrac{dx}{dt}=3\cos t$$

    $$y=3\cos t$$
    $$\dfrac{dy}{dt}=-3\sin t$$

    Thus, $$\dfrac{dy}{dx}=-\tan t$$

    At $$t=\dfrac{\pi}{3}$$,

    $$\dfrac{dy}{dx}=-\sqrt3$$
  • Question 8
    1 / -0
    Differentiate the following function with respect to x.
    If for $$f(x)=\lambda x^2+\mu x+12$$, $$f'(14)=15$$ and $$f'(2)=11$$, then find $$\lambda$$ and $$\mu$$.
    Solution
    Given function 

    $$f(x)=\lambda x^2 +\mu x +12$$

    $$\implies f^{\prime}(x)=\dfrac d{dx}(\lambda x^2+\mu x+12)$$


    $$f^{\prime}(x)=2\lambda x+\mu$$

    $$f^{\prime}(2)=2\lambda (2)+\mu$$

    $$\implies 4\lambda +\mu=11\cdots (1)$$

    $$f^{\prime}(4)=2\lambda (4)+\mu $$

    $$\implies 8\lambda+\mu=15 \cdots (2)$$


    $$(2)-(1)$$

    $$8\lambda-4\lambda+\mu-\mu=15-11$$

    $$4\lambda =4$$

    $$\implies \lambda=1$$


    Put it in $$(1)$$

    $$\implies 4(4)+\mu =11$$

    $$\mu=11-4$$

    $$\mu =7$$
  • Question 9
    1 / -0
    If $$f(9) = 9, f'(9) = 0$$, then $$\underset{x\to 9}{\lim} \dfrac{\sqrt{f(x)}-3}{\sqrt{x}-3}$$ is equal to 
    Solution
    $$\underset{x\to 9}{\lim} \dfrac{\sqrt{f(x)} -3}{\sqrt{x} -3} $$    $$\left[\dfrac{0}{0}from\right]$$

    $$= \underset{x\to 9}{\lim} \dfrac{\dfrac{f'(x)}{2\sqrt{f(x)}}}{\dfrac{1}{2\sqrt{x}}}$$

    $$= \dfrac{\sqrt{x}f'(x)}{\sqrt{f(x)}}$$

    $$= \dfrac{\sqrt{9}f'(9)}{\sqrt{f(9)}}$$

    $$= \dfrac{3\times 0}{3} = 0$$
  • Question 10
    1 / -0
    Let $$f : R \rightarrow R$$ be a continuous function such that $$f(x^2) = f(x^3)$$ for all $$ x \in R$$. Consider the following statements.
    I. f is an odd function.
    II. f is an even function.
    III. f is differentiable everywhere
    Solution
    $$f:R\rightarrow\,R$$  be a continuous function such that
    $$f\left({x}^{2}\right)=f\left({x}^{3}\right)$$          ..........$$(1)$$ for all $$x\in R$$
    Put $$x=-x$$
    $$f\left({x}^{2}\right)=f\left(-{x}^{3}\right)$$ 
    From $$(1)$$ we have $$f\left({x}^{3}\right)=f\left(-{x}^{3}\right)$$
    Put $${x}^{3}=t$$ we have $$f\left(t\right)=f\left(-t\right)$$
    $$\Rightarrow\,f\left(x\right)$$ is an even function.
    $$(ii)$$Now take $${x}^{3}=t$$
    $$\Rightarrow\,f\left({t}^{\frac{2}{3}}\right)=f\left(t\right)$$
    Put $$t={t}^{\frac{2}{3}}\Rightarrow\,f\left({\left({t}^{\frac{2}{3}}\right)}^{2}\right)=f\left(t\right)$$
    $$\Rightarrow\,f\left(t\right)=f\left({t}^{\frac{2}{3}}\right)=f\left({\left({t}^{\frac{2}{3}}\right)}^{2}\right)=f\left({\left({t}^{\frac{2}{3}}\right)}^{3}\right)=....=f\left({\left({t}^{\frac{2}{3}}\right)}^{n}\right)$$
    This is true for all $$t\in R$$ and any $$n\in I$$
    Hence if we take $$n\rightarrow\,\infty,\,{\left(\dfrac{2}{3}\right)}^{n}\rightarrow\,0$$
    Then $$f\left(t\right)=f\left({t}^{\circ}\right)=1$$
    $$\Rightarrow\,f\left(x\right)$$  is a constant function, hence it is differentiable everywhere.
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