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Limits and Derivatives Test - 33

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Limits and Derivatives Test - 33
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  • Question 1
    1 / -0
    For x>yx>y, limx0[(sinx)1/x+(1x)sinx]\displaystyle\lim_{x\rightarrow 0}{\left[\left(\sin{x}\right)^{1/x}+\left(\cfrac{1}{x}\right)^{\sin{x}}\right]} is :
  • Question 2
    1 / -0
    If f:R(0,)f : R \rightarrow (0, \infty) is an increasing function and if limx2018f(3x)f(x)=1\displaystyle\lim_{x \rightarrow 2018} \dfrac{f(3x)}{f(x)} = 1, then limx2018f(2x)f(x)\displaystyle\lim_{x \rightarrow 2018} \dfrac{f(2x)}{f(x)} is equal to 
    Solution
    Given f:R(0,)f : R \rightarrow (0, \infty) is an increasing function.
    And limx2018f(3x)f(x)=1\underset{x \rightarrow 2018}{\lim} \dfrac{f(3x)}{f(x)} = 1
    So, limx2018f(3x)=limx2018f(x)\underset{x \rightarrow 2018}{\lim} f(3x) = \underset{x \rightarrow 2018}{\lim} f(x)
    f(x)=\Rightarrow f(x) = constant.
    Therefore,limx2018f(2x)f(x)=1 \underset{x \rightarrow 2018}{\lim} \dfrac{f(2x)}{f(x)} = 1
  • Question 3
    1 / -0
    If ff is differentiable at x=1x = 1 and limh01hf(1+h)=5,f(1)=\underset{h \rightarrow 0}{\lim} \dfrac{1}{h} f (1 + h) = 5, f'(1) =
    Solution
    f(1)=limh0f(1+h)f(1)hf'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)-f(1)}{h}; function is differentiable.
     
    and limh0f(1+h)h=5\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5 [Given function is continuous]
    f(1)=0\Rightarrow f(1)=0

    Hence, f(1)=limh0f(1+h)h=5f'(1)=\underset{h\rightarrow 0}{lim} \dfrac{f(1+h)}{h}=5
  • Question 4
    1 / -0
    If y=sinx+sinx+sinx+.....y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+}}}.....\infty then dydx=?\dfrac{dy}{dx}=?
    Solution

  • Question 5
    1 / -0
    If y=(tanx)cotxy=(\tan x)^{\cot x} then dydx=?\dfrac{dy}{dx}=?
    Solution

  • Question 6
    1 / -0
    If x=a(cosθ+θsinθ)x=a(\cos\theta+\theta\sin\theta) and y=a(sinθθcosθ)y=a(\sin\theta-\theta\cos\theta) then dydx=?\dfrac{dy}{dx}=?
    Solution

  • Question 7
    1 / -0
    If x=asecθ,y=btanθx=a\sec\theta, y=b\tan\theta then dydx=\dfrac{dy}{dx}=
    Solution

  • Question 8
    1 / -0
    If y=xsinxy=\sqrt{x\sin x} then dydx=?\dfrac{dy}{dx}=?
    Solution

  • Question 9
    1 / -0
    If y=1+sinx1sinxy=\sqrt{\dfrac{1+\sin x}{1-\sin x}} then dydx=?\dfrac{dy}{dx}=?
    Solution
    y={1+cos(π2x)1cos(π2x)}12={2cos2(π4x2)2sin2(π4x2)}12=cot(π4x2)y=\left\{\dfrac{1+\cos \left(\dfrac{\pi}{2}-x\right)}{1-\cos \left(\dfrac{\pi}{2}-x\right)}\right\}^{\dfrac{1}{2}}=\left\{\dfrac{2\cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}{2\sin^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}\right\}^{\dfrac{1}{2}}=\cot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right).
    dydx=csc2(π4x2)×dydx(π4x2)\Rightarrow \dfrac{dy}{dx}=-\csc^2\left({\dfrac{\pi}{4}-\dfrac{x}{2}}\right)\times\dfrac{dy}{dx}\left({\dfrac{\pi}{4}-\dfrac{x}{2}}\right)
    dydx=12csc2(π4x2)\therefore \dfrac{dy}{dx}=\dfrac12\csc^2\left({\dfrac{\pi}{4}-\dfrac{x}{2}}\right)
  • Question 10
    1 / -0
    The value of limxπ1+cos3xsin2x \displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x}  is
    Solution
    We have limxπ1+cos3xsin2x \displaystyle \lim _{x \rightarrow \pi} \dfrac{1+\cos ^{3} x}{\sin ^{2} x}
    =limxπ(1+cosx)(1cosx+cos2x)(1cosx)(1+cosx)=\displaystyle \lim _{x \rightarrow \pi} \dfrac{(1+\cos x)\left(1-\cos x+\cos ^{2} x\right)}{(1-\cos x)(1+\cos x)}
    =limxπ1cosx+cos2x1cosx=1+1+11+1=32=\displaystyle \lim _{x \rightarrow \pi} \dfrac{1-\cos x+\cos ^{2} x}{1-\cos x}=\dfrac{1+1+1}{1+1}=\dfrac{3}{2}
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