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Differentiation Test 34

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Differentiation Test 34
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  • Question 1
    1 / -0
    How many committee of five persons with a chairperson can be selected from $$12$$ persons.
    Solution
    No of Persons $$=5+1$$(Chairperson)

    $$\therefore$$ No of committees that can be selected $$={}^{12}C_6=924$$
  • Question 2
    1 / -0
    Find the value of ?. 

    Solution
    $${\left( {3 + 1} \right)^2} = {\left( 4 \right)^2} = 16$$

    $${\left( {15 + 6} \right)^2} = {\left( {21} \right)^2} = 441$$

    $${\left( {10 + 5} \right)^2} = {\left( {15} \right)^2} = 225$$

    $$so,\,{\left( {12 + 13} \right)^2} = {\left( {25} \right)^2} = 625$$

    therefore the answer is $$625$$
  • Question 3
    1 / -0
    The shortest distance between line $$y-x=1$$ and curve $$x={y}^{2}$$ is
    Solution
    Line is $$y-x=1$$ and curve $$x=y^2$$
    $$y^2=x$$
    $$2y\dfrac{dy}{dx}=1$$
    $$\dfrac{dy}{dx}=\dfrac{1}{2y}=1$$
    $$\therefore$$  $$y=\dfrac{1}{2}$$
    $$x=\dfrac{1}{4}$$
    Tangent at $$\left(\dfrac{1}{4},\,\dfrac{1}{2}\right)$$
    $$\dfrac{1}{2}y=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)$$
    $$y=x+\dfrac{1}{4}$$
    Distance $$=\left|\dfrac{1-\dfrac{1}{4}}{\sqrt{2}}\right|=\dfrac{3}{4\sqrt{2}}=\dfrac{3\sqrt{2}}{8}$$
  • Question 4
    1 / -0
    How many $$3$$-digit even numbers can be formed from the digits $$1, 2, 3, 4, 5, 6$$ if the digits can be repeated?
    Solution
    Only $$3$$ numbers are possible at units place $$(2,4,6)$$ as we need even numbers. But at $$10's$$ place and $$100's$$ place all $$6$$ are possible.
    No. of digit even no's$$=3\times 6\times 6=108$$

  • Question 5
    1 / -0
    Let $$y=a\cos t+b\sin t$$ then $$\dfrac{d^2y}{dt^2}=$$
    Solution
    Given,
    $$y=a\cos t+b\sin t$$......(1).
    Now differentiating both sides with respect to $$t$$ we get,
    $$\dfrac{dy}{dt}=-a\sin t+b\cos t$$
    Again differentiating both sides with respect to $$t$$ we get,
    $$\dfrac{d^2y}{dt^2}=-a\cos t-b\sin t$$
    or, $$\dfrac{d^2y}{dt^2}=-y$$. [ Using (1)]
  • Question 6
    1 / -0
    Find the area of the triangle whose vertices are $$(-5, -1), (3, -5), (5, 2)$$
    Solution

    Area of a triangle whose vertices are  $$A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$$ is given as,

    $$Area=(\dfrac{1}{2})[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] $$

    Let $$A$$ be the required area

    $$A=(\dfrac{1}{2})[(-5)(-5-2)+3(2+1)+5(-1+5)]\\ (\dfrac{1}{2})[35+9+20]\\=32sq\>unit$$

    So, the correct option is (B)

  • Question 7
    1 / -0
    Let $$f(x)=\dfrac{1}{ax+b}$$ then $$f''(0)=$$
    Solution
    Given,
    $$f(x)=\dfrac{1}{ax+b}$$
    Now differentiating both sides with respect to $$x$$ we get,
    $$f'(x)=-\dfrac{a}{(ax+b)^2}$$
    Again differentiating both sides with respect to $$x$$ we get,
    $$f''(x)=\dfrac{2a^2}{(ax+b)^3}$$
    So $$f''(0)=\dfrac{2a^2}{b^3}$$.
  • Question 8
    1 / -0
    Ten students of the physics department decided to go on a educational trip.They hired a mini bus for the trip, but the bus can only carry eight students at a time and each student goes at least once. Find the minimum number of trips the bus has to make so that each students can go for equal number of trips.
    Solution
    Ten students $$\longrightarrow$$ Eight students in bus $$2$$ students are excluded/not taken on a trip so we have exclude every students once, so that no. of trips made by each students are equal so no. of trips $$= \dfrac{10}{2} =5$$ 
    $$A$$ is correct.
  • Question 9
    1 / -0
    The sides AB, BC, CA of a triangle ABC have $$3,4$$ and $$5$$ interior points respectively on them. The number of triangles that can be constructed using these points as vertices is-
    Solution
    $$\rightarrow $$Suppose we had $$12$$ points, none of them collinear, number of possible triangles $$=^{ 12 }{ C }_{ 3 }$$
    $$\rightarrow $$If out of these $$3,4$$ and $$5$$ points were chosen separately to lie on each side of a triangle ABC, then number of possible triangles
    $$=^{ 12 }{ C }_{ 3 }-^{ 3 }{ C }_{ 3 }-^{ 5 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 }$$
    $$=220-1-10-4$$
    $$=205$$
  • Question 10
    1 / -0
    If $$y=(x^{x})^{x}$$ then $$\dfrac {dy}{dx}=$$
    Solution

    We have,

    $$ y={{\left( {{x}^{x}} \right)}^{x}} $$

    $$ y={{x}^{{{x}^{2}}}} $$

     

    On taking $$\log $$ both sides, we get

    $$\log y={{x}^{2}}\log x$$                                               …… (1)

     

    On differentiating w.r.t $$x$$, we get

    $$ \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{x}+\log x\left( 2x \right) $$

    $$ \dfrac{1}{y}\dfrac{dy}{dx}=x+2x\log x $$

    $$ \dfrac{dy}{dx}=y\left( x+2x\log x \right) $$

    $$ \dfrac{dy}{dx}={{\left( {{x}^{x}} \right)}^{x}}\left( x+2x\log x \right) $$

    $$ \dfrac{dy}{dx}=x{{\left( {{x}^{x}} \right)}^{x}}\left( 1+2\log x \right) $$

     

    Hence, this is the answer.

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