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Differentiation Test 45

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Differentiation Test 45
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  • Question 1
    1 / -0
    The number of rectangles that can be obtained by joining four of the twelve vertices of a $$12$$ sided regular polygon is
    Solution
    The first vertex can be choosed in $$12$$ ways and diagonally opposite to it is $$1$$ vertex. Now for $$3rd$$ vertex we have $$10$$ choices and for $$4th$$ $$1.$$
    However, each rectangle is counted $$8$$ times.
    $$\therefore$$ No. of ways $$=\dfrac{12\times1\times10\times1}{8}$$  $$=15$$ ways.
    Hence, the answer is $$15.$$
  • Question 2
    1 / -0
    Three vertices are chosen randomly from the seven vertices of a regular $$7$$ -sided polygon. The probability that they form the vertices of an isosceles triangle is
    Solution
    A regular 3 sided polygon is nothing else but a heptagon for creating isosceles triangle we need to choose adjacent sides only. 
    No. of $$\triangle 's$$ formed $$=7{ { C }_{ 3 } }$$
    While number of isosceles triangle formed $$=$$ No. of points $$\times $$ points available $$=7\times 3.$$
    $$\Rightarrow$$ So, probability $$=\dfrac { 7\times 3 }{ 7{ { C }_{ 3 } } } = \dfrac { 21 }{ 35 } =\dfrac { 3 }{ 5 } .$$
    Hence, the answer is $$\dfrac { 3 }{ 5 }.$$

  • Question 3
    1 / -0
    Let f and g be differentiable function such that $${f}'\left ( x \right )=2g\left ( x \right )$$ and $${g}'\left ( x \right )=-f\left ( x \right )$$, and let $$T\left ( x \right )=\left ( f\left ( x \right ) \right )^{2}-\left ( g\left ( x \right ) \right )^{2}$$. Then $${T}'\left ( x \right )$$ is equal to
    Solution
    The given equation is:
    $$T(x) = f(x)^2 - g(x)^2$$

    $$\Rightarrow T(x) = (f(x) - g(x))(f(x) + g(x))$$

    Differentiating once w.r.t to x we get,

    $$\Rightarrow T'(x) = (f'(x) + g'(x))(f(x)-g(x)) + (f'(x)-g'(x))(f(x)+g(x))$$

    $$\Rightarrow T'(x) = (2g(x)-f(x))(f(x)-g(x)) + (2g(x)+f(x))(f(x)+ g(x))$$

    $$\Rightarrow T'(x) = 2g(x)f(x) - 2g(x)^2 -f(x)^2 + f(x)g(x) +2g(x)f(x) + 2g(x)^2 +f(x)^2 + f(x)g(x)$$

    $$\Rightarrow T'(x)= 6f(x)g(x)$$     .....Answer


  • Question 4
    1 / -0
    A college offers $$7$$ courses in the morning and $$5$$ courses in the evening. Find the number of ways a student can select exactly one course either in the morning or in the evening.
    Solution
    $$7$$ Courses in morning 
    $$5$$ courses in evening 
    Total number of courses $$=12$$
    Selecting any one of the course 
    Number of ways $$^{ 12 }{ C }_{ 1 }$$
    $$=\cfrac { 12! }{ 1!\times 1! } $$ 
    $$=12$$ Ways 
    Therefore total ways $$=12$$
  • Question 5
    1 / -0
    The derivative of $$\displaystyle (\tan x)^{x}$$ is equal to-
    Solution
    $$\displaystyle y=(\tan x)^{x}$$

    $$\Rightarrow \log y = x \log \tan x$$ 

    $$\displaystyle \Rightarrow \frac{1}{y}\frac{dy}{dx}=\log { \tan { x }  } +\frac { x\times 1 }{ \tan  x } \times \sec ^{ 2 } x$$

    $$\Rightarrow \displaystyle \frac{dy}{dx}=y\left [ \log \tan x+x\sec x \, cosec x \right ]$$

    $$\Rightarrow \displaystyle \frac{dy}{dx}=(\tan x)^{x}\left [ \log \tan x+x\sec x\, cosec x \right ]$$
  • Question 6
    1 / -0
    The sides of a quadrilateral are all positive integers and three of them are $$5, 10, 20.$$ How many possible value are there for the fourth side?
    Solution

  • Question 7
    1 / -0
    Numbers can be classified into two categories,depending on their divisible conditions.
    They are (i) Even numbers $$(2p) \vee p \epsilon N$$ (ii) odd numbers $$(2p + 1) \vee p \epsilon N$$
    a.    $$a_1, a_2 ...... a_{2013}$$ are integers, not necessarily distinct.
    $$x = (-1)^{a_1}+(-1)^{a_2}+.....+(-1)^{a_{1006}}$$
    $$y = (-1)^{a_{1007}}+(-1)^{a_{1008}}+......+(-1)^{a_{2013}}$$

    Then which of the following is true?
    Solution

  • Question 8
    1 / -0
    If $$ { _{  }^{ n }{ C } }_{ 4 },{ _{  }^{ n }{ C } }_{ 5 }$$ and $$ { _{  }^{ n }{ C } }_{ 6 }$$ are in AP, then $$n$$ is
    Solution
    $$\textbf{Step 1: Find the value of n}$$

                    $$\text{If a,b,c are in A.P,then}$$

                    $$2b=a+c$$

                    $$\text{So applying the same conditions, we have}$$

                    $$\Rightarrow$$ $$2 \times\left({ }^{n} c_{5}\right)=\left({ }^{n} C_{6}\right)+\left({ }^{n} C_{4}\right)$$

                    $$\text{We know that,}$$

                    $${ }^{n} C_{r} = \dfrac{ n !}{r !(n-r) !}$$

                    $$\Rightarrow$$$$\dfrac{2 \times n !}{5 !(n-5) !}=\dfrac{n !}{6 !(n-6) !}+\dfrac{n !}{4 !(n-4) !}$$

                    $$\Rightarrow \dfrac{2}{5 !(n-5) !}=\dfrac{1}{6(n-6) !}+\dfrac{1}{4 !(n-4) !} $$         

                    $$\Rightarrow \dfrac{2}{5 !(n-5)(n-6) !}=\dfrac{1}{6 !(n-6) !}+\dfrac{1}{4 !(n-4)(n-5)(n-6)!} $$

                    $$ \Rightarrow \dfrac{2}{5 !(n-5)}=\dfrac{1}{6 !}+\dfrac{1}{4 !(n-4)(n-5)} $$               

                    $$\Rightarrow \dfrac{2}{5 \times 4 ! \times(n-5)}=\dfrac{1}{6 \times 5 \times 4 !}+\dfrac{1}{4 !(n-4)(n-5)} $$

                    $$=\dfrac{2}{5 \times(n-5)}=\dfrac{1}{6 \times 5}+\dfrac{1}{(n-4)(n-5)}$$                  

                    $$\Rightarrow \dfrac{2}{5 \times(n-5)}-\dfrac{1}{(n-4)(n-5)}=\dfrac{1}{6 \times 5}$$      

                    $$\Rightarrow 12 n-78=n^{2}-9 n+20$$

                    $$\Rightarrow n^2-21n+98=0$$

                    $$\Rightarrow (n-7)(n-14)=0$$

                    $$\therefore n=14$$ $$\text{(or)}$$ $$n=7$$

    $$\textbf{Hence, option A is correct.}$$
  • Question 9
    1 / -0
    If j, k, and n are consecutive integers such that $$0 < j < k < n$$ and the units (ones) digit of the product jn is 9, what is the units digit of k ? 
    Solution

    There are only a few ways you can make a product have a units digit of $$9$$. Either both numbers you’re multiplying have to be $$3$$, or one has to be $$1$$ and one has to be $$9$$. Since we’re dealing with consecutive integers, $$j$$ and $$n$$ can’t both end in $$3$$, so they’re going to have to end in $$1$$ and $$9$$. 

    It’s important to remember that although we only really care about the units digits in this problem, we’re dealing with numbers that might (or in fact, must) be $$2$$ or more digits. That’s why you can have $$k's$$ units digit be $$0$$. 

    Say $$j= 19$$, $$k= 20$$, and $$n= 21$$. Then $$jn = (19)(21) = 399$$ (units digit is $$9$$), and the units digit of $$k = 0$$.

    Hence option A is correct. 

  • Question 10
    1 / -0
    Two classrooms A and B having capacity of $$25$$ and $$(n-25)$$ seats respectively. $$A_n$$ denotes the number of possible seating arrangements of room $$'A'$$, when 'n' students are to be seated in these rooms, starting from room $$'A'$$ which is to be filled up to its capacity. If $$A_n-A_{n-1}=25!(^{49}C_{25})$$ then 'n' equals:
    Solution
    Given $$A_n=nC_{25} \cdot 25!$$

    $$A_{n-1}={n-1}C_{25} \cdot 25!$$

    Hence $$nC_{25} \cdot 25! - (n-1)C_{25} \cdot 25!=25! 49C_{25}$$

    $$\Rightarrow (n-1)C_{25}+(n-1)C_{24}-(n-1)C_{25}=49C_{24}$$

    $$\Rightarrow n-1=49$$

    $$\Rightarrow n=50$$
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