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Differentiation Test 41

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Differentiation Test 41
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  • Question 1
    1 / -0
    How many integers are there such that $$2 \le n \le 100$$ and the highest common factor of $$n$$ and $$36$$ is $$1$$?
    Solution
    $$36.2^{2}.3^{2}$$
    Since HCF (36,n) =1
    Therefore, n sholud not be a multiple of 2 or 3.
    $$ 2 \leq n\leq 100$$
    Total numbers = T= 1000.
    Number of numbers divisible by 2= $$N_{2}$$
    Number of numbers divisible by 3= $$N_{3}$$
    Number of numbers divisible by 6= $$N_{6}$$
    Therefore, there are $$T-N_{2}-N_{3}+N_{6}$$ total integers
    a=2
    l=1000
    d=2
    We know, $$l=a+(N_{2}-1)d.$$
    $$N_{2}=\frac{1000-2}{2}+1$$
    $$N_{2}=\frac{998}{2}+1$$
    $$N_{2}=499+1=500$$
    Similarly,
    $$N_{3}=\frac{999-3}{3}+1=333$$
    $$N_{6}=\frac{999-6}{3}+1=166$$
    Answer
    = 999-500-333+166
    = 332

  • Question 2
    1 / -0
    If area of a triangle is $$35$$ square units with vertices $$\left( {2, - 6} \right),\,\,\left( {5,\,\,4} \right)$$ and $$({k},\,\,4)$$ then $${k}$$ is :
    Solution

  • Question 3
    1 / -0
    The number of ways in which $$9$$ persons can be divided into three equal groups, is
    Solution

  • Question 4
    1 / -0
    What is the next number in the series $$2,12,36,80,150?$$
    Solution
    $$12 - 2 = 10$$.
    $$36 - 12 = 24$$.
    $$80 - 36 = 44$$.
    $$150 - 80 = 70$$.
    Now,
    $$24 - 10 = 14$$.
    $$44 - 24 = 20$$.
    $$70 - 44 = 26$$.
    Then next difference of the difference will be $$26 + 6 = 32$$.
    Hence answer is $$150 + 70 + 32 = 252$$.
  • Question 5
    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 6
    1 / -0
    The number of permutations which can be formed out of the letters of the word "SERIES" three letters together, is:
    Solution

  • Question 7
    1 / -0
    The coefficient of $$x^{18}$$ in the expansion of $$(1+x)(1-x)^{10}\{(1+x+x^2)^9\}$$ is?
    Solution
    Coefficient of $$x^{18}$$ in $$(1+x)(1-x)^{10}(1+x+x^2)^9$$
    $$=$$Coefficient of $$x^{18}$$ in $$(1-x)^2\{(1-x)(1+x+x^2)\}^9$$
    $$=$$Coefficient of $$x^{18}$$ in $$(1-x^2)(1-x^3)^9$$
    $$={^9C_6}=0=84$$.
  • Question 8
    1 / -0
    Differentiate the following function with respect to x.
    $$\dfrac{x^3}{3}-2\sqrt{x}+\dfrac{5}{x^2}$$.
    Solution

  • Question 9
    1 / -0
    Differentiate the following function with respect to x.
    $$3^x+x^3+3^3$$.
    Solution

  • Question 10
    1 / -0
     Let $$ u(x) $$ and $$ v(x) $$ be differentiable functions such that $$ \dfrac{u(x)}{v(x)}=7 . $$ If $$ \dfrac{u^{\prime}(x)}{v^{\prime}(x)}=p $$ and $$ \left(\dfrac{u(x)}{v(x)}\right)^{\prime}=q, $$ then $$ \dfrac{p+q}{p-q} $$ has the value equal to
    Solution
    $$ u(x)=7 v(x) \Rightarrow u^{\prime}(x)=7 v^{\prime}(x) \Rightarrow p=7 $$ (given)

    $$\operatorname{Again} \dfrac{u(x)}{v(x)}=7 \Rightarrow\left(\dfrac{u(x)}{v(x)}\right)^{\prime}=0 \Rightarrow q=0 $$

    $$\text { Now } \dfrac{p+q}{p-q}=\dfrac{7+0}{7-0}=1$$
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