Differentiation Test 30

Result

Differentiation Test 30

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Find the $$4^{th}$$ term of $${ \left( 9x-\cfrac { 1 }{ 3\sqrt { x } } \right) }^{ 18 }$$.

A
$$16500$$

B
$$18564$$

C
$$16540$$

D
$$32600$$

Solution

We know that the $$r^{th}$$ term in expansion of $$(a+b)^n$$ is given by $$^nC_{r-1}a^{n+1-r}b^{r-1}$$
Here $$a=9x, \ n=18 \ \And \ b=-\cfrac{1}{3\sqrt{x}}$$
$$\therefore$$ the fourth term in expansion of $$\left (9x-\cfrac{1}{3\sqrt{x}}\right)^{18}$$ is $$^{18}C_{12} {x}^{6}\left (-\cfrac{1}{3\sqrt{x}}\right)^{12} =18564$$
Question 2
1 / -0

Find the area (in square units) of the triangle whose vertices are $$(a, b+c), (a, b-c) $$ and $$(-a, c). $$

A
$$2ac$$

B
$$2bc$$

C
$$b(a+c)$$

D
$$c(a-h)$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times |[ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] |$$
Given the vertices of triangle,
$$A (a,b+c) ,B(a,b-c) ,C(-a,c)$$
Therefore, area is given by
$$=\dfrac { 1 }{ 2 } |\left[ a\left[ b-c-c \right] +a\left[ c-b-c \right] +(-a)\left[ b+c-b+c \right] \right]| \\ =\dfrac { 1 }{ 2 } |\left[ a(b-2c)+a(-b)-a(2c) \right]| \\ =\dfrac { 1 }{ 2 } |\left[ ab-2ac-ab-2ac \right]| =\left| \dfrac { -4ac }{ 2 } \right| =2ac$$
Question 3
1 / -0

13, 74, 290, 650,.......

A
$$1248$$

B
$$1370$$

C
$$1346$$

D
$$1452$$

E
$$1625$$

Solution

Here we observe the first term 13 can be written as: $$2^2+3^2$$
Similarly other numbers can be written as:
$$74: 5^2+7^2$$
$$290: 11^2+13^2$$
$$650: 17^2+19^2$$
Here, we observe that all above term of series can be written as sum of squares of two prime numbers which comes next to it's previous pair.
Like: $$(2,3);(5,7);(11,13);(17,19)$$ So, other pair of prime number will be $$ (23,29).$$
Hence next term of given series will be:
$$\Rightarrow 23^2+29^2=1370$$
So, correct answer is $$1370$$.
Question 4
1 / -0

If $$|x| < 1$$, then $$\dfrac{d}{dx}\left[1+\dfrac{p}{q}x+\dfrac{p(p+q)}{2!}\left(\dfrac{x}{q}\right)^2+\dfrac{p(p+q)(p+2q)}{3!}\left(\dfrac{x}{q}\right)^3....\infty\right]=$$

A
$$\dfrac{p}{q(1-x)^{\frac{p}{q}+1}}$$

B
$$\dfrac{p}{q(1-x)^{\frac{p}{q}}}$$

C
$$(1-x)^{-pq-1}$$

D
$$(1-x)^{pq+1}$$

Solution

$$\dfrac{d}{dx}\left\{1+\dfrac{px}{q}+\dfrac{p(p+q)}{2!}\dfrac{x^2}{q^2}+\dots\right\}=\dfrac{p}{q}+\dfrac{p(p+q)}{2!}\dfrac{2x}{q^2}+\dfrac{p(p+q)(p+2q)}{3!}\dfrac{3x^2}{q^3}+\dots$$
$$=\dfrac{p}{q}\left(1+\dfrac{(p+q)}{1!}\dfrac{x}{q}+\dfrac{(p+q)(p+2q)}{2!}\dfrac{x^2}{q^2}+\dots\right)$$
Now, consider $$(1-a)^n=1-na+\dfrac{n(n-1)}{2!}x^2+\dots$$
Put $$a=x$$ and $$n=-\left(1+\dfrac{p}{q}\right)$$
$$\therefore (1-x)^{-\left(1+\frac{p}{q}\right)}=1+\dfrac{(p+q)}{q}x+\left[-\dfrac{(p+q)}{q}\left(-\dfrac{(p+q)}{q}-1\right)\right]\dfrac{x^2}{2!}\dots$$
$$=1+\dfrac{(p+q)}{q}x+\left[\dfrac{(p+q)}{q}\left(\dfrac{(p+q+q)}{q}\right)\right]\dfrac{x^2}{2!}\dots$$
$$=1+\dfrac{(p+q)}{q}x+\dfrac{(p+q)(p+2q)}{q^2}\dfrac{x^2}{2!}\dots$$
Hence,
$$\dfrac{d}{dx}\left\{1+\dfrac{px}{q}+\dfrac{p(p+q)}{2!}\dfrac{x^2}{q^2}+\dots\right\}=\dfrac{p}{q}(1-x)^{-\left(1+\frac{p}{q}\right)}$$
This is the required answer.
Question 5
1 / -0

Figures $$1$$ and $$2$$ are related in a particular manner. Establish the same relationship between figures $$3$$ and $$4$$ by choosing a figure from amongst the options.

A

B

C

D

Solution

Hexagon is rotating in anti-clockwise direction two times to get the next figure. (1 to 2).
Same for getting the figure (3 to 4) rotate the figure in an anti-clockwise direction (two times) we will get the desired figure.
Question 6
1 / -0

Select the INCORRECT match

A
$$3249-MMMCCXLIX$$

B
$$1667-MDCLXVII$$

C
$$207-CCXVII$$

D
$$499-CDXCIX$$

Solution

(a) $$MMMCCXLIX=3000+200+40+9=3249$$
(b) $$MDCLXVII=1000+500+100+60+7=1667$$
(c) $$CCXVII=217=200+17$$
(d) $$CDXCIX=400+90+9=499$$
Hence the correct match is option C.
Question 7
1 / -0

If D (3, -1), E (2, 6) and F (5, 7) are the vettices of the sides of $$\Delta DEF$$, the area of triangle DEF is sq. units.

A
$$11$$

B
$$22$$

C
$$48$$

D
$$50$$

Solution

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

Area of triangle $$DEF=|\dfrac{1}{2}(3(6-7)+2(7+1)+5(-1-6))|$$

$$=|\dfrac{1}{2}(-3+16-35)|$$

$$=11$$ sq. Units.
Question 8
1 / -0

A telephone number $$d_1d_2d_3d_4d_5d_6d_7$$ is called memorable if the prefix sequence $$d_1d_2d_3$$ is exactly the same as either of the sequence $$d_4d_5d_6$$ or $$d_5d_6d_7$$(or possibly both). If each $$d_1\epsilon\{x|0\leq x\leq 9, x\epsilon W\}$$, then number of distinct memorable telephone number is(are).

A
$$19810$$

B
$$19,910$$

C
$$19,990$$

D
$$20,000$$

Solution
Question 9
1 / -0

If $$(f(x))^{g(y)} = e^{f(x) - g(y)}$$ then $$\dfrac {dy}{dx} =$$.

A
$$\dfrac {f^{1}(x)\log f(x)}{g^{1}(y) (1 + \log f(x))^{2}}$$

B
$$\dfrac {f^{1}(x)\log f(x)}{g^{1}(y) (1 + \log f(x))^{3}}$$

C
$$\dfrac {f^{1}(x).\log f(x)}{g^{-1}(y) (1 - \log f(x))^{2}}$$

D
$$\dfrac {f^{1}(x)\log f(x)}{g(y) (1 + \log f(x))^{2}}$$

Solution

$${ f\left( x \right) }^{ g\left( y \right) }={ e }^{ f\left( x \right)-g\left( y \right) }$$

$$\log { { f\left( x \right) }^{ g\left( y \right)} }={ f\left( x \right) -g\left( y \right)}$$

$${ g\left( y \right) }\log { { f\left( x \right) } } ={ f\left( x \right) -g\left( y \right) }$$

$${ f\left( x \right) } = { g\left( y \right) }\left[ 1+\log { { f\left( x \right) } }\right]$$ ...(1)

$$f^{ 1 }\left( x \right) ={ g\left( y \right) }\left[ \frac { f^{ 1 }\left( x \right) }{ f\left( x \right) } \right] +g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right) } } \right] \times \frac { dy }{ dx } $$

Rearranging,

$$f^{ 1 }\left( x \right) \left[ \frac { f\left( x \right) -g\left( y \right) }{ f\left( x \right) } \right] =g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right) } } \right] \times \frac { dy }{ dx } $$

from eqn. (1)

$$f^{ 1 }\left( x \right) \left[ \frac { g\left( y \right) \times \log { f\left( x \right) } }{ f\left( x \right) } \right] =g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right) } } \right] \times \frac { dy }{ dx } $$

again, from eqn (1),

$$f^{ 1 }\left( x \right) \left[ \frac { \log { f\left( x \right) } }{ \left[ 1+\log { { f\left( x \right) } } \right]} \right] =g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right) } } \right] \times \frac { dy }{ dx } $$

$$\frac { dy }{ dx } =\frac { f^{ 1 }\left( x \right) }{ g^{ 1 }\left( y \right) } \times \left[ \frac { \log { f\left( x \right) } }{ { \left[ 1+\log { { f\left( x \right) } } \right] }^{ 2 } } \right] $$
Question 10
1 / -0

If $$15! =2^{\alpha}\cdot 3^{\beta}\cdot 5^{\gamma}\cdot 7^{\delta}\cdot 11^{\theta}\cdot 13^{\Phi}$$, then the value of expression $$\alpha -\beta +\gamma -\delta +\theta -\Phi$$ is

A
$$4$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution

$$15!=1\times 2\times 3\times 4 \times 5\times 6\times 7\times 8\times 9\times 10\times 11\times 12\times 13\times 14\times 15$$

$$15!=2\times 3\times 2^2 \times 5\times 3\times 2 \times 7\times 2^3\times 3^2\times 2\times 5\times 11\times 3\times 2^2 \times 13\times 2\times 7 \times 3\times 5$$

$$15!=2^{11}\times 3^6\times 5^3\times 7^ 2\times 11\times 13 $$

$$\implies \alpha =11$$

$$\implies \beta =6$$

$$\implies \gamma=3$$

$$\implies \delta =2$$

$$\implies \theta =1$$

$$\implies \Phi=1$$

$$\alpha-\beta+\gamma-\delta+\theta-\Phi=11-6+3-2+1-1=6$$