Differentiation Test 15

Result

Differentiation Test 15

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

The points $$A (2, 9), B (a, 5)$$ and $$C (5, 5) $$ are the vertices of a triangle $$ABC$$ right angled at $$B$$. Find the values of $$a$$ and hence the area of $$\Delta $$ $$ABC$$.

A
$$a = -1$$ , Area $$=$$ $$15$$ sq. unit

B
$$a = 2$$ , Area $$=$$ $$6$$ sq. unit

C
$$a = 0 $$ , Area $$=$$ $$8$$ sq. unit

D
$$a = 1$$ , Area $$=$$ $$12$$ sq. unit

Solution

Given that the vertices of the $$\Delta ABC$$ are $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,9 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( a,5 \right) $$ and $$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 5,5 \right) and \angle B={ 90 }^{ o }$$.
So, $$AC$$ is the hypotenuse.
$$ \therefore$$ by Pythagoras theorem, we have
$$ { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$$ .........(i)
Now, by distance formula
$$d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } $$
So, $$AB=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 2-a \right) }^{ 2 }+{ \left( 9-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-4a+20 } $$,
$$BC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 5-a \right) }^{ 2 }+{ \left( 5-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-10a+25 } $$ and
$$AC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( 5-2 \right) }^{ 2 }+{ \left( 5-9 \right) }^{ 2 } } $$ units $$=\sqrt { 25 } $$ units.
$$ \therefore$$ by (i), we get
$${ a }^{ 2 }-4a+20{ +a }^{ 2 }-10a+25=25$$
$$ \Rightarrow { a }^{ 2 }-7a+10=0$$
$$ \Rightarrow a=5,2$$
We reject $$a=5$$, since $$B$$ and $$C$$ will coincide in that case and $$\Delta ABC$$ will collapse.
So, $$a=2. i.e. AB=\sqrt { { a }^{ 2 }-4a+20 } =\sqrt { 4-8+20 } 3$$ units $$=43$$ units
$$ BC=\sqrt { { a }^{ 2 }-10a+25 } =\sqrt { 4-20+25 } 3$$ units $$=3$$ units.
Now,
We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Therefore, required area is $$6$$ square units.
Question 2
1 / -0

The area of triangle ABC (in sq. units) is :

A
$$15$$ sq. units

B
$$10$$ sq. units

C
$$7.5$$ sq. units

D
$$2.5$$ sq. units

Solution

The area of a triangle is given as:
Area $$=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] $$
the given points are $$ A(1,3),~B(-1,0) $$and $$C(4,0)$$
by Substituting ,
$$\text{Area} =\dfrac { 1 }{ 2 } \left[ 1(0-0)+(-1)(0-3)+4(3-0) \right] \\ =\dfrac { 1 }{ 2 } \left[ 3+12 \right] =\dfrac { 15 }{ 2 } =7.5$$
$$\therefore$$ Area of the given triangle ABC is $$7.5$$ sq. units
Question 3
1 / -0

Given: $$\dfrac {20!}{18!}=380$$

A
True

B
False

C
Either

D
Neither

Solution

$$\cfrac { 20! }{ 18! } =?\\ 20!=20\times 19\times 18\times ........3\times 2\times 1\\ =20\times 19\times (18!)\\ \cfrac { 20! }{ 18! } =\cfrac { 20\times 19\times (18!) }{ 18! } =20\times 19=380$$
Hence the value matches so the equation is true.
Question 4
1 / -0

Area of the triangle formed by the points P(-1.5, 3), Q(6, -2) and R(-3, 4) is 0.

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Given $$P(-1.5, 3), Q(6, -2)$$ and $$R(-3, 4)$$
Therefore, area is given by
$$= \frac{1}{2}\times[(-1.5)(-2-4) + 6(4-3)+(-3)(3+2)]$$
$$= \frac{1}{2}\times(9 +6-15)$$
$$= 0$$
Question 5
1 / -0

If $$y^x=x^y$$, then find $$\displaystyle\frac{dy}{dx}$$.

A
$$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$$

B
$$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$

C
$${y(x\log{y}-y)}$$

D
$$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$$

Solution

$$y^x=x^y$$

Taking log on both sides

$$\Rightarrow$$ $$\log{y^x}=\log{x^y}$$

$$\Rightarrow$$ $$x\log{y}=y\log{x}$$

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

$$\Rightarrow$$ $$\displaystyle\left(\frac{x}{y}-\log{x}\right)\frac{dy}{dx}=\frac{y}{x}-\log{y}$$

$$\Rightarrow$$ $$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{y}{x}-\log{y}}{\displaystyle\frac{x}{y}-\log{x}}=\frac{y(y-x\log{y})}{x(x-y\log{x})}=\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$
Question 6
1 / -0

If $$y=x^{-\tfrac12}+\log_5x+\displaystyle \frac {\sin x}{\cos x}+2^x$$, then find $$\dfrac {dy}{dx}$$

A
$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

B
$$\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

C
$$-\displaystyle \frac {3}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

D
$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\cos^2x+2^x\log 2$$

Solution

Here, we have function $$y=x^{-1/2}+\log _5x+\tan x+2^x$$
On differentiating w.r.t x, we get
$$\dfrac {dy}{dx}=\dfrac {d}{dx}(x)^{-1/2}+\dfrac {d}{dx}(\log _5x)+\dfrac {d}{dx} \tan x+\dfrac {d}{dx}(2^x)$$

$$=-\dfrac {1}{2}(x)^{-1/2-1}+\dfrac {1}{x \log _e5}+\sec ^2x+2^x \log 2$$

$$=-\dfrac {1}{2}x^{-3/2}+\dfrac {1}{x\log _e5}+\sec ^2x+2^x\log 2$$
Question 7
1 / -0

$$\displaystyle\frac{dy}{dx}$$ for $$y=x^x$$ is

A
$$x^x(1-\log{x})$$

B
$$x^x(1-\log{y})$$

C
$$x^x(1+\log{y})$$

D
$$x^x(1+\log{x})$$

Solution

$$y=x^x$$
Taking $$\log $$ on both the sides
$$\log { y } =\log { \left( { x }^{ x } \right) } $$
$$\log { y } =x\log { x } $$
Differentiate w.r to $$x$$
$$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =x\frac { d }{ dx } \left( \log { x } \right) +\left( \log { x } \right) \frac { d }{ dx } \left( x \right) $$
$$\displaystyle \frac { dy }{ dx } =y\left[ \frac { x }{ x } +\log { x } \right] $$
$$\displaystyle \frac { dy }{ dx } ={ x }^{ x }\left[ 1+\log { x } \right] $$
Question 8
1 / -0

For the function $$f(x) = \displaystyle \frac{x^{100}}{100} + \frac{x^{99}}{99} + ........... + \frac{x^2}{2} + x+1$$, $$f'(1) =$$

A
$$x^{100}$$

B
$$100$$

C
$$101$$

D
None of these

Solution

$$Given:\cfrac { d }{ dx } (x^{ n })=nx^{ n-1 }\\ \therefore for\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...............+\cfrac { x^{ 2 } }{ 2 } +1\\ f'(x)=\cfrac { 100x^{ 99 } }{ 100 } +99\cfrac { x^{ 98 } }{ 99 } +.............+\cfrac { 2x }{ 2 } +1\\ Here\; f'(1)=1+1+.........to\; 100term=\; 100\\ Hence\; f(x)=\cfrac { x^{ 100 } }{ 100 } +\cfrac { x^{ 99 } }{ 99 } +...........+\cfrac { x^{ 2 } }{ 2 } +x+1=100$$
Question 9
1 / -0

If $$y=x^2+sin^{-1}x+log_ex$$, find $$\dfrac {dy}{dx}$$

A
$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

B
$$\displaystyle \frac {dy}{dx}=x+\displaystyle \frac{1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

C
$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}-\displaystyle \frac {1}{x}$$

D
$$\frac {dy}{dx}=2x-\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

Solution

$$y=x^2+sin^{-1}+log_ex$$
On differentiating, we get
$$\dfrac {dy}{dx}=\dfrac {d}{dx}(x^2)+\dfrac {d}{dx}sin^{-1}x)+\dfrac {d}{dx}(log_ex)$$
or $$\dfrac {dy}{dx}=2(x)^{2-1}+\dfrac {1}{\sqrt {1-x^2}}+\dfrac {d}{dx}(log_ex)$$
$$\dfrac {dy}{dx}=2x+\dfrac {1}{\sqrt {1-x^2}}+\dfrac {1}{x}$$
Question 10
1 / -0

A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is agraph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any other point through a sequence of edges. The number of edges "e" in the graph must satisfy the condition

A
$$11 \leq e \leq 66$$

B
$$10 \leq e \leq 66$$

C
$$11 \leq e \leq 65$$

D
$$0 \leq e \leq 11$$

Solution

(A) Since every edge connects a pair of points, the given 12 points have to be joined using lines. We may have minimum number of edges if all the 12 points are collinear.
No. of edges in this particular case
$$=12-1=11$$
Maximum number of edges are possible when all the 12 points are non-collinear. In this particular case number of different straight lines that can be formed using 12 points which is equal to $$^12C_{2}$$
$$=\frac{12\times 11}{2}=66$$
Therefore, following inequality holds for "e"
$$11 \leq e \leq 66$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Differentiation Test 15

Result

Differentiation Test 15

Score

-

Rank

-

SHARING IS CARING

Question 1

$$a = -1$$ , Area $$=$$ $$15$$ sq. unit

$$a = 2$$ , Area $$=$$ $$6$$ sq. unit

$$a = 0 $$ , Area $$=$$ $$8$$ sq. unit

$$a = 1$$ , Area $$=$$ $$12$$ sq. unit

Solution

Question 2

$$15$$ sq. units

$$10$$ sq. units

$$7.5$$ sq. units

$$2.5$$ sq. units

Solution

Question 3

True

False

Either

Neither

Solution

Question 4

True

False

Ambiguous

Data insufficient

Solution

Question 5

$$\displaystyle\frac{x(y\log{y}-y)}{y(x\log{x}-x)}$$

$$\displaystyle\frac{y(x\log{y}-y)}{x(y\log{x}-x)}$$

$${y(x\log{y}-y)}$$

$$\displaystyle\frac{y(x\log{x}-y)}{x(y\log{y}-x)}$$

Solution

Question 6

$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

$$\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

$$-\displaystyle \frac {3}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\sec^2x+2^x\log 2$$

$$-\displaystyle \frac {1}{2}x^{-3/2}+\displaystyle \frac {1}{x\log_e5}+\cos^2x+2^x\log 2$$

Solution

Question 7

$$x^x(1-\log{x})$$

$$x^x(1-\log{y})$$

$$x^x(1+\log{y})$$

$$x^x(1+\log{x})$$

Solution

Question 8

$$x^{100}$$

$$100$$

$$101$$

None of these

Solution

Question 9

$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

$$\displaystyle \frac {dy}{dx}=x+\displaystyle \frac{1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

$$\displaystyle \frac {dy}{dx}=2x+\displaystyle \frac {1}{\sqrt {1-x^2}}-\displaystyle \frac {1}{x}$$

$$\frac {dy}{dx}=2x-\displaystyle \frac {1}{\sqrt {1-x^2}}+\displaystyle \frac {1}{x}$$

Solution

Question 10

$$11 \leq e \leq 66$$

$$10 \leq e \leq 66$$

$$11 \leq e \leq 65$$

$$0 \leq e \leq 11$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.