Differentiation Test 42

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Differentiation Test 42

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Question 1

1 / -0

The letters of word 'ZENITH' are written in all positive ways. If all these words are written in the order of a dictionary, then the rank of the word 'ZENITH' is

$$716$$

$$692$$

$$698$$

$$616$$

Solution

The total number of words is $$6! = 720$$. Let us write the letters of word ZENITH alphabetically, i.e, EHINTZ.

For ZENITH word starts with	Word starting with	Number of words
$$Z$$	$$E$$	$$5!$$
	$$H$$	$$5!$$
	$$I$$	$$5!$$
	$$N$$	$$5!$$
	$$T$$	$$5!$$
ZEN	ZEH	$$3!$$
	ZEI	$$3!$$
ZENI	ZENH	$$2$$
ZENIT	ZENIH	$$1$$
	Total number of words before ZERNITH	$$615$$

Hence, there are $$615$$ words before ZENITH, so the rank of ZENITH is $$616$$.

Question 2
1 / -0

$$ f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2) $$ and $$ g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x) $$
The value of $$ f(3) $$ is

A
1

B
0

C
-1

D
-2

Solution

$$ f(x)=x^{2}+x g^{\prime}(1)+g^{\prime \prime}(2) $$ and $$ g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x) $$

$$\text { Here put } g^{\prime}(1)=a, g^{\prime \prime}(2)=b \quad\quad\quad(1)$$

$$\text { Then } f(x)=x^{2}+a x+b, f(1)=1+a+b \Rightarrow f^{\prime}(x)=2 x+a$$

$$\quad f^{\prime \prime}(x)=2 $$

$$\therefore g(x)=(1+a+b) x^{2}+(2 x+a) x+2=x^{2}(3+a+b)+a x\quad+2$$

$$\Rightarrow g^{\prime}(x)=2 x(3+a+b)+a \text { and } g^{\prime \prime}(x)=2(3+a+b)$$

$$\text { Hence, } g^{\prime}(1)=2(3+a+b)+a \quad\quad\quad(2)$$

$$g^{\prime \prime} (2)=2(3+a+b)\quad\quad\quad(3)$$

From (1),(2) and (3) we have

$$ a=2(3+a+b)+a $$

$$\text { and } b=2(3+a+b) $$

$$\Rightarrow 3+a+b=0 \text { and } b+2 a+6=0$$

Hence, $$ b=0 $$ and $$ a=-3 . $$ So, $$ f(x)=x^{2}-3 x $$ and $$ g(x)=-3 x+2 $$

$$ f(x)=x^{2}-3 x \Rightarrow f(3)=9-9=0 $$
Question 3
1 / -0

The next number in the pattern 62, 37, 12 ____ is

A
25

B
13

C
0

D
-13

Solution

Option (b) is correct; here, the given series has a common difference of +25.
$$\begin{array}{l}\Rightarrow-62+25=-37,-37+25=-12 \\\therefore-12+25=13\end{array}$$
Question 4
1 / -0

the area of a triangle with vertices $$ (-3, 0),(3,0) $$ and $$ (0, k) $$ is $$ 9 $$ sq units. then the value of $$ k $$ will be

A
$$ 9 $$

B
$$ 3 $$

C
$$ -9 $$

D
$$ 6 $$

Solution

We know that , area of a triangle with vertices $$ (a_1,y_1),(x_2,y_2) $$ and $$ (x_3,y_3) $$ is given by
$$ \Delta = \frac {1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| $$
$$ \therefore \Delta = \frac {1}{2} \left| \begin{matrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} \right| $$
Expanding along $$ R_1 $$
$$ 9 = \frac {1}{2} [ -3 ( -k) - 0 +1 ( 3k) ] $$
$$ \Rightarrow 18 = 3k + 3k = 6 k $$
$$ \therefore K = \frac {18 }{6} = 3 $$
Question 5
1 / -0

Derivative of $${\log{x}}^{\displaystyle\cos{x}}$$ with respect to $$x$$ is

A
$$\displaystyle{\log{x}}^{\displaystyle\cos{x}}\left[\frac{\cos{x}}{x\log{x}}-\sin{x}\log{(\log{x})}\right]$$

B
$$\displaystyle{\log{x}}^{\displaystyle\cos{x}}\left[\frac{\cos{x}}{x\log{x}}-\cos{x}\log{(\log{x})}\right]$$

C
$$\displaystyle{\log{x}}^{\displaystyle\sin{x}}\left[\frac{\sin{x}}{x\log{x}}-\cos{x}\log{(\log{x})}\right]$$

D
None of these

Solution

$$y={ \log { x } }^{ \cos { x } }$$
Taking log on both sides, we get
$$\log { y } =\log { \left( { \log { x } }^{ \cos { x } } \right) } $$
$$\log { y } =\cos { x } .\left[ \log { \left( \log { x } \right) } \right] $$
Differentiate w.r. to $$x$$
$$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =\cos { x } \frac { d }{ dx } \left[ \log { \left( \log { x } \right) } \right] +\left[ \log { \left( \log { x } \right) } \right] \frac { d }{ dx } \cos { x } $$

$$\displaystyle \frac { 1 }{ y } \frac { dy }{ dx } =\cos { x } .\frac { 1 }{ \log { x } } \frac { d }{ dx } \left( \log { x } \right) -\left[ \log { \left( \log { x } \right) } \right] \sin { x } $$

$$\displaystyle \frac { dy }{ dx } =y\left[ \cos { x.\frac { 1 }{ x\log { x } } - } \left[ \log { \left( \log { x } \right) } \right] \sin { x } \right] $$

$$\therefore \displaystyle \frac { dy }{ dx } ={ \log { x } }^{ \cos { x } }\left[ \cos { x.\frac { 1 }{ x\log { x } } - } \left[ \log { \left( \log { x } \right) } \right] \sin { x } \right] $$
Question 6
1 / -0

lf $${f}'({x})={g}({x})$$ and $${g}'({x})=-{f}({x})$$ for all $$x$$ and $${f}(2)=4= {g}(2)$$, then $${f}^{2}(24)+{g}^{2}(24)$$ is

A
$$32$$

B
$$24$$

C
$$64$$

D
$$48$$

Solution

$$f(x) = -g'(x)$$ (given)
By multiply both sides with $$f'(x)$$, we get
$$f(x)f'(x) = -g'(x)f'(x)$$ ...(1)
$$\because f'(x) = g(x)$$
$$\therefore$$ Eqn (1) becomes
$$f(x)f'(x) = -g(x)g'(x)$$
Now integrating both sides, we get
$$f^2(x) + g^2(x) = c$$
Where, $$c$$ is the constant of integration.
Now, it is given that $$f(2) = g(2) = 4$$
$$\therefore c = 32$$
$$\implies f^2(24) + g^2(24) = 32$$

Hence, option A is correct.
Question 7
1 / -0

Assertion(A): Let $${ f }({ x })$$ be twice differentiable function such that $$f^{ '' }(x)=-{ f }({ x })$$ and $$f^{ ' }(x)={ g }({ x })$$. lf $${ h }({ x })=[{ f }({ x })]^{ 2 }+[{ g }({ x })]^{ 2 }$$ and $${ h }(1)=8$$, then $${ h }(2)=8$$

Reason (R): Derivative of a constant function is zero.

A
Both A and R are true R is correct reason of A

B
Both A and R are true R is not correct reason of A

C
A is true but R is false

D
A is false but R is true

Solution

Let $$f(x)=c$$
$${f}'(x)=0$$ So Reason is True

Now, $$h(x) = [f(x)]^2 + [g(x)]^2$$
$${h}'(x)=2f(x){f}'(x)+2g(x){g}'(x)$$
$$=2f(x)g(x)+2g(x){f}''(x)$$ ...since $$f'(x) = g(x) \Rightarrow f''(x) = g'(x)$$
$$=2g(x)[f(x) + f''(x)]$$
$$=2g(x)(f(x)-f(x))$$
$$\Rightarrow {h}'(x)=0$$
means $$h(x)=C$$ (a constant function)
Given $$h(1)=8$$
$$\Rightarrow h(x)=8$$
Hence, $$f(2)= 8$$
So, both A and R are correct and R is the correct explanation of A.
Question 8
1 / -0

$$f(x)=|x-1|+|x-3|$$ then $$f^{'}(2)=$$

A
$$-2$$

B
$$2$$

C
$$0$$

D
$$1$$

Solution

$$f(x) = |x-1|+|x-3| =\begin{cases} -(x-1)-(x-3), x\le 1\\(x-1)-(x-3), 1< x\le 3\\x-1+(x-3), x>3\end{cases}$$
$$\Rightarrow f(x) =\begin{cases} 4-2x, x\le 1\\2, 1< x\le 3\\2x-4, x>3\end{cases}$$
$$\therefore f'(x) =\begin{cases} -2, x\le 1\\0, 1< x\le 3\\2, x>3\end{cases}$$
Hence $$f'(2) =0$$
Question 9
1 / -0

Assertion (A): lf $$f(x)=\cos^{2}x+\cos^{2}\left(x+\dfrac{\pi}3\right)- \cos x \cos \left(x+\dfrac{\pi}3\right)$$ then $$f'(x)=0$$

Reason(R): Derivative of constant function is zero

A
Both A & R are true, R is correct explanation for A

B
Both A & R are true,R is not correct explanation for A

C
A is true but R is false

D
A is false but R is true

Solution

$$f(x)=\cos ^{2}x+\cos ^{2}(x+\cfrac{\pi }{3})-2\cos x \cos (x+\cfrac{\pi }{3})+\cos x \cos (x+\cfrac{\pi }{3})$$
$$=(\cos x - \cos (x+\cfrac{\pi }{3}))^{2} + \cos x \cos (x+\cfrac{\pi }{3}) $$
$$=(\cos x - (\cfrac{1}{2}\cos x - \cfrac{\sqrt{3}}{2} \sin x ))^{2} + \cos x \cos (x+\cfrac{\pi }{3})$$
$$=(\cfrac{\cos x }{2} + \cfrac{\sqrt{3}}{2} \sin x )^{2} + \cos x \cos (x+\cfrac{\pi }{3})$$
$$=\cfrac{\cos^2x}4 + \cfrac{3\sin^2x}4 +\cfrac{\sqrt3}2 \sin x\cos x + \cos x\left(\cfrac{\cos x}2 - \cfrac{\sqrt3}2\sin x\right)$$
$$f(x)=\cfrac{3}{4}$$
$$f{}'(x)=0$$
Question 10
1 / -0

Given that $$f (x)$$
is a differentiable function of $$ x$$ and that $$f(x)$$ . $$f (y)$$ = $$f (x) $$+ $$f (y)$$ + $$f (xy) -2$$ and that
$$f (2) =5$$.
Then $$f (3)$$ is equal to?

A
$$6$$

B
$$24$$

C
$$15$$

D
$$19$$

Solution

Given, $$f\left( x \right) .f\left( y \right) =f\left( x \right) +f\left( y \right) +f\left( xy \right) -2\quad -(1)$$
$$f\left( 2 \right) =5$$
$$f\left( x \right) ={ x }^{ 2 }+1$$
Equation $$1 \Rightarrow $$ $$\left( { x }^{ 2 }+1 \right) \left( { y }^{ 2 }+1 \right) ={ x }^{ 2 }+1+{ y }^{ 2 }+1+{ x }^{ 2 }{ y }^{ 2 }+1-2$$
$${ x }^{ 2 }{ y }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 }+1={ x }^{ 2 }+{ y }^{ 2 }+{ x }^{ 2 }{ y }^{ 2 }+1$$
$$\therefore f\left( 3 \right) ={ x }^{ 2 }+1={ 3 }^{ 2 }+1=10$$

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Re-Attempt Weekly Quiz Competition

Differentiation Test 42

Result

Differentiation Test 42

Score

-

Rank

-

SHARING IS CARING

Question 1

$$716$$

$$692$$

$$698$$

$$616$$

Solution

Question 2

1

0

-1

-2

Solution

Question 3

25

13

0

-13

Solution

Question 4

$$ 9 $$

$$ 3 $$

$$ -9 $$

$$ 6 $$

Solution

Question 5

$$\displaystyle{\log{x}}^{\displaystyle\cos{x}}\left[\frac{\cos{x}}{x\log{x}}-\sin{x}\log{(\log{x})}\right]$$

$$\displaystyle{\log{x}}^{\displaystyle\cos{x}}\left[\frac{\cos{x}}{x\log{x}}-\cos{x}\log{(\log{x})}\right]$$

$$\displaystyle{\log{x}}^{\displaystyle\sin{x}}\left[\frac{\sin{x}}{x\log{x}}-\cos{x}\log{(\log{x})}\right]$$

None of these

Solution

Question 6

$$32$$

$$24$$

$$64$$

$$48$$

Solution

Question 7

Both A and R are true R is correct reason of A

Both A and R are true R is not correct reason of A

A is true but R is false

A is false but R is true

Solution

Question 8

$$-2$$

$$2$$

$$0$$

$$1$$

Solution

Question 9

Both A & R are true, R is correct explanation for A

Both A & R are true,R is not correct explanation for A

A is true but R is false

A is false but R is true

Solution

Question 10

$$6$$

$$24$$

$$15$$

$$19$$

Solution

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