Limits and Derivatives Test

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Limits and Derivatives Test - 38

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Question 1
1 / -0

Let $$f(x)$$ be a polynomial function of second degree.If $$f(1)=f(-1)$$ and $$a,b,c$$ are in A.P $$ f'(a)$$,$$f'(b)$$,$$f'(c)$$ are in

A
G.P.

B
H.P.

C
A.G.P

D
A.P.

Solution

Let $$f\left( x \right) ={ px }^{ 2 }+qx+r$$
Differentiatin g w.r. to $$x$$, we get
$$f^{ ' }\left( x \right) =2px+q$$
Now,
$$f\left( -1 \right) =f\left( 1 \right) $$
$$\therefore p-q+r=p+q+r$$
$$\therefore q=0$$
$$\therefore f^{ ' }\left( x \right) =2px$$
$$f^{ ' }\left( a \right) =2pa.f^{ ' }\left( b \right) =2pb,f^{ ' }\left( c \right) =2pc$$
Since $$a,b,c $$ are in $$A.P.$$, therefore
$$2b=a+c$$
$$4pb=2pa+2pc$$
$$2f^{ ' }\left( b \right) =f^{ ' }\left( a \right) +f^{ ' }\left( c \right) $$
$$\therefore f^{ ' }\left( a \right) ,f^{ ' }\left( b \right) ,f^{ ' }\left( c \right) $$ are in A.P
Question 2
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If $$\phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}$$, then

A
$$\phi (x) = g(x)$$ for all x $$\in$$ R

B
$$\phi (x) = f(x)$$ for all x $$\in$$ R

C
$$\left\{\begin{matrix}g(x) & for -1 < x < 1\\ f(x) & for |x| \geq 1\end{matrix}\right.$$

D
$$\left\{\begin{matrix}g(x) & for |x| < 1\\ f(x) & for |x| > 1 \\\displaystyle \frac{f(x) + g(x)}{2} & for |x| = 1\end{matrix}\right.$$

Solution

We have,
$$\displaystyle \lim_{n \rightarrow \infty} x^{2n} = \left\{\begin{matrix}0,& if |x| < 1\\ \infty, & if |x| > 1 \\ 1, & if |x| = 1\end{matrix}\right.$$
Thus, we have the following cases:
CASE I When $$- 1 < x < 1$$
In this case, we have $$\displaystyle \lim_{n \rightarrow \infty} x^{2n} = 0$$
$$\therefore \phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}} = g(x)$$
CASE II When $$|x| > 1$$
In this case, we have $$\displaystyle \lim_{n \rightarrow \infty} \displaystyle \frac{1}{x^{2n}} = 0$$
$$\therefore \phi (x) = \displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}$$
$$\Rightarrow \phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{f(x) + \displaystyle \frac{g(x)}{x^{2n}}}{1 + \displaystyle \frac{1}{x^{2n}}} = \displaystyle \frac{f(x) + 0}{1 + 0} = f(x)$$
CASE III When $$|x| = 1$$
In this case, we have $$x^{2n} = 1\Rightarrow \displaystyle \lim_{n \rightarrow \infty } x^{2n}=1$$.
$$\therefore \phi (x) = \displaystyle \frac{f(x) + g(x)}{2}$$
Question 3
1 / -0

Let $$f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}, 1 \le x \le 26$$ be a real valued function, then $$f'(x)$$ for $$1 < x < 26$$ is

A
$$0$$

B
$$\displaystyle\frac{1}{\sqrt{x - 1}}$$

C
$$2\sqrt{x - 1}$$

D
$$1$$

Solution

We have,
$$f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}}, 1 < x < 26$$

Rearrange the terms so as to get $${ \left( a-b \right) }^{ 2 }=a^2-2ab+b^2$$

$$f(x) = \sqrt{x - 1} + \sqrt{(x - 1) + 25 - 10\sqrt{x - 1}}$$

$$\therefore f(x) = \sqrt{x - 1} + \sqrt{(\sqrt {x - 1} - 5)^2}$$

$$f(x) = \sqrt{x - 1} + \left|\sqrt{x - 1} - 5\right|$$

$$

\therefore f(x) = \sqrt{x - 1} - \left(\sqrt{x - 1} - 5\right) \quad\quad

[\because\sqrt{x - 1} - 5 < 0 \mbox{ for } 1 < x < 26]$$
Now, differentiating w.r. to $$x$$, we get
$$\therefore f'(x) = 0$$ for all $$x \in (1, 26)$$
Question 4
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Let $$f$$ be a differentiable function satisfying $$f(x) + f(y) + f(z) + f(x)f(y)f(z) = 14$$ for all $$x,\space y,\space z \in R$$
Then,

A
$$f'(x) < 0$$ for all $$x \in R$$

B
$$f'(x) = 0$$ for all $$x \in R$$

C
$$f'(x) > 0$$ for all $$x \in R$$

D
none of these

Solution

We have,
$$f(x) + f(y) + f(z) + f(x)f(y)f(z) = 14$$ for all $$x,\space y,\space z \in R \quad\quad ...(i)$$
Putting $$x = y = z = 0$$, we get,
$$3f(0) + \left\{f(0)\right\}^3 = 14$$
$$\left\{f(0)\right\}^3 + 3f(0) - 14 = 0$$
$$f(0) = 2.$$
Now, putting $$y = z = x$$ in $$(i)$$, we get
$$3f'(x) + 3\left\{f(x)\right\}^2f'(x) = 0 \quad$$ for all $$x \in R$$
$$\left\{\left\{f(x)\right\}^2 + 1\right\}f'(x) = 0 \quad$$ for all $$x \in R$$
$$ f'(x) = 0 \quad$$ for all $$x \in R$$
Question 5
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Given $$f(x)=-\displaystyle \frac{x^3}{3}+x^2\sin 1.5a-x\sin a.\sin 2a-5 \arcsin (a^2-8a+17)$$ then :

A
$$f(x)$$ is not defined at $$x=\sin 8$$

B
$${f}'(\sin 8)>0$$

C
$$f'(x)$$ is not defined at $$x=\sin 8$$

D
$${f}'(\sin 8)<0$$

Solution

Given:
$$f(x) = \dfrac{-x^3}{3} + x^2 \sin 1.5\, a - x \sin a . \sin 2a - 5\, arc \sin (a^2 - 8a + 17)$$

$$f(x) = \dfrac{-x^3}{3} + x^2\sin 1.5\, a - x \sin a \times \sin 2a-5\sin^{-1}(a^2-8a+17)$$ $$\left(\sin c + \sin d = 2\sin \dfrac{c+d}{2} \cos \dfrac{c-d}{2}\right)$$

For the function to be defined

$$-1\le a^2 - 8a + 17 \le 1$$

$$-1 \le (a-4)^2 + 1 \le 1$$

$$-1-1\le (a-4)^2 \le 1-1$$

$$-2 \le (a-4)^2 \le 0$$

$$a -4 = 0 \Rightarrow a = 4$$

Putting value of '$$a$$' in the +ve $$f(x)$$

$$f(x) = \dfrac{-x^3}{3} + x^2 \sin 6 - x\sin 4 \times \sin 8 - 5\dfrac{\pi}{2}$$ $$\left[\because \sin^{-1}1 = \dfrac{\pi}{2}\right]$$

domain $$x \in R \because $$ polynomial function

$$f'(x) = -\dfrac{3x^2}{3} + 2x\sin 6 - \sin 4 \times \sin 8$$

$$= -x^2 + 2\sin 6x - \sin 4 \times \sin 8$$

Putting $$x = \sin 8$$

$$f'(\sin 8) = -\sin^28 + 2\sin 6 \sin 8 - \sin 4 \sin 8$$

$$= \sin 8 [2\sin 6 - \sin 4 - \sin 8]$$

$$=\sin 8 [ 2\sin 6 - (\sin 4 + \sin 8)]$$

$$=\sin 8[2\sin 6 - 2\sin 6\cos 2]$$

$$= 2\sin 6 \sin 8 [ 1 - \cos 2]$$

Value of $$\cos 2 = 0.999$$

$$\therefore 1 - \cos 2 = +ve$$

$$6\, rad \approx 344^o$$

$$344^o$$ is in $$IV^{th}$$ quadrant

$$\therefore \sin 6$$ is (-ve)

$$8\, rad \approx 548.4^o$$

$$548.4^o$$ is in $$II^{th}$$ quadrant

$$\therefore \sin 8$$ is (+ve)

$$\therefore f'(\sin 8) = 2\,\underbrace{\sin 6}_{-ve} \,\underbrace{\sin 8}_{+ve} \,\underbrace{(1-\cos 2)}_{+ve}$$

$$\therefore f'(\sin 8)$$ is +ve

$$\therefore f'(\sin 8) < 0$$ $$\to$$ option D
Question 6
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Suppose, $$A=\displaystyle \frac {dy}{dx}$$ of $$x^2+y^2=4$$ at $$(\sqrt 2, \sqrt 2), B=\displaystyle \frac {dy}{dx}$$ of $$sin y+sin x=sin x\cdot sin y$$ at $$(\pi, \pi)$$ and $$C=\displaystyle \frac {dy}{dx}$$ of $$2e^{xy}+e^xe^y-e^x=e^{xy+1}$$ at $$(1, 1)$$, then $$(A-B-C)$$ has the value equal to .....

A
$$\displaystyle \frac { 1 }{ 2 }$$

B
$$\displaystyle \frac { 1 }{ 3 }$$

C
$$1$$

D
$$2$$

Solution

$$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$$ at $$(\sqrt 2, \sqrt 2))$$
$$2x+2y\displaystyle \frac { dy }{ dx } =0$$
$$\displaystyle \frac { dy }{ dx } =-\displaystyle \frac { x }{ y } =-\displaystyle \frac { \sqrt { 2 } }{ \sqrt { 2 } } =-1$$
$$ B: \displaystyle \frac {d}{dx}(\sin y+\sin x=\sin x\cdot \sin y)$$ at $$(\pi, \pi)$$
$$\cos { y\displaystyle \frac { dy }{ dx } } +\cos { x } =\sin { x } \cos { y\displaystyle \frac { dy }{ dx } +\sin { y } \cos { x } } $$
$$\displaystyle \frac { dy }{ dx } \left( \cos { y } -\sin { x } \cos { y } \right) =\sin { y\cos { x-\cos { x } } } $$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { \cos { x\left( \sin { y-1 } \right) } }{ \cos { y\left( 1-\sin { x } \right) } } $$
$$B=\displaystyle \frac { -1\left( 0-1 \right) }{ -1\left( 1-0 \right) } =-1$$
$$C: \displaystyle \frac {d}{dx}(2e^{xy}+e^xe^y-e^x=e^{xy+1})$$ at $$(1, 1)$$
$$\left[ 2{ e }^{ xy }\left( x\displaystyle \frac { dy }{ dx } +y \right) \right] +{ e }^{ x }{ e }^{ y }\displaystyle \frac { dy }{ dx } +{ e }^{ y }{ e }^{ x }-{ e }^{ x }={ e }^{ xy+1 }\left( x\displaystyle \frac { dy }{ dx } +y \right) $$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { y{ e }^{ xy+1 }-2y{ e }^{ xy }-{ e }^{ x+y }+{ e }^{ x } }{ 2x{ e }^{ xy }+{ e }^{ x+y }-x{ e }^{ xy+1 } } $$.......... since $${ e }^{ x }{ e }^{ y }={ e }^{ x+y }$$
$$C=\displaystyle \frac { { e }^{ 2 }-2{ e }^{ 1 }-{ e }^{ 2 }+{ e }^{ 1 } }{ 2{ e }^{ 1 }+{ e }^{ 2 }-{ e }^{ 2 } } $$
$$=\displaystyle \frac { -{ e }^{ 1 } }{ 2{ e }^{ 1 } } =-\displaystyle \frac { 1 }{ 2 } $$
Therefore $$A-B-C=\displaystyle \frac { 1 }{ 2 }$$
Question 7
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Suppose $$A=\displaystyle \frac{dy}{dx}$$ when $$x^2+y^2=4$$ at $$(\sqrt{2},\sqrt{2})$$,$$ B=\displaystyle \frac{dy}{dx}$$ when $$\sin y+ \sin x=\sin x-\sin y$$ at $$(\pi,\pi)$$ and $$C=\displaystyle \frac{dy}{dx}$$ when $$2e^{xy}+e^x e^y-e^x-e^y=e^{xy+1}$$ at $$(1,1)$$, then $$(A+B+C)$$ has the value equal to

A
$$-1$$

B
$$e$$

C
$$-3$$

D
$$0$$

Solution

$$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$$ at $$(\sqrt 2, \sqrt 2))$$
$$2x+2y\displaystyle \frac { dy }{ dx } =0$$
$$\displaystyle \frac { dy }{ dx } =-\displaystyle \frac { x }{ y } =-\displaystyle \frac { \sqrt { 2 } }{ \sqrt { 2 } } =-1$$
$$ B: \displaystyle \frac {d}{dx}(\sin y+\sin x=\sin x\cdot \sin y)$$ at $$(\pi, \pi)$$
$$\cos

{ y\displaystyle \frac { dy }{ dx } } +\cos { x } =\sin { x } \cos {

y\displaystyle \frac { dy }{ dx } +\sin { y } \cos { x } } $$
$$\displaystyle \frac { dy }{ dx } \left( \cos { y } -\sin { x } \cos { y } \right) =\sin { y\cos { x-\cos { x } } } $$
$$\displaystyle

\frac { dy }{ dx } =\displaystyle \frac { \cos { x\left( \sin { y-1 }

\right) } }{ \cos { y\left( 1-\sin { x } \right) } } $$
$$B=\displaystyle \frac { -1\left( 0-1 \right) }{ -1\left( 1-0 \right) } =-1$$
$$C: \displaystyle \frac {d}{dx}(2e^{xy}+e^xe^y-e^x-{ e }^{ y }=e^{xy+1})$$ at $$(1, 1)$$
$$\left[

2{ e }^{ xy }\left( x\displaystyle \frac { dy }{ dx } +y \right)

\right] +{ e }^{ x }{ e }^{ y }\displaystyle \frac { dy }{ dx } +{ e }^{

y }{ e }^{ x }-{ e }^{ x } -{ e }^{ y }\frac { dy }{ dx } ={ e }^{ xy+1 }\left( x\displaystyle \frac {

dy }{ dx } +y \right) $$
$$\displaystyle \frac { dy }{ dx }

=\displaystyle \frac { y{ e }^{ xy+1 }-2y{ e }^{ xy }-{ e }^{ x+y }+{ e

}^{ x } }{ 2x{ e }^{ xy }+{ e }^{ x+y }-x{ e }^{ xy+1 }-{ e }^{ y } } $$..........

since $${ e }^{ x }{ e }^{ y }={ e }^{ x+y }$$
$$C=\displaystyle \frac { { e }^{ 2 }-2{ e }^{ 1 }-{ e }^{ 2 }+{ e }^{ 1 } }{ 2{ e }^{ 1 }+{ e }^{ 2 }-{ e }^{ 2 }-{ e }^{ 1 } }=-{1} $$
$$A+B+C=-1-1-1=-3$$
Question 8
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If $$\displaystyle\lim_{x\rightarrow a}{(f(x)+g(x))}=2$$ and $$\displaystyle\lim_{x\rightarrow a}{(f(x)-g(x))}=1$$,
then the value of $$\displaystyle\lim_{x\rightarrow a}{f(x)g(x)}$$ is?

A
Does not exist

B
Exists and is $$\displaystyle\frac{3}{4}$$

C
Exists and is $$\displaystyle-\frac{3}{4}$$

D
Exists and is $$\displaystyle\frac{4}{3}$$

Solution

Given, $$\displaystyle\lim_{x\rightarrow a}{(f(x)+g(x))}=2$$ and $$\displaystyle\lim_{x\rightarrow a}{(f(x)-g(x))}=1$$,

Lets, assume $$F(x) = (f(x) + g(x))$$ and $$ G(x) = (f(x) - g(x))$$
$$\because$$ Limits of both $$F(x)$$ and $$G(x)$$ exists as $$x\rightarrow a$$, we can say that $$\displaystyle\lim_{x\rightarrow a}{(F(x)+G(x))}$$ and $$\displaystyle\lim_{x\rightarrow a}{(F(x)-G(x))}$$ also exists.

$$\displaystyle\therefore\lim_{x\rightarrow a}{(F(x)+G(x))} = 3$$
$$\displaystyle\Rightarrow 2\times\lim_{x\rightarrow a}{f(x)} = 3$$
$$\Rightarrow \lim_{x\rightarrow a}{f(x)} = \dfrac{3}{2}$$
Similarly,
$$\displaystyle\lim_{x\rightarrow a}{(F(x)-G(x))} = 1$$
$$\displaystyle\Rightarrow 2\times\lim_{x\rightarrow a}{g(x)} = 1$$
$$\Rightarrow \lim_{x\rightarrow a}{g(x)} = \dfrac{1}{2}$$
Hence,
$$\displaystyle\lim_{x\rightarrow a}{(f(x)\cdot g(x))} = \lim_{x\rightarrow a}{f(x)}\cdot\lim_{x\rightarrow a}{g(x)} = \dfrac{3}{2}\times \dfrac{1}{2} = \dfrac{3}{4}$$

Hence, option B.
Question 9
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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

A
$$T(x)$$

B
$$0$$

C
$$2f(x)g(x)$$

D
$$6f(x)g(x)$$

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 10
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The value of $$\displaystyle \underset { n\rightarrow \infty }{ lim } \left( \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 6n } \right) $$ is

A
$$\displaystyle \log { 2 } $$

B
$$\displaystyle \log { 6 } $$

C
$$\displaystyle 1$$

D
$$\displaystyle \log { 3 } $$

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Re-Attempt Weekly Quiz Competition

Limits and Derivatives Test - 38

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Limits and Derivatives Test - 38

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SHARING IS CARING

Question 1

G.P.

H.P.

A.G.P

A.P.

Solution

Question 2

$$\phi (x) = g(x)$$ for all x $$\in$$ R

$$\phi (x) = f(x)$$ for all x $$\in$$ R

$$\left\{\begin{matrix}g(x) & for -1 < x < 1\\ f(x) & for |x| \geq 1\end{matrix}\right.$$

$$\left\{\begin{matrix}g(x) & for |x| < 1\\ f(x) & for |x| > 1 \\\displaystyle \frac{f(x) + g(x)}{2} & for |x| = 1\end{matrix}\right.$$

Solution

Question 3

$$0$$

$$\displaystyle\frac{1}{\sqrt{x - 1}}$$

$$2\sqrt{x - 1}$$

$$1$$

Solution

Question 4

$$f'(x) < 0$$ for all $$x \in R$$

$$f'(x) = 0$$ for all $$x \in R$$

$$f'(x) > 0$$ for all $$x \in R$$

none of these

Solution

Question 5

$$f(x)$$ is not defined at $$x=\sin 8$$

$${f}'(\sin 8)>0$$

$$f'(x)$$ is not defined at $$x=\sin 8$$

$${f}'(\sin 8)<0$$

Solution

Question 6

$$\displaystyle \frac { 1 }{ 2 }$$

$$\displaystyle \frac { 1 }{ 3 }$$

$$1$$

$$2$$

Solution

Question 7

$$-1$$

$$e$$

$$-3$$

$$0$$

Solution

Question 8

Does not exist

Exists and is $$\displaystyle\frac{3}{4}$$

Exists and is $$\displaystyle-\frac{3}{4}$$

Exists and is $$\displaystyle\frac{4}{3}$$

Solution

Question 9

$$T(x)$$

$$0$$

$$2f(x)g(x)$$

$$6f(x)g(x)$$

Solution

Question 10

$$\displaystyle \log { 2 } $$

$$\displaystyle \log { 6 } $$

$$\displaystyle 1$$

$$\displaystyle \log { 3 } $$

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