CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 11 - MCQExams.com

$$\displaystyle \lim_{\rightarrow \infty} x \sin \dfrac {\pi}{4x}\cos \dfrac {\pi}{4x}$$
$$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {2\sin \dfrac {\pi}{4x} \cos \dfrac {\pi}{4x}}{\left (\dfrac {\pi}{2x}\right )} \times \dfrac {1}{2} \times \dfrac {\pi}{2x}$$
$$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin \left (2\times \dfrac {\pi}{4x}\right )}{\left (\dfrac {\pi}{2x}\right )} \times \dfrac {\pi}{4}$$
$$= \dfrac {\pi}{4}\times \displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin \left (\dfrac {\pi}{2x}\right )}{\left (\dfrac {\pi}{2x}\right )} = \dfrac {\pi}{4}\times 1 = \dfrac {\pi}{4}$$
Hence, option (A) is correct.

Given, $$f\left( x \right) .f\left( y \right) =f\left( x \right) +f\left( y \right) +f\left( xy \right) -2\quad -(1)$$

$$f(x) = -g'(x)$$     (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.

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

Since, $$f\left( x \right)$$ is continuous at $$x=0$$ , therefore, $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } =f\left( 0 \right) =0$$

Given $$y=f(x)$$ has a unique tangent at the point $$(e^a,0)$$ . 
So, as $$x\rightarrow e^{a}$$ , $$f(x)\rightarrow 0$$

Now, $$\displaystyle \lim_{x\rightarrow e^{a}}\frac{log_{e}\{1+7f(x)\}-\sin f(x)}{3f(x)}$$

$$\displaystyle \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 3f(x) } -\frac { \sin f(x) }{ 3f(x) }$$

$$\displaystyle =\frac { 7 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { log_{ e }(1+7f(x)) }{ 7f(x) } -\frac { 1 }{ 3 } \lim _{ x\rightarrow e^{ a } } \frac { \sin f(x) }{ f(x) }$$

$$=\dfrac{7-1}{3}=2$$

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

If $$f(x)$$ is continuous at $$x=0$$ , then $$\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =f(0)$$
$$\displaystyle LHL = \lim _{ h\rightarrow 0 }{ \left[ (0-h)\sin { \left( \frac { 1 }{0- h }  \right)  }  \right]  } =\lim _{ h\rightarrow 0 }{ h\sin { \frac { 1 }{ h }  }  } =0\times \; number\; bet.\; 0 \; and \; 1=0$$
$$\displaystyle RHL= \lim _{ h\rightarrow 0 }{ h\sin { \frac { 1 }{ h }  }  } =0\times \; number=0$$
Hence $$LHL=RHL=f(0)$$ . The given function is continuous at $$x=0$$
.
$$\displaystyle f'\left( x \right) =\lim _{ h\rightarrow 0 }{ \frac { h\sin { \frac { 1 }{ h }  } -f\left( 0 \right)  }{ h }  } =\lim _{ h\rightarrow 0 }{ \sin { \frac { 1 }{ h }  }  } -0=\lim _{ h\rightarrow 0 }{ \sin { \frac { 1 }{ h }  }  }$$
Since this limit does not exist, the given function is not differentiable at $$x=0$$

Let $$f(x) =ax^2+bx+c$$
According to the  given condition $$a>0, b^2-4ac <0$$ .........(i)
$$\therefore g(x) = ax^2+bx+c+2ax+b+2a=ax^2+(b+2a)x+(b+c+2a)$$
Now discriminant of $$g(x)$$ is $$D=(b+2a)^2-4a(2a+c+b)=b^2+4a^2+4ab-4ab-4ac-8a^2=(b^2-4ac)-4a^2< 0$$ using (i)
Hence $$g(x)> 0$$ for any $$x \in R$$

Given $$h'\left( x \right) ={ \left[ f\left( x \right)  \right]  }^{ 2 }+{ \left[ g\left( x \right)  \right]  }^{ 2 }$$

We have,
$$y=e^{\displaystyle\sqrt{x}}+e^{\displaystyle -\sqrt{x}}$$
Differentiate w.r. to $$x$$
$$\displaystyle\frac{dy}{dx}=\frac{e^{\displaystyle\sqrt{x}}}{2\sqrt{x}}-\frac{e^{\displaystyle -\sqrt{x}}}{2\sqrt{x}}=\frac{e^{\displaystyle\sqrt{x}}-e^{\displaystyle -\sqrt{x}}}{2\sqrt{x}}$$ ....................option (A)

We can write $${ e }^{\displaystyle \sqrt { x }  }-{ e }^{\displaystyle  -\sqrt { x }  }=\sqrt { { \left( { e }^{\displaystyle \sqrt { x }  }+{ e }^{\displaystyle -\sqrt { x }  } \right)  }^{ 2 }-4 }$$

$$\therefore \displaystyle \frac { dy }{ dx } \displaystyle  =\frac{\sqrt{{\left(e^{\displaystyle\sqrt{x}}+e^{\displaystyle -\sqrt{x}}\right)}^2-4}}{2\sqrt{x}}=\frac{\sqrt{y^2-4}}{2\sqrt{x}}$$ .................option (C)

Here, $$f(x)+f(y)+f(x)\cdot f(y)=1$$ .....(i)
Substitute $$x=y=0$$ , we get

$$2f(0)+\left \{f(0)\right \}^2=1\Rightarrow \left \{f(0)\right \}^2+2f(0)-1=0$$

$$f(0)=\frac {-2\pm \sqrt {4+4}}{2}=-1-\sqrt 2$$ and $$-1+\sqrt 2$$

As $$f(0) > 0\Rightarrow f(0)=\sqrt 2-1$$ [neglecting $$f(0)=-1-\sqrt 2$$ as f(0) is positive]

Again, putting $$y=x$$ is Eq, (i), $$2f(x)+\left \{f(x)\right \}^2=1$$

On differentiating w.r.t. x, $$2f'(x)+2f(x)f'(x)=0$$
$$2f'(x)\left \{1+f(x)\right \}=0\Rightarrow f'(x)=0$$  because $$f(x) > 0$$
Thus, $$f'(x)=0$$ when $$f(x) > 0$$

Let $$y=e^x+x$$
on differentiating w.r.t y,
$$1=(e^x+1)\dfrac {dx}{dy}\Rightarrow \dfrac {dx}{dy}=\dfrac {1}{1+e^x}$$
$$\therefore \left (\dfrac {dx}{dy}\right )_{x=log_e2}=\dfrac {1}{1+e^{log 2}}=\dfrac {1}{3}$$
or $$\left [\dfrac {d}{dx}(f^{-1}(x))\right ]_{x=log 2}=\dfrac {1}{3}$$

$$f(x)=-\displaystyle \frac {x^3}{3}+x^2 \sin 1.5 a-x \sin a\cdot \sin 2a-5 arc \sin (a^2-8a+17)$$
$$f(x)=\displaystyle \frac { -{ x }^{ 3 } }{ 3 } +{ x }^{ 2 }\sin { 6 } -x\sin { 4\sin { 8 } -5\sin ^{ -1 }{ \left( { \left( a-4 \right)  }^{ 2 }+1 \right)  }  }$$
$$f'(x)=-{ x }^{ 2 }+2x\sin { 6-\sin { 4.\sin { 8 }  }  }$$ .......( $$a=4$$ considering the domain of inverse sine function) 
$$f'\left( \sin { \quad 8 }  \right) =-\sin ^{ 2 }{ 8 } +2\sin { 6 } \sin { 8 } -\sin { 4 } \sin { 8 }$$
$$=\sin { 8\left( -\sin { 8+2\sin { 6-\sin { 4 }  }  }  \right)  }$$
$$=-\sin { 8 } \left( \sin { 8 } +\sin { 4-2\sin { 6 }  }  \right)$$
$$=-\sin { 8\left( 2\sin { 6\cos { 2-2\sin { 6 }  }  }  \right)  }$$
$$=2\sin { 8 } \sin { 6\left( 1-\cos { 2 }  \right)  }$$
Now, 
$$sin 8 > 0$$           (Since, $$2\pi  <  8 < 3\pi$$ )
$$sin 6 < 0$$          (Since, $$\pi < 6 < 2\pi$$ )
$$(1-\cos 2) > 0$$      
Hence, $$f'(\sin { 8 } )<0$$

$$\sqrt {y+x}+\sqrt {y-x}=c$$
Squaring both sides, we get
$${ \left( \sqrt { y+x } +\sqrt { y-x }  \right)  }^{ 2 }={ c }^{ 2 }$$
$$2y+2\sqrt { { y }^{ 2 }-{ x }^{ 2 } } ={ c }^{ 2 }$$
$$y+\sqrt { { y }^{ 2 }-{ x }^{ 2 } } =\displaystyle \frac { { c }^{ 2 } }{ 2 }$$ ...........(1)
Differentiating bothe the sides, we get
$$\displaystyle \frac { dy }{ dx } +\displaystyle \frac { 1 }{ 2\sqrt { { y }^{ 2 }-{ x }^{ 2 } }  } \left( 2y\displaystyle \frac { dy }{ dx } -2x \right) =0$$
$$\displaystyle \frac { dy }{ dx } +\displaystyle \frac { y }{ \sqrt { { y }^{ 2 }-{ x }^{ 2 } }  } \displaystyle \frac { dy }{ dx } -\displaystyle \frac { x }{ \sqrt { { y }^{ 2 }-{ x }^{ 2 } }  } =0$$
$$\displaystyle \frac { dy }{ dx } \left( 1+\displaystyle \frac { y }{ \sqrt { { y }^{ 2 }-{ x }^{ 2 } }  }  \right) =\displaystyle \frac { x }{ \sqrt { { y }^{ 2 }-{ x }^{ 2 } }  }$$
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { x }{ y+\sqrt { { y }^{ 2 }-{ x }^{ 2 } }  }$$ ..................Option (b)
On rationalizing we get,
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { x\left( y-\sqrt { { y }^{ 2 }-{ x }^{ 2 } }  \right)  }{ { x }^{ 2 } }$$
$$=\displaystyle \frac { y-\sqrt { { y }^{ 2 }-{ x }^{ 2 } }  }{ x }$$ ........Option (c)
Putting $$y+\sqrt { { y }^{ 2 }-{ x }^{ 2 } } ={ c }^{ 2 }$$ . we get
$$\displaystyle \frac { dy }{ dx } =\displaystyle \frac { x }{ \displaystyle \frac { { c }^{ 2 } }{ 2 }  } =\displaystyle \frac { 2x }{ { c }^{ 2 } }$$ ......... Option (a)

$$h'(x) = 2f(x) \times f'(x) + 2 g(x) g'(x)$$

Let $$\phi (x)=\left ( x+tan^{-1} x\right )-\dfrac{x}{1+x^{2}}$$
$$\therefore$$   $$\displaystyle \phi'(\mathrm{x})=1+\frac{1}{1+\mathrm{x}^{2}}-\frac{1-\mathrm{x}^{2}}{(1+\mathrm{x}^{2})^{2}}=1+\frac{2\mathrm{x}^{2}}{(1+\mathrm{x}^{2})^{2}}>0 \ for$$ all $$\mathrm{x}$$
$$\therefore$$ f(x) is increasing in the domain 
$$\phi(0)=0$$
$$\therefore$$     $$\displaystyle \mathrm{x}+t\mathrm{a}\mathrm{n}^{-1}\mathrm{x}-\frac{\mathrm{x}}{1+\mathrm{x}^{2}}>0, \mathrm{x}>0$$  
$$\Rightarrow \mathrm{f}(\mathrm{x})>g(\mathrm{x}), \mathrm{x}>0$$

We can write $$|x|$$ as $$\sqrt{x^{2}}$$
so, $$\frac{\mathrm{d} (\sqrt{x^{2}})}{\mathrm{d} x}=\frac{2x}{2\sqrt{x^{2}}}=\frac{x}{|x|}$$
and, $$\frac{\mathrm{d} (a_{0}x^{n})}{\mathrm{d} x}=na_{0}x^{n-1}$$
Similarly, $$\frac{\mathrm{d} (a_{n-1}x)}{\mathrm{d} x}=a_{n-1}$$ and $$\frac{\mathrm{d} (a_{n})}{\mathrm{d} x}=0$$
Hence, option A is the correct option. 

$$f(x)=\sqrt {x+2\sqrt {2x-4}}+\sqrt {x-2\sqrt {2x-4}}$$
$$f(x)=\sqrt { { \left( \sqrt { x-2 } +\sqrt { 2 }  \right)  }^{ 2 } } +\sqrt { { \left( \sqrt { x-2 } -\sqrt { 2 }  \right)  }^{ 2 } }$$
$$f(x)=\left| \sqrt { x-2 } +\sqrt { 2 }  \right| +\left| \sqrt { x-2 } -\sqrt { 2 }  \right| \\ $$
For $$\sqrt { x-2 }$$   to exist, $$x\ge 2$$
Also $$\sqrt { x-1 } +\sqrt { 2 } >0$$ .............(always true)
But $$\sqrt { x-1 } -\sqrt { 2 } \ge 0\quad$$ only if $$x\ge 4$$
                                         $$<0$$ only if $$x<4$$
Now $$f(x)$$ becomes
$$f(x)=\sqrt { x-2 } +\sqrt { 2 } -\sqrt { x-2 } +\sqrt { 2 }$$ for $$2\le x<4$$
$$f(x)=\sqrt { x-2 } +\sqrt { 2 } +\sqrt { x-2 } -\sqrt { 2 }$$ for $$x\ge 4$$
$$f(x)=2\sqrt { 2, }$$ for $$2\le x<4$$
$$f(x)=2\sqrt { x-2 }$$ for $$4\le x<\infty$$

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

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 $$x^{2n} = 1\Rightarrow \displaystyle \lim_{n \rightarrow \infty } x^{2n}=1$$ .
$$\therefore \phi (x) = \displaystyle \frac{f(x) + g(x)}{2}$$

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

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 $$x \in R$$

Given:

$$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$$ 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 $$A-B-C=\displaystyle \frac { 1 }{ 2 }$$

$$A: \displaystyle \frac {d}{dx} (x^2+y^2=4)$$ 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 $${ e }^{ x }{ e }^{ y }={ e }^{ x+y }$$
$$C=\displaystyle \frac { { e }^{ 2 }-2{ e }^{ 1 }-{ e }^{ 2 }+{ e }^{ 1 } }{ 2{ e }^{ 1 }+{ e }^{ 2 }-{ e }^{ 2 }-{ e }^{ 1 } }=-{1}$$
$$A+B+C=-1-1-1=-3$$

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.