f(x)=\(\frac{4x+5 sinx}{3x+7 cos x}\)
\(f^{'}(x)=\frac{d}{dx}[\frac{4x+5sinx}{3x+7 cos x}]\)
=\(\frac{(3x+7cos x)\frac{d}{dx}(4x+5sinx)-(4x+5sinx\frac{d}{dx}(3x+7 cos x)}{(3x+7cos x)^{2}}\)
=\(\frac{(4x+5 sinx)(3-7sinx)}{(3x+7 cosx)^{2}}\)
=\(\frac{12x+15x cosx+28 cos x+35 cos^{2}x-12x+28 x sinx-15 sinx+35 sin^{2}x}{(3x+7 cos)^{2}}\)
=\(\frac{15x cosx+28 cosx+28x sinx-15 sinx+35(cos^{2}x+sin^{2}x)}{(3x+7 cos x)^{2}}\)
=\(\frac{28x sinx-15 sin x+35}{(3x+7 cos x)^{2}}\)

Focus = (0, –3); directrix y = 3
Since the focus lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form x² = 4ay or x² = – 4ay.
It is also seen that the directrix, y = 3 is above the x-axis, while the focus (0, –3) is below the x-axis. Hence, the parabola is of the form x² = –4ay.
Here, a = 3
Thus, the equation of the parabola is x² = –12y.

\(\lim_ { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }2x }{ { \left[ sin\quad 2(2x) \right] }^{ 2 } } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }2x }{ { \left( 2sin\quad 2x\quad cos\quad 2x \right) }^{ 2 } } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }2x }{ { 4sin }^{ 2 }2x\quad { cos }^{ 2 }2x } =\lim_ { x\rightarrow 0 }{ lim } \frac { 1 }{ 4{ cos }^{ 2 }2x } \)
Ans:1

The point (-3,4,2) lies in II octant and the point (-3,1,4) lies in VI octant.

Given, centre (h, k) = (1, 2)
and circle touches on X-axis.
\(\therefore\) Radius (r) = y-coordinate of centre = 2
So, equation of circle is
\((x-1)^{2}+(y-2)^{2}=2^{2} \quad\left[\because(x-h)^{2}+(y-k)^{2}=r^{2}\right] \)
\(\Rightarrow x^{2}-2 x+1+y^{2}-4 y+4=4 \)
\(\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \)
\(\Rightarrow x^{2}+y^{2}-2 x-4 y+1=0\)
which is the required equation of circle.

\(LHL=\lim _{ x\rightarrow { 1 }^{ - } }{ a+bx)= } \lim _{ h\rightarrow { 0 } }{ [a+b(1-h)]=a+b } \)
\(RHL=\lim _{ x\rightarrow { 1 }^{ + } }{ (b-ax) } =\lim _{ h\rightarrow { 0 } }{ [b-a(1+h)]=b-a } \)
\(\because LHL=RHL=f\left( 1 \right) \Rightarrow A+B=B-A=4\)
Ans. A = 0, B = 4

Let  A(-1, 2, 1), B(1, -2, 5), C(4, -7, 8) and D(2, -3, 4) be the vertices of a quadrilateral ABCD.

\(Then\quad the\quad mid\quad point\quad AC\)
\(=\left( \frac { -1+4 }{ 2 } ,\frac { 2-7 }{ 2 } ,\frac { 1+8 }{ 2 } \right) =\left( \frac { -3 }{ 2 } ,\frac { -5 }{ 2 } ,\frac { 9 }{ 2 } \right) \)
\(\left[ \because coordinates\quad of\quad mid-point=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { { y }_{ 1 }+ }{ y }_{ 2 } }{ 2 } ,\frac { { z }_{ 1 }+{ z }_{ 2 } }{ 2 } \right) \right] \)
\(and\quad mid\quad point\quad BD=\left( \frac { 1+2 }{ 2 } ,\frac { -2-3 }{ 2 } ,\frac { 5+4 }{ 2 } \right)\)
\(=\left( \frac { 3 }{ 2 } ,\frac { -5 }{ 2 } ,\frac { 9 }{ 2 } \right) \)
Here, mid-point of both the diagonals are same i.e they bisect each other. Hence ABCD is a parallelogram

Given equation of parabola is 2y²+3y-4x-3=0
\(2\left[ y^{ 2 }+2.y.\frac { 3 }{ 4 } +\left( \frac { 3 }{ 4 } \right) ^{ 2 }-\left( \frac { 3 }{ 4 } \right) ^{ 2 } \right] =4x+3\)
\(\Rightarrow 2\left( y+\frac { 3 }{ 4 } \right) =4x+3+\frac { 9 }{ 8 } \)
\(\Rightarrow \left( y+\frac { 3 }{ 4 } \right) ^{ 2 }=2\left( x+\frac { 33 }{ 32 } \right) \)
\(This\quad is\quad of\quad the\quad form\quad Y^{ 2 }=\quad 4aX,\quad Where\quad Y=\quad y+\frac { 3 }{ 4 } ,X=x+\frac { 33 }{ 32 } \)
Ans \(\left( \frac { -33 }{ 32 } ,\frac { -3 }{ 4 } \right) \left( \frac { -17 }{ 32 } ,\frac { -3 }{ 4 } \right) ,x=\frac { -49 }{ 32 } ,2\)

\(\overset{Lt}{x\rightarrow 0}\frac{log(3+x)-log(3-x)}{x}\)=\(\overset{Lt}{x\rightarrow 0}\frac{log[3(1+\frac{x}{3})]-log[3(1-\frac{x}{3})]}{x}\)
⇒ \(\overset{Lt}{x\rightarrow 0}\frac{log3+log(1+\frac{x}{3}-log 3-log(1-\frac{x}{3})}{x} \Rightarrow \overset{Lt}{x\rightarrow 0}\frac{log(1+\frac{x}{3})}{x}-\frac{log(1-\frac{x}{3})}{x}\)
⇒ \(\overset{Lt}{x\rightarrow 0} \frac{log(1+\frac{x}{3})}{3\times \frac{x}{3}}-\frac{log(1-\frac{x}{3})}{\frac{-x}{3}\times -3} \Rightarrow \overset{Lt}{\frac{-x}{3}\times -3} \Rightarrow \overset{Lt}{\frac{x}{3}\rightarrow 0} \frac{1}{3}[\frac{log(1+\frac{x}{3})}{\frac{x}{3}}]+\overset{Lt}{-\frac{x}{3}\rightarrow 0}\frac{1}{3}[\frac{log(1-\frac{x}{3})}{-\frac{x}{3}}]\)
\(\Rightarrow \frac{1}{3}\times 1+\frac{1}{3}\times 1 \Rightarrow\frac{1}{3}+\frac{1}{3}\Rightarrow \frac{2}{3}\)
Hence \(\overset{Lt}{x\rightarrow 0}\frac{log(3+x)-log(3-x)}{x}=\frac{2}{3}\)

Let the equation of the required circle be (x – h)² + (y – k)² = r².
Since the circle passes through (0, 0),
(0 – h)² + (0 – k)² = r²
⇒ h² + k² = r²
The equation of the circle now becomes (x – h)² + (y – k)² = h² + k².
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,
(a – h)² + (0 – k)² = h² + k² … (1)
(0 – h)² + (b – k)² = h² + k² … (2)
From equation (1), we obtain
a² – 2ah + h² + k² = h² + k²
⇒ a² – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
\(\text { However, } a \neq 0 ; \text { hence, }(a-2 h)=0 \Rightarrow h=\frac{a}{2} \text { . }\)
\(\text { From equation (2), we abtain }\)
\(h^{2}+b^{2}-2 b k+k^{2}=h^{2}+k^{2}\)
\(\Rightarrow b^{2}-2 b k=0\)
\(\Rightarrow b(b-2 k)=0\)
\(\Rightarrow b=0 \text { or }(b-2 k)=0\)
\(\text { However, } b \neq 0 \text { ; hence, }(b-2 k)=0 \Rightarrow k=\frac{b}{2} \text { . }\)


Thus, the equation of the required circle is
\(\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2} \)
\(\Rightarrow\left(\frac{2 x-a}{2}\right)^{2}+\left(\frac{2 y-b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{4} \)
\(\Rightarrow 4 x^{2}-4 a x+a^{2}+4 y^{2}-4 b y+b^{2}=a^{2}+b^{2} \)
\(\Rightarrow 4 x^{2}+4 y^{2}-4 a x-4 b y=0 \)
⇒ x² + y²- ax - by = 0
which is required equation of circle.

Let A(2, 0, 3), B(O, 3, 2) and C(O, 0,1) be given points.
Let P(x, y, 0) be any point in XY plane such that PA = PB = PC.
Now PA = PB => PA2 = PB2
\(\therefore \) (x - 2)² + (y - 0)² + (- 3)²
= (x - 0)² + (y - 3)² + (- 2)²
=> x²+4-4x+y²+9
=x²+y²+9-6y+4
=> 4x - 6y = 0 => 2x - 3y = 0 ....(i)
Also PB = PC => PB² = PC²
(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 2)²
\(\therefore \)(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 1)²
=> x²+ y² + 9 - 6y + 4
=x²+y²+1
=> 6y = 12 => y = 2
Putting value of y in (i), we have
2x-3x2=0 => x=3
Thus co-ordinates ofrequired point are (3, 2, 0).