Given equation can be written as yt² ? 2ct + x = 0 ——–> eq(1)
The envelope of At² + Bt + C = 0 is B² ? 4AC = 0 ———> eq(2)
From eq(1), A = y, B= -2c, C = x
Putting values in eq(2),
(-2c)² ? 4(y)(x) = 0
4c² ? 4xy = 0
xy = c².

From equation, it is clear that radius of curve (circle) is 5 units. The curvature is equal to reciprocal of radius. Hence, curvature = 1/5 = 0.2.

lt_{x → 0}\((\frac{ln(1+x^4)}{x}) = lt_{x\rightarrow 0} (\frac{1}{x}) \times (\frac{x^4}{1}-\frac{x^8}{2}+\frac{x^{12}}{3}-…\infty)\)
lt_{x → 0}\((\frac{x^3}{1}-\frac{x^7}{2}+\frac{x^{11}}{3}-…\infty)\)
= 0.

Using formula for Curvature in parametric form
k=\(\left |\frac{y”x’-y’x”}{((x’)^2+(y’)^2)^{\frac{3}{2}}} \right |\)
Equating y”x’-y’x” to zero (given curvature function is zero) we get
\(\frac{y”}{y’}=\frac{x”}{x’}\)
Put x=k₁e^a₁t:y=k₂e^a₂t
\(\frac{k_1(a_1)^2.e^{a_1t}}{k_1.a_1.e^{a_1t}}=\frac{k_2(a_2)^2.e^{a_2t}}{k_2.a_2.e^{a_2t}}\)
a₁=a₂

\(y=\lim_{x\rightarrow\infty}\left [\frac{x-1}{x-2} \right ]^x\)
\(ln y=\lim_{x\rightarrow\infty}xln\left [\frac{x-1}{x-2} \right ]\)
=>\(\lim_{x\rightarrow\infty}\frac{\left [\frac{x-1}{x-2} \right ]}{\frac{1}{x}}\)
By putting 1=1/y, we get
=>\(\lim_{y\rightarrow 0}\frac{ln\left [\frac{x-1}{x-2} \right ]}{y}=[ln(1/2)]/0\) (i.e indeterminate)
Hence by applying L’Hospital rule
=>\(\lim_{y\rightarrow 0}\frac{ln\left [\frac{x-1}{x-2} \right ]}{y}=\lim_{y\rightarrow 0}\frac{\frac{2-y-1+y}{(2-y)^2}}{\frac{1-y}{2-y}}=\lim_{y\rightarrow 0}\frac{\frac{1}{2-y}}{\frac{1-y}{1}}=\lim_{y\rightarrow 0}(\frac{1}{(2-y)(1-y)})\)=1/2

We know that series expansion of cos(x) is
cos(t)=1-\(\frac{t^2}{2!}+\frac{t^4}{4!}….\infty\)
Now substituting t=sin(x) we have
cos(sin(x))=\(1-\frac{1}{2!}\times (\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}+…\infty)^2+\frac{1}{4!}\times (\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}+…\infty)^4+..\infty\)
Observe that every term has odd powered series raised to an even term.
Thus, we must have only even powered terms in above series expansion. The coefficient of any odd powered term is zero.

Applying L hospitals rule
\(=lt_{x\rightarrow 0}\frac{a.cos(ax)+b.cos(bx)-2c.cos(cx)}{a.cos(ax)+2b.cos(bx)-3c.cos(cx)}=lt_{x\rightarrow 0}\frac{a+b-2c}{a+2b-3c}\)
Assume now that
a + b + 2c = 0 and a + 2b – 3c = 0
We must have
a = c = b but given a ≠ c ≠ b
Thus, our assumption is false and a finite limit exists after first round of differentiation.
Hence, 2 is right answer.

The Taylor series has no effect whatsoever on nature of polynomial. This is because Taylor series is itself a polynomial and hence no change takes place.
?_a(f(x)) = f(x).

=\(lt_{x\rightarrow 0}(\frac{1}{(\frac{1-cos(2x)}{2})})=lt_{x\rightarrow 0}(\frac{2}{1-cos(2x)})\)
=\((\frac{2}{1-cos(0)})=\frac{2}{0} \rightarrow \infty\)

Correct forms of Trigonometric Derivative Functions

• \(\frac{d(sinx)}{dx}=cosx \)
• \(\frac{d(cosx)}{dx}=-sinx\)
• \(\frac{d(secx)}{dx}=secxtanx\)
• \(\frac{d(tanx)}{dx}=sec^2 x \)