Solve the Following Question.(5 Marks)

Question 15 Marks

Determine all real values of p and q that ensure the function
f(x) = px + q, for x ≤ 1
= tan $\frac{\pi x}{4}$for 1 < x < 2
is differentiable at x = 1.

Answer

f(x) is differentiable at x = 1.
∴ f(x) is continuous at x = 1.
Continuity at x= 1:
$f(x) \text { is continuous at } x=1$
$\therefore \quad \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
$\therefore \quad \lim _{x \rightarrow 1^{-}}(p x+q)=\lim _{x \rightarrow 1^{+}} \tan \left(\frac{\pi x}{4}\right)$
$\therefore \quad p+q=\tan \frac{\pi}{4}=1$
Differentiability at $x=1$ :
$\operatorname{Lf}^{\prime}(1)=\lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0^{-}} \frac{p(1+h)+q-(p+q)}{h}$
$=\lim _{ h \rightarrow 0} \frac{ ph }{ h }$
$=p \quad \ldots[\because h \rightarrow 0, \therefore h \neq 0]$
$R f^{\prime}(1)=\lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0^{+}} \frac{\tan \frac{\pi}{4}(1+h)-(p+q)}{h}$
$=\lim _{h \rightarrow 0^{+}} \frac{\tan \left(\frac{\pi}{4}+\frac{\pi h}{4}\right)-1}{h} \ldots[\because p+q=1]$
$=\lim _{h \rightarrow 0^{+}}\left[\frac{\frac{1+\tan \frac{\pi h}{4}}{1-\tan \frac{\pi h}{4}}-1}{h}\right]$
$=\lim _{h \rightarrow 0^{+}}\left[\frac{2 \tan \frac{\pi h}{4}}{h\left(1-\tan \frac{\pi h}{4}\right)}\right]$
$=\frac{2 \pi}{4} \lim _{ h \rightarrow 0^{+}} \frac{\tan \left(\frac{\pi h}{4}\right)}{\left(\frac{\pi h}{4}\right)}\left(\frac{1}{1-\tan \frac{\pi h}{4}}\right)$
$=\frac{\pi}{2}(1)\left(\frac{1}{1-0}\right)=\frac{\pi}{2}$
$f (x)$ is differentiable at $x=1$.
$\therefore \quad Lf ^{\prime}(1)= Rf ^{\prime}(1)$
$\therefore \quad p =\frac{\pi}{2}$
Substituting $p =\frac{\pi}{2}$ in (i), we get
$ \frac{\pi}{2}+q=1$
$\therefore \quad q=1-\frac{\pi}{2}=\frac{2-\pi}{2}$

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Question 25 Marks

Determine the values of $p$ and $q$ that make the function $f(x)$ differentiable on $R$ where
$f(x) = px^3,$ for $x < 2$
$= x^2 + q,$ for $x \geq 2$

Answer

$\therefore f ( x )$ is differentiable at $x =2$.
$\therefore f ( x )$ is continuous at $x =2$.
Continuity at $x=2$ :
$f(x)$ is continuous at $x=2$.
$\therefore \quad \lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2^{+}} f (x)$
$\therefore \quad \lim _{x \rightarrow 2^{-}} p x^3=\lim _{x \rightarrow 2^{+}}\left(x^2+ q \right)$
$\therefore \quad 8 p =4+ q \text { }$
$\therefore \quad 8 p - q =4$
$\text { Differentiability at } x= 2 \text { : }$
$L f^{\prime}(2)=\lim _{h \rightarrow 0^{-}} \frac{f(2+h)-f(2)}{h}$
$=\lim _{h \rightarrow 0^{-}} \frac{p(2+h)^3-\left(2^2+q\right)}{h}$
$=\lim _{b \rightarrow 0^{-}}\left(\frac{ ph ^3+6 ph ^2+12 hp +8 p -(4+ q )}{ h }\right)$
$=\lim _{h \rightarrow 0^{-}} \frac{h\left( ph ^2+6 ph +12 p \right)}{ h }$
$\ldots[\because 8 p =4+q]$
$=\lim _{h \rightarrow 0^{-}}\left( ph ^2+6 ph +12 p \right)$
$\ldots[\because h \rightarrow 0, \therefore h \neq 0]$
$=12 p$
$R f^{\prime}(2)=\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(2)}{h}$
$=\lim _{h \rightarrow 0^{+}} \frac{(2+h)^2+q-\left(2^2+q\right)}{h}$
$=\lim _{h \rightarrow 0^{+}} \frac{h^2+4 h+4+q-(4+q)}{h}$
$=\lim _{h \rightarrow 0}\left(\frac{h^2+4 h}{h}\right)$
$=\lim _{h \rightarrow 0}(h+4) \quad \ldots[\because h \rightarrow 0, \therefore h \neq 0]$
$=4$
$f(x)$ is differentiable at $x=2$.
$\therefore Lf ^{\prime}(2)= Rf ^{\prime}(2)$
$\therefore 12 p =4$
$\therefore p =\frac{1}{3}$
Substituting $p=\frac{1}{3}$ in (i), we get
$8\left(\frac{1}{3}-q=4\right.$
$\therefore q=\frac{8}{3}-4=\frac{-4}{3}$

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Question 35 Marks

If $f(x)=a \sin x-b \cos x, f^{\prime}\left(\frac{\pi}{4}\right)=\sqrt{2}$ and $f^{\prime}\left(\frac{\pi}{6}\right)=2$, then find $f(x)$.

Answer

$f(x)=a \sin x-b \cos x$
Differentiating w.r.t. $x$, we get
$f ^{\prime}( x )= a \cos x - b (-\sin x )$
$\therefore f ^{\prime}( x )= a \cos x + b \sin x$
$\therefore \quad f ^{\prime}\left(\frac{\pi}{4}\right)= a \cos \left(\frac{\pi}{4}\right)+ b \sin \left(\frac{\pi}{4}\right)$
But, $f ^{\prime}\left(\frac{\pi}{4}\right)=\sqrt{2}$
...(given)
$\therefore a \cos \frac{\pi}{4}+b \sin \frac{\pi}{4}=\sqrt{2}$
$\therefore a\left(\frac{1}{\sqrt{2}}\right)+b\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2}$
$\therefore \frac{a+b}{\sqrt{2}}=\sqrt{2}$
$\therefore a+b=2$
Also, $f^{\prime}\left(\frac{\pi}{6}\right)=a \cos \frac{\pi}{6}+b \sin \frac{\pi}{6}$
But, $f^{\prime}\left(\frac{\pi}{6}\right)=2$
...(given)
$\therefore \quad a \cos \frac{\pi}{6}+b \sin \frac{\pi}{6}=2$
$\therefore \quad a\left(\frac{\sqrt{3}}{2}\right)+b\left(\frac{1}{2}\right)=2$
$\therefore \quad \sqrt{3} a+b=4$
equation (ii) - equation (i), we get
$\sqrt{3} a - a =2$
$\therefore \quad a(\sqrt{3}-1)=2$
$\therefore \quad a=\frac{2}{\sqrt{3}-1}=\frac{2(\sqrt{3}+1)}{3-1}$
$\therefore \quad a =\sqrt{3}+1$
Substituting $a=\sqrt{3}+1$ in equation (i), we get
$\sqrt{3}+1+b=2$
$\therefore \quad b =1-\sqrt{3}$
Now, $f(x)=a \sin x-b \cos x$
$\therefore f(x)=(\sqrt{ } 3+1) \sin x+(\sqrt{ } 3-1) \cos x$

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Question 45 Marks

If $f(x)$ is a quadratic polynomial such that $f(0) = 3, f'(2) = 2$ and $f'(3) = 12$, then find $f(x).$

Answer

Let $f(x) = ax^2 + bx + c …..(i)$
$\therefore f(0) = a(0)^2 + b(0) + c$
$\therefore f(0) = c$
But, $f(0) = 3 …..($given$)$
$\therefore c = 3 …..(ii)$
Differentiating (i) w.r.t. $x,$ we get
$f'(x) = 2ax + b$
$\therefore f'(2) = 2a(2) + b$
$\therefore f'(2) = 4a + b$
But, $f'(2) = 2 …..($given$)$
$\therefore 4a + b = 2 …..(iii)$
Also, $f'(3) = 2a(3) + b$
$\therefore f'(3) = 6a + b$
But, $f'(3) = 12 …..($given$)$
$\therefore 6a + b = 12 …..(iv)$
equation $(iv) –$ equation $(iii),$ we get
$2a = 10$
$\therefore a = 5$
Substituting $a = 5 $ in $(iii),$ we get
$4(5) + b = 2$
$\therefore b = -18$
$\therefore a = 5, b = -18, c = 3$
$\therefore f(x) = 5x^2 – 18x + 3$ Check:
If $f(0) = 3, f'(2) = 2$ and $f'(3) = 12,$ then our answer is correct.
$f(x) = 5x^2 – 18x + 3$ and $f'(x) = 10x – 18$
$f(0) = 5(0)^2 – 18(0) + 3 = 3$
$f'(2) = 10(2) – 18 = 2$
$f'(3) = 10(3) – 18 = 12$
Thus, our answer is correct.

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Question 55 Marks

If f(x) = sin x – cos x if x ≤ $\frac{\pi}{2}$
= 2x – π + 1 if x > $\frac{\pi}{2}$
Test the continuity and differentiability of f at x =$\frac{\pi}{2}$

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Question 65 Marks

Test the continuity and differentiability of
$f(x) = 3x + 2$ if $x > 2$
$= 12 – x^2 $ if $x \leq 2\ at\ x = 2.$

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Question 75 Marks

Discuss the continuity and differentiability of :

f(x) = (2x + 3) |2x + 3| at x = $\frac{-3}{2}$

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Question 85 Marks

Find the derivatives of the following w.r.t. x. at the points indicated against them by using the method of the first principle:
$\cos x$ at $x=\frac{5 \pi}{4}$

Answer

Let $f (x)=\cos x$
$f\left(\frac{5 \pi}{4}\right)=\cos \left(\frac{5 \pi}{4}\right)$
and $f \left(\frac{5 \pi}{4}+ h \right)=\cos \left(\frac{5 \pi}{4}+ h \right)$
By first principle, we get
$f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$
$f^{\prime}\left(\frac{5 \pi}{4}\right)=\lim _{h \rightarrow 0} \frac{f\left(\frac{5 \pi}{4}+h\right)-f\left(\frac{5 \pi}{4}\right)}{h}$
$=\lim _{h \rightarrow 0} \frac{\cos \left(\frac{5 \pi}{4}+h\right)-\cos \left(\frac{5 \pi}{4}\right)}{h}$
$=\lim _{h \rightarrow 0} \frac{-2 \sin \left(\frac{\frac{10 \pi}{4}+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$
$\ldots\left[\because \cos C-\cos D=-2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]$
$=\lim _{h \rightarrow 0} \frac{-2 \sin \left(\frac{5 \pi}{4}+\frac{h}{2}\right) \sin \frac{h}{2}}{h}$
$=-2 \lim _{h \rightarrow 0} \sin \left(\frac{5 \pi}{4}+\frac{h}{2}\right) \cdot \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2} \times 2}$
$=-2 \lim _{h \rightarrow 0} \sin \left(\frac{5 \pi}{4}+\frac{h}{2}\right)\left(\frac{1}{2}\right) \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$=-2 \sin \left(\frac{5 \pi}{4}+0\right) \cdot \frac{1}{2}(1)$
$\ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right]$
$=-\sin \frac{5 \pi}{4}$
$=-\sin \left(\pi+\frac{\pi}{4}\right)$
$=\sin \frac{\pi}{4} \quad \ldots[\because \sin (\pi+\theta)=-\sin \theta]$
$=\frac{1}{\sqrt{2}}$

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Question 95 Marks

Find the derivatives of the following w.r.t. x. at the points indicated against them by using the method of the first principle:
$\tan x$ at $x =\frac{\pi}{4}$

Answer

Let $f (x)=\tan x$
$\therefore \quad f \left(\frac{\pi}{4}\right)=\tan \frac{\pi}{4}=1 \text { and } f \left(\frac{\pi}{4}+ h \right)=\tan \left(\frac{\pi}{4}+ h \right)$
By first principle, we get
$f^{\prime}(a)=\lim _{b \rightarrow 0} \frac{f(a+h)-f(a)}{h}$
$\therefore \quad f^{\prime}\left(\frac{\pi}{4}\right)=\lim _{h \rightarrow 0} \frac{f\left(\frac{\pi}{4}+h\right)-f\left(\frac{\pi}{4}\right)}{h}$
$=\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}+h\right)-1}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{\sin \left(\frac{\pi}{4}+h\right)}{\cos \left(\frac{\pi}{4}+h\right)}-1}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{4}+h\right)-\cos \left(\frac{\pi}{4}+h\right)}{h \cos \left(\frac{\pi}{4}+h\right)}$
$=\lim _{h \rightarrow 0} \frac{\left[\left(\sin \frac{\pi}{4} \cos h+\cos \frac{\pi}{4} \sin h\right)-\left(\cos \frac{\pi}{4} \cos h-\sin \frac{\pi}{4} \sin h\right)\right]}{h \cos \left(\frac{\pi}{4}+h\right)}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{\sqrt{2}} \cos h+\frac{1}{\sqrt{2}} \sin h-\frac{1}{\sqrt{2}} \cos h+\frac{1}{\sqrt{2}} \sin h}{h \cos \left(\frac{\pi}{4}+h\right)}$
$=\lim _{h \rightarrow 0} \frac{\frac{2}{\sqrt{2}} \sin h}{h \cos \left(\frac{\pi}{4}+h\right)}$
$=\frac{2}{\sqrt{2}} \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right) \cdot \lim _{h \rightarrow 0} \frac{1}{\cos \left(\frac{\pi}{4}+h\right)}$
$=\sqrt{2}(1) \cdot \frac{1}{\cos \left(\frac{\pi}{4}+0\right)} \quad \cdots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=\sqrt{2} \times \frac{1}{\sqrt{2}}$
$= 2$

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Maths Solve the Following Question.(5 Marks)

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