A function f : R ® R satisfies the equation
Question: A functionf: R R satisfies the equationf(x+y) =f(x)f(y) for allx,yÎR,f(x) 0. Suppose that the function is differentiable atx= 0 andf (0) = 2. Prove thatf(x) = 2f(x). Solution: Given, f: R R satisfies the equationf(x+y) =f(x)f(y) for allx,yÎR,f(x) 0 Let us take any point x = 0 at which the function f(x) is differentiable. So, $f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$ $2=\lim _{h \rightarrow 0} \frac{f(0) \cdot f(h)-f(0)}{h} \quad[\because f(0)=f(h)]$ .....$\ldots(i)$...
Read More →Show that f (x) = |x – 5| is continuous
Question: Show thatf(x) = |x 5| is continuous but not differentiable atx= 5. Solution: Given, $f(x)=|x-5|$ $\Rightarrow \quad f(x)=\left\{\begin{array}{r}-(x-5) \text { if } x-50 \text { or } x5 \\ x-5 \text { if } x-50 \text { or } x5\end{array}\right.$ For continuity at $x=5$ L.H.L. $\lim _{h \rightarrow 5} f(x)=-(x-5)$ $=\lim _{h \rightarrow 0}-(5-h-5)=\lim _{h \rightarrow 0} h=0$ R.H.L. $\lim _{x \rightarrow 5^{+}} f(x)=x-5$ $=\lim _{h \rightarrow 0}(5+h-5)=\lim _{h \rightarrow 0} h=0$ L.H.L...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \mathrm{e}^{2 \mathrm{x}}\left(\frac{1+\sin 2 \mathrm{x}}{1+\cos 2 \mathrm{x}}\right) \mathrm{dx}$ Solution: Put $2 x=t d x=d t / 2$ $\frac{1}{2} \int e^{t}\left(\frac{1+\sin t}{1+\cos t}\right) d t=\frac{1}{2} \int\left(e^{t} \tan \frac{t}{2}+\frac{1}{2} e^{t} \sec ^{2} \frac{t}{2}\right) d t$ $=\frac{1}{2} \int\left(e^{t} \tan \frac{t}{2}\right) d t+\frac{1}{4} \int \mathrm{e}^{\mathrm{t}} \sec ^{2} \frac{t}{2} d t$ $=\frac{1}{2} \int\left(e^{t} \tan \frac{t}{2}\right)...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{x \sin ^{-1} x}{\left(1-x^{2}\right)^{3 / 2}} d x$ Solution: $\int \frac{x \sin ^{-1} x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}} d x$ we can put $\sin ^{-1} x=t ; d x /\left(1-x^{2}\right)^{1 / 2}=d t ;\left(1-x^{2}\right)=\cos ^{2} t$ and $x=\sin t$. $\int \frac{t \sin t}{\cos ^{2} t} d t=\int t$ tant sect $d t$ By by parts, $\int t$ tant $\sec t d t=t \sec t-\int \operatorname{sect} d t \ldots \ldots$ $\because \int \operatorname{sect} \tan t d t=\int \frac{\sin t}{...
Read More →Prove the following identities.
Question: $f(x)=\left\{\begin{array}{lll}1+x \text {, if } x \leq 2 \\ 5-x \text {, if } x2\end{array}\right.$ at $x=2$ Solution: We know that, f(x) is differentiable at x = 2 if $L f^{\prime}(2)=R f^{\prime}(2)$ NoW, $L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$ $=\lim _{h \rightarrow 0} \frac{(1+2-h)-(1+2)}{-h}=\lim _{h \rightarrow 0} \frac{3-h-3}{-h}=\frac{-h}{-h}=1$ $R f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$ $=\lim _{h \rightarrow 0} \frac{[5-(2+h)]-(1...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \cos ^{-1}\left(1-2 \mathrm{x}^{2}\right) \mathrm{dx}$ Solution: Put $x=\sin t$ $; d x=\cos t d t$ $\int \cos ^{-1}\left(1-2 x^{2}\right) d x=\int \cos ^{-1}\left(1-2 \sin ^{2} t\right) \cos t d t=\int \cos ^{-1}\left(1-\sin ^{2} t-\sin ^{2} t\right) \cos t d t$ $\int \cos ^{-1}\left(\cos ^{2} t-\sin ^{2} t\right) \cos t d t=\int \cos ^{-1}(\cos 2 t) \cos t d t$ $2 \int t \cos t d t=2[t \sin t+\cos t]+c$ Ans $=2 \mathrm{x} \sin ^{-1} x+2 \sqrt{1-x^{2}}+c$...
Read More →The value of
Question: $f(x)= \begin{cases}x^{2} \sin \frac{1}{x} \text {, if } x \neq 0 \\ 0 \text {, if } x=0\end{cases}$ at $x=0$ Solution: Given, $f(x)=\left\{\begin{array}{c}x^{2} \sin \frac{1}{x}, \text { if } x \neq 0 \\ 0, \quad \text { if } x=0\end{array}\right.$ at $x=0$ For differentiability we know that: $\mathrm{L} f^{\prime}(c)=\mathrm{R} f^{\prime}(c)$ $\therefore \mathrm{L} f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$ $=\lim _{h \rightarrow 0} \frac{(0-h)^{2} \sin \frac{1}{(0...
Read More →Evaluate the following integrals:
Question: Evaluate $\int\left(\sin ^{-1} x\right)^{3} d x$ Solution: $\int\left(\sin ^{-1} x\right)^{3} d x$ Put $x=\sin t$ $d x=\cos t d t$ $\int\left(\sin ^{-1} x\right)^{3} d x=\int\left(\sin ^{-1}(\sin t)\right)^{3} \cos t d t$ $\int t^{3} \cos t d=\left[t^{3} \sin t-3 \int t^{2} \sin t d t\right]=\left[t^{3} \sin t-3\left[-t^{2} \cos t+2 \int t \cos t d t\right]\right]$ $=\left[t^{3} \sin t+3 t^{2} \cos t-6 \int t \cos t d t\right]=\left[t^{3} \sin t+3 t^{2} \cos t-6[t \sin t+\cos t]\right]...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sin ^{-1}\left(3 x-4 x^{3}\right) d x$ Solution: Put $x=\sin t ; d x=\cos t d t$ $\int \sin ^{-1}\left(3 x-4 x^{3}\right) d x=\int \sin ^{-1}\left(3 \sin t-4 \sin ^{3} t\right) \cos t d t \ldots\left(3 \sin t-4 \sin ^{3} t\right)=\sin 3 t$ $=\int \sin ^{-1}(\sin 3 t) \cos t d t=\int 3 t \cos t d t$ $=3 \int t \cos t d t$ By by parts, $=3\left[t \sin t-\int \sin t d t\right]+c$ $=3[t \sin t+\cos t]+c$ $=3 x \sin ^{-1} x+3 \sqrt{1-x^{2}}+c$...
Read More →Prove the following
Question: $f(x)=\left\{\begin{array}{l}x[x], \quad \text {, if } 0 \leq x2 \\ (x-1) x, \quad \text { if } 2 \leq x3\end{array}\right.$ at $\mathrm{x}=2$ Solution: We know that, a function f is differentiable at a point a in its domain if Lf(c) = Rf(c) where L $f^{\prime}(c)=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}$ and $\mathrm{R} f^{\prime}(c)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ Here, $f(x)=\left\{\begin{array}{r}x[x], \text { if } 0 \leq x2 \\ (x-1) x, \text { if } 2 \leq x2\e...
Read More →Show that the function f (x) = |sin x + cos x|
Question: Show that the functionf(x) = |sinx+ cosx|is continuous atx= p. Examine the differentiability off, wherefis defined by Solution: Given, f(x) = |sin x + cos x| at x = Now, put g(x) = sin x + cos x and h(x) = |x| Hence, h[g(x)] = h(sin x + cos x) = |sin x + cos x| Now, g(x) = sin x + cos x is a continuous function since sin x and cos x are two continuous functions at x = . We know that, every modulus function is a common function is a continuous function everywhere. Therefore, f(x) = |sin...
Read More →Find all points of discontinuity of the function
Question: Find all points of discontinuity of the function $f(t)=\frac{1}{t^{2}+t-2}$, where $t=\frac{1}{x-1}$ Solution: Given, $\quad f(t)=\frac{1}{t^{2}+t-2}$ $\Rightarrow f(t)=\frac{1}{\frac{1}{(x-1)^{2}}+\frac{1}{(x-1)}-2} \quad\left[\right.$ putting $\left.t=\frac{1}{x-1}\right]$ $=\frac{1}{\frac{1+x-1-2(x-1)^{2}}{(x-1)^{2}}}=\frac{(x-1)^{2}}{x-2 x^{2}-2+4 x}$ $=\frac{(x-1)^{2}}{-2 x^{2}+5 x-2}=\frac{(x-1)^{2}}{-\left(2 x^{2}-5 x+2\right)}$ $=\frac{(x-1)^{2}}{-\left[2 x^{2}-4 x-x+2\right]}=...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$ Solution: $\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$ Put $x=\operatorname{atan}^{2} t ; d x=2 a \cdot \tan t \cdot \sec ^{2} t d t$ $=\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=\int \sin ^{-1} \sqrt{\frac{\mathrm{a} \tan ^{2} t}{a+\mathrm{a} \tan ^{2} t}} 2 \mathrm{a} \cdot \operatorname{tant} \cdot \sec ^{2} t \mathrm{dt}=\int \mathrm{t} \cdot 2 \mathrm{a} \cdot \tan \mathrm{t} \cdot \sec ^{2} t \mathrm{dt}$ $=2 a \int \mathrm{t} \cdot...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$ Solution: Put $x=\cos 2 t ; d x=-2 \sin 2 t$ $=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} \mathrm{~d} x=\int \tan ^{-1} \sqrt{\frac{1-\cos 2 t}{1+\cos 2 t}}(-2 \sin 2 t) d t$ $=\int \tan ^{-1} \sqrt{\frac{1-\cos 2 t}{1+\cos 2 t}}(-2 \sin 2 t) d t$ $=-2 \int \tan ^{-1} t \tan t \sin 2 t d t$ $=-2 \int t \sin 2 t d t$ $=-2\left[-\frac{t \cos 2 t}{2}+\frac{1}{2} \int \cos 2 t d t\right]$ $=t \cos 2 t-\frac{\sin 2 t}{2}+c$ $=\frac{x \cos ^{...
Read More →Given the function f (x) = 1/(x + 2) .
Question: Given the functionf(x) = 1/(x+ 2) . Find the points of discontinuity of the composite function y=f(f(x)). Solution: Given, $f(x)=\frac{1}{x+2}$ $f[f(x)]=\frac{1}{f(x)+2}=\frac{1}{\frac{1}{x+2}+2}=\frac{1}{\frac{1+2 x+4}{x+2}}=\frac{x+2}{2 x+5}$ $\therefore f[f(x)]=\frac{x+2}{2 x+5}$ Now, the function will not be defined and continuous where 2x + 5 = 0 ⇒ x = -5/2 Therefore, x = -5/2 is the point of discontinuity....
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sec ^{-1} \sqrt{x} d x$ Solution: $\int \sec ^{-1} \sqrt{x} d x$ $\int u \cdot d v=u v-\int v d u$ Choose $u$ in these order LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG, E-EXPONENTIAL) Here $u=\sec ^{-1} \sqrt{x}$ and $v=1$ $\int \sec ^{-1} \sqrt{x} d x=x \sec ^{-1} x-\int \frac{x d x}{2 x \sqrt{x-1}}$ $=x \sec ^{-1} x-\int \frac{d x}{2 \sqrt{x-1}}$ Put $x-1=t d x=d t$ $=x \sec ^{-1} x-\int \frac{d t}{2 \sqrt{t}}$ $=x \sec ^{-1} x-\frac{2}{2}(\sqrt{t})+c$ $=x \sec ^{-1} x...
Read More →Find the values of a and b such that
Question: Find the values ofaandbsuch that the functionfdefined by $f(x)= \begin{cases}\frac{x-4}{|x-4|}+a, \text { if } x4 \\ a+b , \text { if } x=4 \\ \frac{x-4}{|x-4|}+b, \text { if } x4\end{cases}$ is a continuous function atx= 4. Solution: Finding the left hand and right hand limit for the given function, we have $\lim _{x \rightarrow 4} f(x)=\frac{x-4}{|x-4|}+a$ $=\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \rightarrow 0} \frac{-h}{h}+a=-1+a$ $\lim _{x \rightarrow 4} f(x)=a+b$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sin ^{-1} \sqrt{x} d x$ Solution: $\int \sin ^{-1} \sqrt{x} d x$ $\int u \cdot d v=u v-\int v d u$ Choose u in these order LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL) $\mathrm{u}=\sin ^{-1} \sqrt{x} \mathrm{v}=1$ $\therefore \int \sin ^{-1} \sqrt{x}=\mathrm{x} \cdot \sin ^{-1} \sqrt{x}-\frac{1}{2} \int \frac{\sqrt{\mathrm{x}}}{\sqrt{1-x}} \mathrm{dx}$ Put $\sqrt{x}=t$ $\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$ $=\mathrm{x} \cdot \sin ^{-1} \sqrt{x}-\int \frac{\...
Read More →Prove that the function f defined by
Question: Prove that the functionfdefined by $f(x)= \begin{cases}\frac{x}{|x|+2 x^{2}}, x \neq 0 \\ k, x=0\end{cases}$ remains discontinuous atx= 0, regardless the choice ofk. Solution: Finding the left hand and right hand limit for the given function, we have $\lim _{x \rightarrow 0^{+}} f(x)=\frac{x}{|x|+2 x^{2}}=\lim _{h \rightarrow 0} \frac{0-h}{|0-h|+2(0-h)^{2}}$ $=\lim _{h \rightarrow 0} \frac{-h}{h+2 h^{2}}=\lim _{h \rightarrow 0} \frac{-h}{h(1+2 h)}$ $=\lim _{h \rightarrow 0} \frac{-1}{1...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \tan ^{-1} \sqrt{\mathrm{x}} \mathrm{dx}$ Solution: $\int \tan ^{-1} \sqrt{x} d x$ $\int u . d v=u v-\int v d u$ Choose $u$ in these odder LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL) Here $\mathrm{u}=\tan ^{-1} \sqrt{x}$ and $\mathrm{v}=1$ $\therefore \int \tan ^{-1} \sqrt{x} d x$ $\therefore x \tan ^{-1} \sqrt{x}-\int x \cdot \frac{d\left(\tan ^{-1} \sqrt{x}\right)}{d x}$ $=x \tan ^{-1} \sqrt{x}-\frac{1}{2} \int \frac{\sqrt{x}}{1+x} d x$ Put $\sqrt{x}=t$ $\...
Read More →Evaluate the following integrals:
Question: Evaluate $\int x^{2} \tan ^{-1} x d x$ Solution: $\int x^{2} \tan ^{-1} x d x$ Here we will use integration by parts, $\int u \cdot d v=u v-\int v d u$ Choose $u$ in these oder LIATE(L-LOGS,I-INVERSE,A-ALGEBRAIC,T-TRIG,E-EXPONENTIAL) So here, $u=\tan ^{-1} x$ $=\tan ^{-1} x \int x^{2} d x-\frac{1}{3} \int x^{3}\left(d\left(\tan ^{-1} x\right)\right) /$ $d x+c$ $\left.\int x^{2} d x=\left(\frac{x^{3}}{3}\right)+c\right)$ $=\left(\frac{x^{3}}{3}\right) \tan ^{-1} x-\frac{1}{3} \int \frac...
Read More →f(x)= {1-coskx/xsinx, if x is not equal to 0
Question: $f(x)=\left\{\begin{array}{ll}\frac{1-\cos k x}{x \sin x}, \text { if } x \neq 0 \\ \frac{1}{2} \text {, if } x=0\end{array}\right.$ at $x=0$ Solution: Finding the left hand and right hand limits for the given function, we have $\lim _{x \rightarrow 0^{-}} f(x)=\frac{1-\cos k x}{x \sin x}$ $=\lim _{h \rightarrow 0} \frac{1-\cos k(0-h)}{(0-h) \sin (0-h)}=\lim _{h \rightarrow 0} \frac{1-\cos (-k h)}{-h \cdot \sin (-h)}$ $=\lim _{h \rightarrow 0} \frac{1-\cos k h}{h \sin h} \quad\left[\be...
Read More →Evaluate
Question: Evaluate $\int \frac{\sin x+\cos x}{\sin ^{4} x+\cos ^{4} x} d x$ Solution: $\int \frac{(\sin x+\cos x)}{\sin ^{4} x+\cos ^{4} x} d x$ $=\int \frac{(\sin x+\cos x)}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x} d x$ $=\int \frac{(\sin x+\cos x)}{1-2 \sin ^{2} x \cos ^{2} x} d x$ $=\int \frac{2(\sin x+\cos x)}{2-4 \sin ^{2} x \cos ^{2} x} d x$ $=\int \frac{2(\sin x+\cos x)}{2-\sin ^{2} 2 x} d x$ Let $\sin x-\cos x=t$, $(\cos x+\sin x) d x=d t$ $=\int \frac{2}{2-\le...
Read More →Evaluate
Question: Evaluate $\int \frac{\sin x+\cos x}{\sin ^{4} x+\cos ^{4} x} d x$ Solution: $\int \frac{(\sin x+\cos x)}{\sin ^{4} x+\cos ^{4} x} d x$ $=\int \frac{(\sin x+\cos x)}{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x} d x$ $=\int \frac{(\sin x+\cos x)}{1-2 \sin ^{2} x \cos ^{2} x} d x$ $=\int \frac{2(\sin x+\cos x)}{2-4 \sin ^{2} x \cos ^{2} x} d x$ $=\int \frac{2(\sin x+\cos x)}{2-\sin ^{2} 2 x} d x$ Let $\sin x-\cos x=t$, $(\cos x+\sin x) d x=d t$ $=\int \frac{2}{2-\le...
Read More →Prove the following
Question: $f(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} , \text { if }-1 \leq x0 \\ \frac{2 x+1}{x-1} , \text { if } 0 \leq x \leq 1\end{array}\right.$ at $x=0$ Solution: Finding the left hand and right hand limits for the given function, we have $\lim _{x \rightarrow 0^{-}} f(x)=\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}$ $=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \times \frac{\sqrt{1+k x}+\sqrt{1-k x}}{\sqrt{1+k x}+\sqrt{1-k x}}$ $=\lim _{x \rightarrow 0^{-}...
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