Prove the following
Question: $f(x)=\left\{\begin{array}{l}3 x+5, \text { if } x \geq 2 \\ x^{2}, \text { if } x2\end{array}\right.$ at $\mathrm{x}=2$ Solution: Checking the continuity of the given function, we have $\lim _{x \rightarrow 2^{-}} f(x)=3 x+5$ $=\lim _{h \rightarrow 0} 3(2+h)+5=11$ $\lim _{x \rightarrow 2} f(x)=3 x+5=3(2)+5=11$ $=\lim _{h \rightarrow 0}(2)^{2}+h^{2}-4 h=(2)^{2}=4$ Now, since $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2} f(x) \neq \lim _{x \rightarrow 2} f(x)$ Thus, f(x) is ...
Read More →Examine the continuity of the function
Question: Examine the continuity of the function f (x) = x3+ 2x2 1 at x = 1 Solution: We know that, y = f(x) will be continuous at x = a if, $\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a^{+}} f(x)$ Given:$f(x)=x^{3}+2 x^{2}-1$ $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0}(1+h)^{3}+2(1+h)^{2}-1=1+2-1=2$ $\lim _{x \rightarrow 1} f(x)=(1)^{3}+2(1)^{2}-1$ $=1+2-1=2$ $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{\rightarrow}(1+h)^{3}+2(1+h)^{2}-1$ $\lim ...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{x+3}{(x+4)^{2}} e^{x} d x=$ A. $\frac{e^{x}}{x+4}+C$ B. $\frac{e^{x}}{x+3}+C$ C. $\frac{1}{(x+4)^{2}}+C$ D. $\frac{\mathrm{e}^{\mathrm{x}}}{(\mathrm{x}+4)^{2}}+\mathrm{C}$ Solution: $\int \frac{x+3}{(x+4)^{2}} e^{x} d x$ $=\int \frac{x+4}{(x+4)^{2}} e^{x} d x-\int \frac{1}{(x+4)^{2}} e^{x} d x$ $=\int e^{x}\left(\frac{1}{x+4} d x-\frac{1}{(x+4)^{2}} d x\right)$ $\left[\because f(x)=\frac{1}{x+4} ; f^{\prime}(x)...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \frac{1}{\mathrm{x}(1+\log \mathrm{x})} \mathrm{dx}$ Solution: Given, $\int \frac{1}{x(1+\log x)} d x$ Let $1+\log x=t$ $\Rightarrow \frac{d}{d x}(1+\log x)=d t$ $\Rightarrow \frac{1}{x} d x=d t$ $=\int \frac{1}{t} d t$ $=\log t+c$ $=\log (1+\log x)+c$...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x$ Solution: Given, $\int \frac{1}{\sin ^{2} x \cdot \cos ^{2} x} d x$ $=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x} d x\left[\right.$ since, $\left.\sin ^{2} x+\cos ^{2} x=1\right]$ $=\int \frac{\sin ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x \cdot \cos ^{2} x} d x$ $=\int \frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x} d x$ $=\int\left(\sec ^{2} x+\operatorname{cosec}^{2} x\right) d x$ $...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \cos ^{-1}(\sin x) d x$ Solution: Given, $\int \cos ^{-1}(\sin x) d x$ Let us consider, $\int \cos ^{-1} \mathrm{dx}$ We know that, $\int f(x) \cdot g(x) d x=f(x) \int g(x) d x-\int\left[f^{\prime}(x) \int g(x)\right] d x$ By comparison, $f(x)=\cos ^{-1} x ; g(x)=1$ $=\cos ^{-1} x x \int 1 d x-\int-\frac{1}{\sqrt{1-x^{2}}} \cdot x d x$ $=x \cos ^{-1} x-\frac{1}{2} \int \frac{1}{\sqrt{1-x^{2}}}(-2 x) d x$ $=x \cos ^{-1} x-\frac{1}{2} \int\left(1-x^{2}\right)^{-\frac{1}{2...
Read More →If A, B and C are angles of a triangle,
Question: If A, B and C are angles of a triangle, then the determinant is equal to (A) 0 (B) -1 (C) 1 (D) None of these Solution: Option (A) 0 Given, $\Delta=\left|\begin{array}{ccc}-1 \cos C \cos B \\ \cos C -1 \cos A \\ \cos B \cos A -1\end{array}\right|$ On expanding the determinant, we get $\Delta=-1+2 \cos A \cos B \cos C+\cos ^{2} A+\cos ^{2} B+\cos ^{2} C$ Now, $2 \cos ^{2} A+2 \cos ^{2} B+2 \cos ^{2} C$ $=1+\cos 2 A+1+\cos 2 B+1+\cos 2 C$ $=3+(\cos 2 A+\cos 2 B+\cos 2 C)$ $=3+(\cos 2 A+\...
Read More →Write the anti-derivative
Question: Write the anti-derivative of $\left(3 \sqrt{x}+\frac{1}{\sqrt{x}}\right)$. Solution: Anti-derivative is nothing but integration Therefore its Anti-derivative can be found by integrating the above given equation. $=\int 3 \sqrt{x}+\frac{1}{\sqrt{x}} d x$ $=\int 3 x^{\frac{1}{2}}+x^{-\frac{1}{2}} d x$ $=3 \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\left[\right.$ since, $\left.\int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$ $=3 \frac{x^{\frac{3}{2}}}{\fr...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \frac{2}{1-\cos 2 x} d x$ Solution: Given, $\int \frac{2}{1-\cos 2 x} d x$ We Know that, $\cos 2 x=1-2 \sin ^{2} x$ $\Rightarrow 1-\cos 2 x=2 \sin ^{2} x$ Substitute this in the given, $=\int \frac{2}{2 \sin ^{2} x} d x$ $=\int \frac{1}{\sin ^{2} x} d x$ $=\int \operatorname{cosec}^{2} x d x$ $=-\cot x+c$...
Read More →The number of distinct real
Question: The number of distinct real roots of = 0 in the interval -/4 x /4 is (A) 0 (B) 2 (C) 1 (D) 3 Solution: Option (C) 1 Given, $\left|\begin{array}{lll}\sin x \cos x \cos x \\ \cos x \sin x \cos x \\ \cos x \cos x \sin x\end{array}\right|=0$ Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$, we get $\left|\begin{array}{lll}2 \cos x+\sin x \cos x \cos x \\ 2 \cos x+\sin x \sin x \cos x \\ 2 \cos x+\sin x \cos x \sin x\end{array}\right|=0$ $(2 \cos x+\sin x)\left|\begin{array}{lll}1 \cos x \cos...
Read More →If the solve the problem
Question: If $\int e^{x}(\tan x+1) \sec x d x=e^{x} f(x)+C$, then write the value $f(x)$. Solution: Given, $\int e^{x}(\tan x+1) \sec x d x$ It is clearly of the form $\int e^{x}\left[f(x)+f^{I}(x)\right] d x=e^{x} f(x)+c$ By comparison, $f(x)=1+\tan x ; f^{\prime}(x)=\sec x$ $=e^{x}(1+\tan x)+C$ Therefore, the value of $f(x)=1+\tan x$...
Read More →If the solve the problem
Question: If $\int\left(\frac{x-1}{x^{2}}\right) e^{x} d x=f(x) e^{x}+C$, then write the value of $f(x)$. Solution: Consider, $\int \frac{x-1}{x^{2}} e^{x} \mathrm{dx}$ $=\int \frac{x}{x^{2}}-\frac{1}{x^{2}} e^{x} d x$ $=\int \frac{1}{x}-\frac{1}{x^{2}} e^{x} d x$ It is clearly of the form, $\int e^{x}\left[f(x)+f^{l}(x)\right] d x=e^{x} f(x)+c$ By comparison, $\mathrm{f}(\mathrm{x})=\frac{1}{x} ; \mathrm{f}^{\prime}(\mathrm{x})=-\frac{1}{x^{2}}$ $=e^{x} \frac{1}{x}+c$ Therefore, the value of $f...
Read More →The determinant equals
Question: The determinant equals (A)abc(bc) (ca) (ab) (B) (bc) (ca) (ab) (C) (a+b+c) (bc) (ca) (ab) (D) None of these $\left|\begin{array}{lll}b^{2}-a b b-c b c-a c \\ a b-a^{2} a-b b^{2}-a b \\ b c-a c c-a a b-a^{2}\end{array}\right|$ Solution: Option (D) Given $\left|\begin{array}{lll}b^{2}-a b b-c b c-a c \\ a b-a^{2} a-b b^{2}-a b \\ b c-a c c-a a b-a^{2}\end{array}\right|$ $=\left|\begin{array}{lll}b(b-a) b-c c(b-a) \\ a(b-a) a-b b(b-a) \\ c(b-a) c-a a(b-a)\end{array}\right|$ Now, [Taking $...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \frac{x+\cos 6 x}{3 x^{2}+\sin 6 x} d x$ Solution: Given, $\int \frac{x+\cos 6 x}{3 x^{2}+\sin 6 x} d x$ Let $3 x^{2}+\sin 6 x=t$ $\Rightarrow \frac{d}{d x}\left(3 x^{2}+\sin 6 x\right)=d t$ $\Rightarrow 6 x+\cos 6 x \cdot 6=d t$ $\Rightarrow x+\cos 6 x=\frac{d t}{6}$ Substituting the values, $=\int \frac{1}{6 t} d t$ $=\frac{1}{6} \log t+c$ $=\frac{1}{6} \log \left(3 x^{2}+\sin 6 x\right)+c$...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int(1-\mathrm{x}) \sqrt{\mathrm{x}} \mathrm{dx}$. Solution: Given, $\int(1-x) \sqrt{x} \mathrm{dx}$ $=\int(\sqrt{x}-x \sqrt{x}) d x$ $=\int\left(x^{\frac{1}{2}}-x \cdot x^{\frac{1}{2}}\right) d x$ $=\int x^{\frac{1}{2}}-x^{\frac{3}{2}} d x$ $=\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+c\left[\right.$ since, $\left.\int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$ $=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+c$ $=\fr...
Read More →The area of a triangle with
Question: The area of a triangle with vertices (3, 0), (3, 0) and (0,k) is 9 sq. units. The value ofkwill be (A) 9 (B) 3 (C) 9 (D) 6 Solution: Option (B) 3 We know that, the area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by Area of triangle with vertices $(-3,0),(3,0)$ and $(0, k)$ is $[-3(-k)-0+1(3 k)]=\pm 18$ $6 k=\pm 18$ Thus, $k=\pm \frac{18}{6}=\pm 3$...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \sec x(\sec x+\tan x) d x$ Solution: Given, $\int \sec x(\sec x+\tan x) d x$ We know that, $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}$ By comparison, $a=4$ $=\frac{1}{4} \tan ^{-1} \frac{x}{4}+c$...
Read More →The value of determinant
Question: The value of determinant (A) a3+ b3+ c3 (B) 3 bc (C)a3+b3+c3 3abc (D) none of these Solution: Option (C)a3+b3+c3 3abc Given, $\Delta=\left|\begin{array}{lll}a-b b+c a \\ b-c c+a b \\ c-a a+b c\end{array}\right|$ [Applying $C_{1} \rightarrow C_{1}-C_{3}$ ] $=\left|\begin{array}{lll}-b b+c a \\ -c c+a b \\ -a a+b c\end{array}\right|$ [Applying $C_{2} \rightarrow C_{2}+C_{1}$ ] $=\left|\begin{array}{ccc}-b c a \\ -c a b \\ -a b c\end{array}\right|=-\left|\begin{array}{lll}b c a \\ c a b \...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \frac{\mathrm{e}^{\tan ^{-1}}}{1+\mathrm{x}^{2}} \mathrm{dx}$ Solution: Given, $\int \frac{e^{\tan ^{-1}}}{1+x^{2}} d x$ Let $\tan ^{-1} x=t$ $\partial \frac{d y}{d x}\left(\operatorname{Tan}^{-1} x\right)=\mathrm{dt}$ $\partial \frac{1}{1+x^{2}} d x=d t$ Now, $\int \frac{e^{\tan ^{-1}}}{1+x^{2}} d x$ $=\int e^{t} d t$ $=e^{t}+c$ $=e^{\tan ^{-1} x+c}$...
Read More →If then, value of x is
Question: If then, value of x is (A) 3 (B) 3 (C) 6 (D) 6 $\left|\begin{array}{cc}2 x 5 \\ 8 x\end{array}\right|=\left|\begin{array}{cc}6 -2 \\ 7 3\end{array}\right|$ Solution: Option (C) 6 From the given, On equating the determinants, we have 2x2 40 = 18 + 14 2x2= 72 x2= 36 Thus, x = 6...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$ Solution: Given, $\int \frac{x^{2}-x^{2}+x-1}{x-1} d x$. $=\int \frac{x^{2}(x-1)+x-1}{x-1} d x$ $=\int \frac{(x-1)\left[x^{2}+1\right]}{x-1} d x$ $=\int\left(x^{2}+1\right) d x\left[\right.$ since, $\left.\int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$ $=\frac{x^{3}}{3}+x+c$...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \frac{x^{3}-1}{x^{2}} d x$ Solution: Given, $\int \frac{x^{2}-1}{x^{2}} d x$ $=\int \frac{x^{3}}{x^{2}}-\frac{1}{x^{2}} d x$ $=\int x-\frac{1}{x^{2}} d x$ $\left[\right.$ since, $\left.\int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$ $=\frac{x^{2}}{2}-\frac{x^{-2+1}}{-2+1}+c$ $=\frac{x^{2}}{2}-\frac{x^{-1}}{-1}+c$ $=\frac{x^{2}}{2}+\frac{1}{x}+c$...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int \frac{1-\sin x}{\cos ^{2} x} d x$ Solution: Given, $\int \frac{1-\sin x}{\cos ^{2} x} d x$ $=\int \frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x} d x$ $=\int \sec ^{2} x-\tan x \cdot \sec x d x\left[\operatorname{since}, \cos x=\frac{1}{\sec x}\right]$ $=\tan x-\sec x+c$...
Read More →Evaluate the following integrals:
Question: Evaluate: $\int 2^{x} d x$ Solution: Given, $\int 2^{x} d x$ $=\frac{2^{x}}{\log 2}+c\left[\operatorname{since}, \int a^{x} d x=\frac{a^{x}}{\log a}\right]$...
Read More →Write a value
Question: Write a value of $\int \frac{1}{1+e^{x}} d x$. Solution: given $\int \frac{1}{1+e^{x}} d x$ $=\int\left(1-\frac{e^{x}}{1+e^{x}}\right) d x$ Let $1+e^{x}=t$ Differentiating on both sides we get, $E^{x} d x=d t$ Substituting above equation in given equation we get, $=\int\left(1-\frac{1}{t}\right) d t$ $=t-\log t+c$ $=1+e^{x}-\log \left(1+e^{x}\right)+c$...
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