Let the function
Question: Let $\vec{a}=\hat{i}+2 \hat{j}+4 \hat{k}, \quad \vec{b}=\hat{i}+\lambda \hat{j}+4 \hat{k}$ and $\overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\left(\lambda^{2}-1\right) \hat{\mathrm{k}}$ be coplanar vectors. Then the non-zero vector $\vec{a} \times \vec{c}$ is :$-14 \hat{i}-5 \hat{j}$$-10 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}$$-10 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}$$-14 \hat{i}+5 \hat{j}$Correct Option: , 3 Solution: $\left[\begin{array}{lll}\overrightarrow{\mathrm{a...
Read More →Solve this following
Question: If a circle $C$ passing through the point $(4,0)$ touches the circle $x^{2}+y^{2}+4 x-6 y=12$ externally at the point $(1,-1)$, then the radius of $C$ is : $\sqrt{57}$4$2 \sqrt{5}$5Correct Option: , 4 Solution: $x^{2}+y^{2}+4 x-6 y-12=0$ Equation of tangent at $(1,-1)$ $x-y+2(x+1)-3(y-1)-12=0$ $3 x-4 y-7=0$ $\therefore$ Equation of circle is $\left(x^{2}+y^{2}+4 x-6 y-12\right)+\lambda(3 x-4 y-7)=0$ It passes through $(4,0)$ : $(16+16-12)+\lambda(12-7)=0$ $\Rightarrow 20+\lambda(5)=0$ ...
Read More →The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of
Question: The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this series is$\frac{4}{9}$$\frac{2}{9}$$\frac{2}{3}$$\frac{1}{3}$Correct Option: , 2 Solution: $\frac{a}{1-r}=3$ ........(1) $\frac{\mathrm{a}^{3}}{1-\mathrm{r}^{3}}=\frac{27}{19} \Rightarrow \frac{27(1-\mathrm{r})^{3}}{1-\mathrm{r}^{3}}=\frac{27}{19}$ $\Rightarrow 6 r^{2}-13 r+6=0$ $\Rightarrow r=\frac{2}{3}$ as $|r|1$...
Read More →The area (in sq. units) bounded by the parabola
Question: The area (in sq. units) bounded by the parabola $y=x^{2}-1$, the tangent at the point $(2,3)$ to it and the $y$-axis is :$\frac{14}{3}$$\frac{56}{3}$$\frac{8}{3}$$\frac{32}{3}$Correct Option: , 3 Solution: Equation of tangent at $(2,3)$ on $y=x^{2}-1$, is $y=(4 x-5)$ ............(i) $\therefore$ Required shaded area $=\operatorname{ar}(\Delta \mathrm{ABC})-\int_{-1}^{3} \sqrt{\mathrm{y}+1} \mathrm{dy}$ $=\frac{1}{2} \cdot(8) \cdot(2)-\frac{2}{3}\left((y+1)^{3 / 2}\right)_{-1}^{3}$ $=8-...
Read More →The outcome of each of 30 items was observed;
Question: The outcome of each of 30 items was observed; 10 items gave an outcome $\frac{1}{2}-\mathrm{d}$ each, 10 items gave outcome $\frac{1}{2}$ each and the remaining 10 items gave outcome $\frac{1}{2}+\mathrm{d}$ each. If the variance of this outcome data is $\frac{4}{3}$ then $|\mathrm{d}|$ equals :-2$\frac{\sqrt{5}}{2}$$\frac{2}{3}$$\sqrt{2}$Correct Option: , 4 Solution: Variance is independent of origin. So we shift the given data by $\frac{1}{2}$. so, $\frac{10 \mathrm{~d}^{2}+10 \times...
Read More →Solve this following
Question: Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a function such that $f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in R$ Then $f(2)$ equal :8$-2$$-4$30Correct Option: , 2 Solution: $f(x)=x^{3}+x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)$ $\Rightarrow f^{\prime}(x)=3 x^{2}+2 x f^{\prime}(1)+f^{\prime \prime}(x)$ $\ldots \ldots(1)$ $\Rightarrow f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$$\ldots \ldots(2)$ $\Rightarr...
Read More →The area (in sq. units) of the region bounded by
Question: The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y 2 :-$\frac{5}{4}$$\frac{9}{8}$$\frac{3}{4}$$\frac{7}{8}$Correct Option: , 2 Solution: $x=4 y-2 \ x^{2}=4 y$ $\Rightarrow x^{2}=x+2 \Rightarrow x^{2}-x-2=0$ $x=2,-1$ So, $\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right) d x=\frac{9}{8}$...
Read More →The area (in sq. units) of the region bounded by
Question: The area (in sq. units) of the region bounded by the curve $x^{2}=4 y$ and the straight line x = 4y 2 :-$\frac{5}{4}$$\frac{9}{8}$$\frac{3}{4}$$\frac{7}{8}$Correct Option: , 2 Solution: $x=4 y-2 \ x^{2}=4 y$ $\Rightarrow x^{2}=x+2 \Rightarrow x^{2}-x-2=0$ $x=2,-1$ So, $\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right) d x=\frac{9}{8}$...
Read More →Consider a triangular plot
Question: Consider a triangular plot $\mathrm{ABC}$ with sides $\mathrm{AB}=7 \mathrm{~m}, \mathrm{BC}=5 \mathrm{~m}$ and $\mathrm{CA}=6 \mathrm{~m}$. A vertical lamp-post at the mid point $\mathrm{D}$ of $\mathrm{AC}$ subtends an angle $30^{\circ}$ at $\mathrm{B}$. The height (in $\mathrm{m}$ ) of the lamp-post is: $7 \sqrt{3}$$\frac{2}{3} \sqrt{21}$$\frac{3}{2} \sqrt{21}$$2 \sqrt{21}$Correct Option: , 2 Solution: $\mathrm{BD}=\mathrm{h} \cot 30^{\circ}=\mathrm{h} \sqrt{3}$ So, $\left.7^{2}+5^{...
Read More →Let the function
Question: Let $A=\left(\begin{array}{ccc}0 2 q r \\ p q -r \\ p -q r\end{array}\right)$. It AAT $A^{T} I_{3}$, then $|p|$ is :$\frac{1}{\sqrt{2}}$$\frac{1}{\sqrt{5}}$$\frac{1}{\sqrt{6}}$$\frac{1}{\sqrt{3}}$Correct Option: 1 Solution: A is orthogonal matrix $\Rightarrow 0^{2}+\mathrm{p}^{2}+\mathrm{p}^{2}=1 \Rightarrow|\mathrm{p}|=\frac{1}{\sqrt{2}}$...
Read More →The value of
Question: $\int \frac{d t}{\sqrt{3 t-2 t^{2}}}$ Solution: Let, $\mathrm{I}=\int \frac{d t}{\sqrt{3 t-2 t^{2}}}=\int \frac{d t}{\sqrt{-2\left(t^{2}-\frac{3}{2} t\right)}}$ $=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{-\left(t^{2}-\frac{3}{2} t+\frac{9}{16}-\frac{9}{16}\right)}} \quad$ [Making perfect $=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{-\left[\left(t-\frac{3}{4}\right)^{2}-\frac{9}{16}\right]}}=\frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{\frac{9}{16}-\left(t-\frac{3}{4}\right)^{2}}}$ $=\frac{1}...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{2} x} d x$ Solution: $y=\int \frac{\sin 2 x}{\left(\sin ^{2} x\right)^{2}+\left(1-\sin ^{2} x\right)^{2}} d x$ Let, $\sin ^{2} x=t$ Differentiating both side with respect to $x$ $\frac{d t}{d x}=2 \sin x \cos x \Rightarrow d t=\sin 2 x d x$ $y=\int \frac{d t}{t^{2}+(1-t)^{2}}$ $y=\int \frac{d t}{2 t^{2}-2 t+1}$ Try to make perfect square in denominator $y=\int \frac{d t}{2 t^{2}-2 t+\frac{1}{2}+\frac{1}{2}}$ $y=\int \frac{d t}{(\sqrt{2}...
Read More →Solve the following systems of equations:
Question: $\int \frac{d x}{\sqrt{16-9 x^{2}}}$ Solution: Let, $\mathrm{I}=\int \frac{d x}{\sqrt{16-9 x^{2}}}$ $=\frac{1}{3} \int \frac{d x}{\sqrt{\frac{16}{9}-x^{2}}}=\frac{1}{3} \int \frac{d x}{\sqrt{\left(\frac{4}{2}\right)^{2}-x^{2}}}$ $=\frac{1}{3} \sin ^{-1} \frac{x}{4 / 3}+C\left[\because \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+C\right]$ $=\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C$ Therefore, $\mathrm{I}=\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C$....
Read More →Prove the following identities.
Question: $\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x$ Solution: Let, $\mathrm{I}=\int \frac{\sqrt{1+x^{2}}}{x^{4}} d x$ $=\int \sqrt{\frac{1+x^{2}}{x^{2}}} \cdot \frac{1}{x^{3}} d x=\int \sqrt{\frac{1}{x^{2}}+1} \cdot \frac{1}{x^{3}} d x$ Taking, $\frac{1}{x^{2}}+1=t^{2}$ So, $\frac{-2}{x^{3}} d x=2 t d t \Rightarrow \frac{d x}{x^{3}}=-t d t$ $\therefore \mathrm{I}=\int t(t d t)=-\int t^{2} d t=-\frac{1}{3} t^{3}+\mathrm{C}$ Therefore, $I=-\frac{1}{3}\left(\frac{1}{x^{2}}+1\right)^{3 / 2}+C$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sqrt{\sin x} \cos ^{3} x d x$ Solution: $y=\int \sqrt{\sin x}\left(1-\sin ^{2} x\right) \cos x d x$ Let, $\sin x=t$ Differentiating both side with respect to $x$ $\frac{d t}{d x}=\cos x \Rightarrow d t=\cos x d x$ $y=\int \sqrt{t}\left(1-t^{2}\right) d t$ $y=\int t^{\frac{1}{2}}-t^{\frac{5}{2}} d t$ Using formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$ $y=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\frac{t^{\frac{7}{2}}}{\frac{7}{2}}+c$ Again, put $t=\sin x$ $y=\frac{\sin x^{\frac...
Read More →Prove the following
Question: $\int \frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}} d x$ (hint : Put $\left.x=z^{4}\right)$ Solution: Let's consider, $I=I_{1}-I_{2}$ Now, $\mathrm{I}_{1}=4 \int t^{2} d t=4 \cdot \frac{t^{3}}{3}+C_{1}=\frac{4}{3} x^{3 / 4}+C_{1}$ $I_{2}=4 \int \frac{t^{2}}{t^{3}+1} d t$ Putting, $t^{3}+1=z \Rightarrow 3 t^{2} d t=d z$ $t^{2} d t=\frac{1}{3} d z$ So, $\quad I_{2}=\frac{4}{3} \int \frac{d z}{z}=\frac{4}{3} \log |z|+C_{2}=\frac{4}{3} \log \left|t^{3}+1\right|+C_{2}$ $=\frac{4}{3} \log \left|...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \cos ^{5} x d x$. Solution: $y=\int\left(1-\sin ^{2} x\right)^{2} \cos x d x$ Let, $\sin x=t$ Differentiating both side with respect to $x$ $\frac{d t}{d x}=\cos x \Rightarrow d t=\cos x d x$ $y=\int\left(1-t^{2}\right)^{2} d t$ $y=\int 1+t^{4}-2 t^{2} d t$ Using formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$ and $\int c d t=c t$ $y=\left(t+\frac{t^{5}}{5}-2 \frac{t^{3}}{3}\right)+c$ Again, put $t=\sin x$ $y=\left(\sin x+\frac{\sin ^{5} x}{5}-2 \frac{\sin ^{3} x}{3}\right)...
Read More →Show that A B=B A in each of the following cases:
Question: Show that $A B=B A$ in each of the following cases: $A=\left[\begin{array}{lll}1 3 -1 \\ 2 2 -1 \\ 3 0 -1\end{array}\right]$ and $B=\left[\begin{array}{rrr}-2 3 -1 \\ -1 2 -1 \\ -6 9 -4\end{array}\right]$ Solution: Given : $A=\left[\begin{array}{ccc}1 3 -1 \\ 2 2 -1 \\ 3 0 -1\end{array}\right]$ and $B=\left[\begin{array}{ccc}-2 3 -1 \\ -1 2 -1 \\ -6 9 -4\end{array}\right]$ Matrix A is of order $3 \times 3$ and Matrix $B$ is of order $3 \times 3$ To show : matrix $\mathrm{AB}=\mathrm{BA...
Read More →Prove the following
Question: $\int \sqrt{\frac{a+x}{a-x}}$ Solution: Let, $I=\int \sqrt{\frac{a+x}{a-x}} d x$ $=\int \sqrt{\frac{a+x}{a-x} \times \frac{a+x}{a+x}} d x=\int \frac{a+x}{\sqrt{(a-x)(a+x)}} d x$ $=\int \frac{a+x}{\sqrt{a^{2}-x^{2}}} d x=\int \frac{a}{\sqrt{a^{2}-x^{2}}} d x+\int \frac{x}{\sqrt{a^{2}-x^{2}}} d x$ Let's consider, $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$ Now, $\mathrm{I}_{1}=\int \frac{a}{\sqrt{a^{2}-x^{2}}} d x=a \cdot \sin ^{-1} \frac{x}{a}+\mathrm{C}_{1}$ And, $\mathrm{I}_{2}=\int \f...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sin ^{5} x d x$ Solution: $y=\int\left(1-\cos ^{2} x\right)^{2} \sin x d x$ Let, $\cos x=\mathrm{t}$ Differentiating both side with respect to $x$ $\frac{d t}{d x}=-\sin x \Rightarrow-d t=\sin x d x$ $y=-\int\left(1-t^{2}\right)^{2} d t$ $y=-\int 1+t^{4}-2 t^{2} d t$ Using formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$ and $\int c d t=c t$ $y=-\left(t+\frac{t^{5}}{5}-2 \frac{t^{3}}{3}\right)+c$ Again, put $t=\cos x$ $y=-\left(\cos x+\frac{\cos ^{5} x}{5}-2 \frac{\cos ^{3}...
Read More →The value of
Question: $\int \frac{x}{\sqrt{x}+1} d x$ (Hint : Put $\sqrt{x}=z)$ Solution: Let, $\quad \mathrm{I}=\int \frac{x}{\sqrt{x}+1} d x$ Put $\sqrt{x}=t \Rightarrow x=t^{2} \quad \therefore d x=2 t . d t$ So, $I=\int \frac{t^{2} \cdot 2 t \cdot d t}{t+1}=2 \int \frac{t^{3}}{t+1} d t=2 \int \frac{t^{3}+1-1}{t+1} d t$ $=2 \int \frac{t^{3}+1}{t+1} d t-2 \int \frac{1}{t+1} d t$ $=2 \int \frac{(t+1)\left(t^{2}-t+1\right)}{t+1} d t-2 \int \frac{1}{t+1} d t$ $=2 \int\left(t^{2}-t+1\right) d t-2 \int \frac{1...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sin ^{2} x \cos ^{2} x d x$ Solution: $y=\int\left(1-\cos ^{2} x\right) \cos ^{4} x \sin x d x$' Let, $\cos x=t$ Differentiating both side with respect to $x$ $\frac{d t}{d x}=-\sin x \Rightarrow-d t=\sin x d x$ $y=\int-\left(1-t^{2}\right) t^{4} d t$ $y=-\int t^{4}-t^{6} d t$ Using formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$ $y=-\left(\frac{t^{5}}{5}-\frac{t^{7}}{7}\right)+c$ Again, put $t=\cos x$ $y=\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}+c$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int x \sin ^{5} x^{2} \cos x^{2} d x$ Solution: Let, $\sin x^{2}=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=\cos x^{2} \times 2 x \Rightarrow \frac{d t}{2}=x \cos x^{2} d x$ $y=\int \frac{t^{5}}{2} d t$ Using formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$ $y=\frac{t^{6}}{2 \times 6}+c$ Again, put $t=\sin x^{2}$ $y=\frac{\sin ^{6} x^{2}}{12}+c$...
Read More →Solve the following systems of equations:
Question: $\int \sqrt{1+\sin x} d x$ Solution: Let $\mathrm{I}=\int \sqrt{1+\sin x} d x$ $=\int \sqrt{\left(\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} d x$ $=\int \sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}} d x=\int\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x$ $=\int \sin \frac{x}{2} d x+\int \cos \frac{x}{2} d x=-2 \cos \frac{x}{2}+2 \sin \frac{x}{2}+C$ $=2\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)+C$, where $C$ is a constant ...
Read More →Prove the following
Question: $\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x$ Solution: Let $\mathrm{I}=\int \frac{\sin x+\cos x}{\sqrt{1+2 \sin x \cos x}} d x$ $=\int \frac{(\sin x+\cos x)}{\sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}} d x$ $=\int \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^{2}}} d x=\int \frac{\sin x+\cos x}{\sin x+\cos x} d x$ $=\int 1 d x$ $=\int 1 d x$ $=x+C$ , where $C$ is a constant Therefore, $\mathrm{I}=\mathrm{x}+\mathrm{C}$...
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