In figure, if PQR is the tangent to a circle at Q
Question: In figure, if PQR is the tangent to a circle at Q whose centre is 0, AB is a chord parallel to PR and BQR = 70, then AQB is equal to (a) 20 (b) 40 (c) 35 (d) 45 Solution: (b) Given. $A B \| P R$ $\therefore$$\angle A B Q=\angle B Q R=70^{\circ}$ [alternate angles] $\triangle \operatorname{lon} O D$ is nernendicular to $A B$ and $Q D$ bisects $A B$. In $\triangle Q D A$ and $\triangle Q D B, \quad \angle Q D A=\angle Q D B$ [each $90^{\circ}$ ] $A D=B D$ $Q D=Q D$ [common side] $\theref...
Read More →Let f (x) = |sin x|. Then,
Question: Let $f(x)=|\sin x|$. Then, (a) $f(x)$ is everywhere differentiable. (b) $f(x)$ is everywhere continuous but not differentiable at $x=n \pi, n \in Z$ (c) $f(x)$ is everywhere continuous but not differentiable at $x=(2 n+1) \frac{\pi}{2}, n \in Z$. (d) none of these Solution: (b) $f(x)$ is everywhere continuous but not differentiable at $x=n \pi, n \in Z$ We have, $f(x)=|\sin x|$ $\Rightarrow f(x)= \begin{cases}0, x=2 n \pi \\ \sin x, 2 n \pix(2 n+1) \pi \\ 0, x=(2 n+1) \pi \\ -\sin x, (...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer 0.0000463 in standard form is (a) 463 107 (b) 4.63 105 (c) 4.63 109 (d) 46.3 106 Solution: (b) 4.63 105 $0.0000463=\frac{463}{10^{7}}=\frac{4.63 \times 10^{2}}{10^{7}}=4.63 \times 10^{(2-7)}=4.63 \times 10^{-5}$...
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Question: Tick (✓) the correct answer 3670000 in standard form is (a) 367 104 (b) 36.7 105 (c) 3.67 106 (d) none of these Solution: (c) 3.67 106 $3670000=367 \times 10^{4}=3.67 \times 100 \times 10^{4}=3.67 \times 10^{2} \times 10^{4}=3.67 \times 10^{(2+4)}=3.67 \times 10^{6}$...
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Question: Let $R=\{(a, b): a, b, \in N$ and $ab\}$. Show that $\mathrm{R}$ is a binary relation on $\mathrm{N}$, which is neither reflexive nor symmetric. Show that $R$ is transitive. Solution: N is the set of all the natural numbers. $N=\{1,2,3,4,5,6,7 \ldots \ldots\}$ $R=\{(a, b): a, b, \in N$ and $ab\}$ $R=\{(1,2),(1,3),(1,4) \ldots(2,3),(2,4),(2,5) \ldots \ldots\}$ For reflexivity, A relation R on N is said to be reflexive if (a, a) є R for all a є N. But, here we see that $\mathrm{a}\mathrm...
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Question: Tick (✓) the correct answer $\left(\frac{-3}{4}\right)^{2}=?$ (a) $\frac{-9}{16}$ (b) $\frac{9}{16}$ (c) $\frac{16}{9}$ (d) $\frac{-16}{9}$ Solution: (b) $\frac{9}{16}$ $\left(\frac{-3}{4}\right)^{2}=\frac{(-3)^{2}}{(4)^{2}}=\frac{9}{16}$...
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Question: Tick (✓) the correct answer $\left(\frac{-1}{2}\right)^{3}=?$ (a) $\frac{-1}{6}$ (b) $\frac{1}{6}$ (c) $\frac{1}{8}$ (d) $\frac{-1}{8}$ Solution: (d) $\frac{-1}{8}$ $\left(\frac{-1}{2}\right)^{3}=\frac{-1^{3}}{2^{3}}=\frac{-1}{8}$...
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Question: Let $f(x)\left\{\begin{array}{ll}a x^{2}+1, x1 \\ x+1 / 2, x \leq 1\end{array}\right.$. Then, $f(x)$ is derivable at $x=1$, if (a) $a=2$ (b) $a=1$ (c) $a=0$ (d) $a=1 / 2$ Solution: (d) $a=1 / 2$ Given: $f(x)= \begin{cases}a x^{2}+1, x1 \\ x+\frac{1}{2}, x \leq 1\end{cases}$ The function is derivable at $x=1$, iff left hand derivative and right hand derivative of the function are equal at $x=1$. $(\mathrm{LHD}$ at $x=1)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$ $(\mathrm{LHD}$ ...
Read More →If two tangents inclined at an angle 60°
Question: If two tangents inclined at an angle 60 are drawn to a circle of radius 3 cm, then the length of each tangent is (a) $\frac{3}{2} \sqrt{3} \mathrm{~cm}$ (b) $6 \mathrm{~cm}$ (c) $3 \mathrm{~cm}$ (d) $3 \sqrt{3} \mathrm{~cm}$ Solution: (d) Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60. Join $O A$ and $O P$. Also, $O P$ is a bisector line of $\angle A P C$. $\therefore \quad \angle A P O=\angle C P O=30^{\circ}$ Also, $...
Read More →Let A = {a, b}. List all relation on A and find their number
Question: Let A = {a, b}. List all relation on A and find their number Solution: Any relation on A is a subset of AA. $A \times A=\{(a, a),(a, b),(b, a),(b, b)\}$ The subsets are. {} empty set $\{(a, a)\}$ $\{(a, b)\}$ $\{(a, a),(a, b)\}$ $\{(b, a)\}$ $\{(b, b)\}$ $\{(b, a),(b, b)\}$ $\{(a, a),(b, a)\}$ $\{(a, b),(b, a)\}$ $\{(a, a),(b, a),(b, b)\}$ $\{(a, a),(b, b)\}$ $\{(a, a),(a, b),(b, a)\}$ $\{(a, a),(a, b),(b, b)\}$ $\{(a, b),(b, a),(b, b)\}$ $\{(a, a),(a, b),(b, a),(b, b)\}$ Thus, there a...
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Question: Tick (✓) the correct answer $\left(\frac{-5}{3}\right)^{-1}=?$ (a) $\frac{5}{3}$ (b) $\frac{3}{5}$ (c) $\frac{-3}{5}$ (d) none of these Solution: (C) $\frac{-3}{5}$ $\left(\frac{-5}{3}\right)^{-1}=\left(\frac{3}{-5}\right)^{1}=\frac{3}{-5}=\frac{3 \times(-1)}{-5 \times(-1)}=\frac{-3}{5}$...
Read More →The function f (x) = x − [x], where [⋅] denotes
Question: The function $f(x)=x-[x]$, where [.] denotes the greatest integer function is (a) continuous everywhere(b) continuous at integer points only(c) continuous at non-integer points only(d) differentiable everywhere Solution: (c) continuous at non-integer points only We have, $f(x)=x-[x]$ Consider $n$ be an integer. $f(x)=x-[x]= \begin{cases}x-(n-1) n-1 \leq xn \\ 0 x=n \\ x-n n \leq xn+1\end{cases}$ Now, $(\mathrm{LHL}$ at $x=n)=\lim _{x \rightarrow n^{-}} f(x)=x-(n-1)=x-n+1$ $(\mathrm{RHL...
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Question: Tick (✓) the correct answer $\left(\frac{2}{3}\right)^{0}=?$ (a) $\frac{3}{2}$ (b) $\frac{2}{3}$ (C) 1 (d) 0 Solution: (C) 1 Using the law of exponents $\left(\frac{a}{b}\right)^{0}=1$ : $\therefore\left(\frac{2}{3}\right)^{0}=1$...
Read More →The function f (x) = x − [x], where [⋅] denotes
Question: The function $f(x)=x-[x]$, where [.] denotes the greatest integer function is (a) continuous everywhere(b) continuous at integer points only(c) continuous at non-integer points only(d) differentiable everywhere Solution: (c) continuous at non-integer points only We have, $f(x)=x-[x]$ Consider $n$ be an integer. $f(x)=x-[x]= \begin{cases}x-(n-1) n-1 \leq xn \\ 0 x=n \\ x-n n \leq xn+1\end{cases}$ Now, $(\mathrm{LHL}$ at $x=n)=\lim _{x \rightarrow n^{-}} f(x)=x-(n-1)=x-n+1$ $(\mathrm{RHL...
Read More →In figure, if PA and PB are tangents
Question: In figure, if PA and PB are tangents to the circle with centre 0 such that APB = 50, then OAB is equal to (a) 25 (b) 30 (c) 40 (d) 50 Solution: (a)Given, PA and PB are tangent lines. $\therefore$$P A=P B$ [since, the length of tangents drawn from an external point to a circle is equal] $\Rightarrow \quad \angle P B A=\angle P A B=\theta \quad$ [say] $\ln \triangle P A B, \quad \angle P+\angle A+\angle B=180^{\circ}$ [since, sum of angles of a triangle $=180^{\circ}$ ] $\Rightarrow \qua...
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Question: Tick (✓) the correct answer If (23x 1+ 10) 7 = 6, thenxis equal to (a) 2 (b) 0 (c) 1 (d) 2 Solution: (d) 2 $\left(2^{3 x-1}+10\right) \div 7=6$ $\Rightarrow \frac{\left(2^{3 x-1}+10\right)}{7}=\frac{6}{1}$ $\left(2^{3 x-1}+10\right) \times 1=6 \times 7=42$ $\Rightarrow 2^{3 x-1}=42-10$ $\Rightarrow 2^{3 x-1}=32$ $\Rightarrow 2^{3 x-1}=2^{5}$ $\Rightarrow 3 x-1=5$ $\Rightarrow 3 x=6$ Therefore, $x=2$...
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Question: Let $f(x)= \begin{cases}\frac{1}{|x|} \text { for }|x| \geq 1 \\ a x^{2}+b \text { for }|x|1\end{cases}$ If $f(x)$ is continuous and differentiable at any point, then (a) $a=\frac{1}{2}, b=-\frac{3}{2}$ (b) $a=-\frac{1}{2}, b=\frac{3}{2}$ (c) $a=1, b=-1$ (d) none of these Solution: (b) $a=-\frac{1}{2}, b=\frac{3}{2}$ We have, $f(x)=\left\{\begin{array}{l}\frac{-1}{x}, \quad x \leq-1 \\ a x^{2}+b, \quad-1x1 \\ \frac{1}{x}, \quad x \geq 1\end{array}\right.$ Given: $f(x)$ is differentiabl...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer The value of $x$ for which $\left(\frac{7}{12}\right)^{-4} \times\left(\frac{7}{12}\right)^{3 x}=\left(\frac{7}{12}\right)^{5}$, is (a) $-1$ (b) 1 (c) 2 (d) 3 Solution: (d) 3 $\left(\frac{7}{12}\right)^{-4} \times\left(\frac{7}{12}\right)^{3 x}=\left(\frac{7}{12}\right)^{5}$ $\Rightarrow\left(\frac{7}{12}\right)^{-4+3 x}=\left(\frac{7}{12}\right)^{5}$ $\Rightarrow 3 x-4=5$ $3 x=9$ or $x=\frac{9}{3}=3$...
Read More →In figure, if 0 is the centre of a circle,
Question: In figure, if 0 is the centre of a circle, PQ is a chord and the tangent PR at P , ; makes an angle of 50 with PQ, then POQ is equal to (a) 100 (b) 80 (c) 90 (d) 75 Solution: (a)Given, QPR = 50 We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact. $\therefore \quad \angle O P R=90^{\circ}$ $\Rightarrow \quad \angle O P Q+\angle Q P R=90^{\circ} \quad$ [from figure] $\Rightarrow \quad \angle O P Q=90^{\circ}-50^{\circ}=40^{\circ}...
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Question: Tick (✓) the correct answer $\left[\left\{\left(-\frac{1}{2}\right)^{2}\right\}^{-2}\right]^{-1}=?$ (a) $\frac{1}{16}$ (b) 16 (c) $-\frac{1}{16}$ (d) $-16$ Solution: (a) $\frac{1}{16}$ $\left[\left\{\left(-\frac{1}{2}\right)^{2}\right\}^{-2}\right]^{-1}$ $=\left[\left\{-\frac{1}{2}\right\}^{-4}\right]^{-1}$ $=\left(-\frac{1}{2}\right)^{(-4 \times-1)}$ $=\left(-\frac{1}{2}\right)^{4}$ $=\frac{1}{16}$...
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Question: Find $\mathrm{R}^{-1}$, when (i) $R=\{(1,2),(1,3),(2,3),(3,2),(4,5)\}$ (ii) $R=\{(x, y): x, y \in N, x+2 y=8\}$. Solution: (i) $R=\{(1,2),(1,3),(2,3),(3,2),(4,5)\}$ $\mathrm{R}^{-1}=\{(2,1),(3,1),(3,2),(2,3),(5,4)\}$ (ii) $R=\{(x, y): x, y \in N, x+2 y=8\}$. $y=\frac{8-x}{2}$ Put $x=2, y=3$ Put $x=4, y=2$ Put $x=6, y=1$ $R=\{(2,3),(4,2),(6,1)\}$ $\mathrm{R}^{-1}=\{(3,2),(2,4),(1,6)\}$...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer $\left\{\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-3}\right\} \div\left(\frac{1}{4}\right)^{-3}=?$ (a) $\frac{19}{64}$ (b) $\frac{27}{16}$ (c) $\frac{64}{19}$ (d) $\frac{16}{25}$ Solution: (a) $\frac{19}{64}$ $\left\{\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-3}\right\} \div\left(\frac{1}{4}\right)^{-3}$ $=\left\{3^{3}-2^{3}\right\} \div 4^{3}$ $=\{27-8\} \div 64$ $=19 \div 64$ $=\frac{19}{64}$...
Read More →In figure, AT is a tangent to the circle
Question: In figure, $\mathrm{AT}$ is a tangent to the circle with centre 0 such that $\mathrm{OT}=4 \mathrm{~cm}$ and $\angle \mathrm{OTA}=30^{\circ}$. Then, $\mathrm{AT}$ is equal to (a) 4 cm (b) 2 cm (c) 23 cm (d) 43 cm Solution: (c)Join OA We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact. $\therefore \quad \angle O A T=90^{\circ}$ In $\triangle O A T$, $\cos 30^{\circ}=\frac{A T}{O T}$ $\Rightarrow$$\frac{\sqrt{3}}{2}=\frac{A T}{4...
Read More →Let R = (x, y) : x, y ϵ Z and x2 + y2 = 25}.
Question: Let $R=(x, y): x, y \in Z$ and $\left.x^{2}+y^{2}=25\right\}$ Express $\mathrm{R}$ and $\mathrm{R}^{-1}$ as sets of ordered pairs. Show that $\mathrm{R}=\mathrm{R}^{-1}$. Solution: $x^{2}+y^{2}=25$ Put $x=0, y=5,0^{2}+5^{2}=25$ Put $x=3, y=4,3^{2}+4^{2}=25$ $R=\{(0,5),(0,-5),(5,0),(-5,0),(3,4),(-3,4),(-3,-4),(3,-4)\}$ Since, $x$ and $y$ get interchanged in the ordered pairs, $R$ and $R^{-1}$ are same....
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Question: Tick (✓) the correct answer $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}=?$ (a) $\frac{61}{144}$ (b) $\frac{144}{61}$ (C) 29 (d) $\frac{1}{29}$ Solution: (C) 29 $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}=\left(\frac{2}{1}\right)^{2}+\left(\frac{3}{1}\right)^{2}+\left(\frac{4}{1}\right)^{2}$ $=2^{2}+3^{2}+4^{2}$ $=4+9+16$ $=29$...
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