The tangent to the circumcircle of an isosceles
Question: The tangent to the circumcircle of an isosceles ΔABC at A, in which AB = AC, is parallel to BC. Solution: TrueLet EAF be tangent to the circumcircle of ΔABC. To prove $F A F \| B C$ $\angle E A B=\angle A B C$ Here. $A B=A C$ $\Rightarrow$ $\angle A B C=\angle A C B$$\ldots(1)$ [angle between tangent anc is chord equal to angle made by chord in the alternate segment] $\therefore$ Also, $\quad \angle E A B=\angle B C A$ ....(ii) From Eqs. (i) and (ii), we get $\angle E A B=\angle A B C$...
Read More →Mark (✓) against the correct answer
Question: $\operatorname{Mark}(\checkmark)$ against the correct answer The value of $\left(\frac{3}{4}\right)^{-3}$ is (a) $\frac{-27}{64}$ (b) $\frac{64}{27}$ (c) $\frac{-9}{4}$ (d) $\frac{27}{64}$ Solution: (b) $\frac{64}{27}$ $\left(\frac{3}{4}\right)^{-3}=\left(\frac{4}{3}\right)^{3}=\frac{4^{3}}{3^{3}}=\frac{64}{27}$...
Read More →The function f (x) = |cos x| is
Question: The functionf(x) = |cosx| is(a) differentiable atx= (2n+ 1)/2,nZ(b) continuous but not differentiable atx= (2n+ 1) /2,nZ(c) neither differentiable nor continuous atx=nZ(d) none of these Solution: (b) continuous but not differentiable at $x=(2 n+1) \pi / 2, n \in Z$ We have, $f(x)=|\cos x|$ $\Rightarrow f(x)=\left\{\begin{array}{rc}\cos x, 2 n \pi \leq x(4 n+1) \frac{\pi}{2} \\ 0, x=(4 n+1) \frac{\pi}{2} \\ -\cos x, (4 n+1) \frac{\pi}{2}x(4 n+3) \frac{\pi}{2} \\ 0, x=(4 n+3) \frac{\pi}{...
Read More →Mark (✓) against the correct answer
Question: Mark (✓) against the correct answer The value of (3)3is (a) 27 (b) 9 (C) $\frac{-1}{27}$ (d) $\frac{1}{27}$ Solution: (c) $\frac{-1}{27}$ $(-3)^{-3}=\left(\frac{1}{-3}\right)^{3}$ $=\frac{1^{3}}{-3^{3}}$ $=\frac{1}{-27}$ $=\frac{1 \times-1}{-27 \times-1}$ $=\frac{-1}{27}$...
Read More →If angle between two tangents drawn
Question: If angle between two tangents drawn from a point P to a circle of radius a and centre 0 is 60, then OP = a3. Solution: True From point $P$, two tangents are drawn. Given, $O T=a$ Also line $O P$ bisects the $\angle R P T$. $\angle T P O=\angle R P O=30^{\circ}$ Also, $O T \perp P T$ In right angled $\triangle O T P$, $\sin 30^{\circ}=\frac{O T}{O P}$ $\Rightarrow$ $\frac{1}{2}=\frac{a}{O P}$ $\Rightarrow$ $O P=2 a$...
Read More →Express each of the following in standard form:
Question: Express each of the following in standard form: (i) 345 (ii) 180000 (iii) 0.000003 (iv) 0.000027 Solution: (i) $345=3.45 \times 100=3.45 \times 10^{2}$ (ii) $180000=18 \times 10000=18 \times 10^{4}=1.8 \times 10 \times 10^{4}=1.8 \times 10^{(1+4)}=1.8 \times 10^{5}$ (iii) $0.000003=\frac{3}{1000000}=3 \times 10^{-6}$ (iv) $0.000027=\frac{27}{100000}=\frac{27}{10^{6}}=\frac{2.7 \times 10}{10^{6}}=2.7 \times 10^{(1-6)}=2.7 \times 10^{-5}$...
Read More →If angle between two tangents drawn
Question: If angle between two tangents drawn from a point Pto a circle of radius a and centre 0 is 90, then OP = a 2. Solution: True From point $P$, two tangents are drawn. Given, $O T=a$ Also, line OP bisects the $\angle R P T$. $\therefore$ $\angle T P O=\angle R P O=45^{\circ}$ Also, $O T \perp P T$ In right angied $\triangle O T P$. $\sin 45^{\circ}=\frac{O T}{O P}$ $\Rightarrow$ $\frac{1}{\sqrt{2}}=\frac{a}{O P} \Rightarrow O P=a \sqrt{2}$...
Read More →Solve this
Question: If $A \subseteq B$, prove that $A \times C=B \times C$ Solution: Given: $A \subseteq B$ Then, A = B at some value Multiplying by C both sides, we get, A C = B C Hence, Proved....
Read More →By what number should
Question: By what number should (3)1be multiplied so that the product becomes 61? Solution: Let the number be $x$. $\therefore(-3)^{-1} \times x=(6)^{-1}$ $\Rightarrow \frac{1}{-3} \times x=\frac{1}{6}$ $\Rightarrow \frac{1 \times-1}{-3 \times-1} \times x=\frac{1}{6}$ $\therefore \frac{-x}{3}=\frac{1}{6}$ On cross multiplying: $-x \times 6=1 \times 3$ $\Rightarrow-6 x=3$ $\Rightarrow 6 x=-3$ $\therefore x=\frac{-3}{6}=\frac{-1}{2}$...
Read More →If A = {3, 4}, B = {4, 5} and C = {5, 6}, find A × (B × C).
Question: If A = {3, 4}, B = {4, 5} and C = {5, 6}, find A (B C). Solution: A = {3, 4}, B = {4, 5} and C = {5, 6} B C = {(4, 5), (4, 6), (5, 5), (5, 6)} A (B C) = {(3, 4, 5), (3, 4, 6), (3, 5, 5), (3, 5, 6), (4, 4, 5), (4, 4, 6), (4, 5, 5), (4, 5, 6)}...
Read More →By what number should
Question: By what number should $\left(\frac{-2}{3}\right)^{-3}$ be divided so that the quotient is $\left(\frac{4}{9}\right)^{-2} ?$ Solution: Let the number be $x$. $\therefore\left(\frac{-2}{3}\right)^{-3} \div x=\left(\frac{4}{9}\right)^{-2}$ $\Rightarrow\left(\frac{3}{-2}\right)^{3} \div x=\left(\frac{9}{4}\right)^{2}$ $\Rightarrow \frac{\left(\frac{3}{-2}\right)^{3}}{x}=\left(\frac{9}{4}\right)^{2}$ $\Rightarrow \frac{3^{3}}{\frac{-2^{3}}{x}}=\frac{9^{2}}{4^{2}}$ $\Rightarrow x=\frac{\left...
Read More →The angle between two tangents
Question: The angle between two tangents to a circle may be 0. Solution: True This may be possible only when both tangent lines coincide or are parallel to each other....
Read More →If A = {2, 3, 4} and B = {4, 5}
Question: If A = {2, 3, 4} and B = {4, 5}, draw an arrow diagram represent (A B}. Solution:...
Read More →Simplify:
Question: Simplify: $\left(3^{-1}+6^{-1}\right) \div\left(\frac{3}{4}\right)^{-1}$. Solution: $\left(3^{-1}+6^{-1}\right) \div\left(\frac{3}{4}\right)^{-1}$ $=\left(\frac{1}{3}+\frac{1}{6}\right) \div\left(\frac{4}{3}\right)^{1}$ $=\left(\left[\frac{1 \times 2}{3 \times 2}\right]+\left[\frac{1 \times 1}{6 \times 1}\right]\right) \div\left(\frac{4}{3}\right)$ $=\left(\frac{2+1}{6}\right) \div\left(\frac{4}{3}\right)$ $=\left(\frac{3}{6}\right) \div\left(\frac{4}{3}\right)$ $=\left(\frac{1}{2}\rig...
Read More →The function f (x) = 1 + |cos x| is
Question: The function $f(x)=1+|\cos x|$ is (a) continuous no where (b) continuous everywhere (c) not differentiable at $x=0$ (d) not differentiable at $x=n \pi, n \in Z$ Solution: (b) continuous everywhere Graph of the function $f(x)=1+|\cos x|$ is as shown below: From the graph, we can see that $f(x)$ is everywhere continuous but not differentiable at $x=(2 n+1) \frac{\pi}{2}, n \in Z$...
Read More →Solve this
Question: If $A=\{1,2\}$, find $A \times A \times A$. Solution: A = {1, 2} $A \times A=\{1,2\} \times\{1,2\}=\{(1,1),(1,2),(2,1),(2,2)\}$ $A \times A \times A=\{1,2\} \times\{(1,1),(1,2),(2,1),(2,2)\}$ Therefore $A \times A \times A=\{(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)\}$...
Read More →The length of tangent from an
Question: The length of tangent from an external point P on a circle with centre 0 is always less than OP. Solution: True $P T$ is a tangent drawn from external point $P$. Join $O T$. So, OPT is a right angled triangle formed. In right angled triangle, hypotenuse is always greater than anv of the two sides of the triangle. $\therefore$$Q PP I$ or $P TO P$...
Read More →Evaluate:
Question: Evaluate: $\left\{\left(\frac{-2}{3}\right)^{3}\right\}^{-2}$ Solution: $\left\{\left(\frac{-2}{3}\right)^{3}\right\}^{-2}=\left(\frac{-2}{3}\right)^{-6}=\left(\frac{3}{-2}\right)^{6}=\frac{3^{6}}{-2^{6}}=\frac{729}{64}$...
Read More →Let f (x) = |cos x|. Then,
Question: Letf(x) = |cosx|. 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: (c) $f(x)$ is everywhere continuous but not differentiable at $x=(2 n+1) \frac{\pi}{2}, n \in Z$. We have, $f(x)=|\cos x|$ $\Rightarrow f(x)=\left\{\begin{array}{rc}\cos x, 2 n \pi \leq x(4 n+1) \frac{\pi}{2} \\ 0, x=(4...
Read More →Evaluate:
Question: Evaluate: (i) $3^{-4}$ (ii) $(-4)^{3}$ (iii) $\left(\frac{3}{4}\right)^{-2}$ (iv) $\left(\frac{-2}{3}\right)^{-5}$ (v) $\left(\frac{5}{7}\right)^{0}$ Solution: (i) $3^{-4}=\frac{1}{3^{4}}=\frac{1}{81}$ (ii) $(-4)^{3}=(-1)^{3} \times(4)^{3}=-1 \times 64=-64$ (iii) $\left(\frac{3}{4}\right)^{-2}=\left(\frac{4}{3}\right)^{2}=\frac{4^{2}}{3^{2}}=\frac{16}{9}$ (iv) $\left(\frac{-2}{3}\right)^{-5}=\left(\frac{3}{-2}\right)^{5}=\frac{3^{5}}{-2^{5}}=\frac{243}{-32}=\frac{243 \times-1}{-32 \tim...
Read More →Evaluate:
Question: Evaluate: (i) $3^{-4}$ (ii) $(-4)^{3}$ (iii) $\left(\frac{3}{4}\right)^{-2}$ (iv) $\left(\frac{-2}{3}\right)^{-5}$ (v) $\left(\frac{5}{7}\right)^{0}$ Solution: (i) $3^{-4}=\frac{1}{3^{4}}=\frac{1}{81}$ (ii) $(-4)^{3}=(-1)^{3} \times(4)^{3}=-1 \times 64=-64$ (iii) $\left(\frac{3}{4}\right)^{-2}=\left(\frac{4}{3}\right)^{2}=\frac{4^{2}}{3^{2}}=\frac{16}{9}$ (iv) $\left(\frac{-2}{3}\right)^{-5}=\left(\frac{3}{-2}\right)^{5}=\frac{3^{5}}{-2^{5}}=\frac{243}{-32}=\frac{243 \times-1}{-32 \tim...
Read More →Let A and B be two sets such that
Question: Let $A$ and $B$ be two sets such that $n(A)=5, n(B)=3$ and $n(A \cap B)=2$. (i) $n(A \cup B)$ (ii) $n(A \times B)$ (iii) $n(A \times B) \cap(B \times A)$ Solution: (i) $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $=5+3-2$ $=6$ (ii) $n(A \times B)=n(A) \times n(B)$ $=5 \times 3$ $=15$ (iii) $n(A \times B) \cap(B \times A)=n(A \times B)+n(B \times A)-$...
Read More →The length of tangent from an external
Question: The length of tangent from an external point P on a circle is always greater than the radius of the circle. Solution: False Because the length of tangent from an external point P on a circle may or may not be greater than the radius of the circle....
Read More →If a chord AB subtends an angle of 60°
Question: If a chord AB subtends an angle of 60 at the centre of a circle, then angle between the tangents at A and B is also 60. Solution: False Since a chord AB subtends an angle of 60 at the centre of a circle. i.e. $\quad \angle A O B=60^{\circ}$ As $O A=O B=$ Radius of the circle $\therefore \quad \angle O A B=\angle O B A=60^{\circ}$ The tangent at points $A$ and $B$ is drawn, which intersect at $C$. We know, $O A \perp A C$ and $O B \perp B C$. $\therefore \quad \angle O A C=90^{\circ}, \...
Read More →Tick (✓) the correct answer
Question: Tick (✓) the correct answer 0.000367 104in usual form is (a) 3.67 (b) 36.7 (c) 0.367 (d) 0.0367 Solution: (a) 3.67 $0.000367 \times 10^{4}=\frac{367}{10^{6}} \times 10^{4}=367 \times 10^{(4-6)}=367 \times 10^{-2}=\frac{367}{10^{2}}=\frac{367}{100}=3.67$...
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