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Question: Let $f(x)=\left|\begin{array}{ccc}\cos x x 1 \\ 2 \sin x x 2 x \\ \sin x x x\end{array}\right|$, then $\lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}$ is equal to (a) 0 (b) $-1$ (c) 2 (d) 3 Solution: $f(x)=\left|\begin{array}{ccc}\cos x x 1 \\ 2 \sin x x 2 x \\ \sin x x x\end{array}\right|$ $=\left|\begin{array}{lll}\cos x x 1 \\ \sin x 0 x \\ \sin x x x\end{array}\right|$ $\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{3}\right]$ $=\left|\begin{array}{ccc}\cos x x 1 \\ \sin x 0...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:9z2x2+ 4xy 4y2 Solution: $9 z^{2}-x^{2}+4 x y-4 y^{2}$ $=9 z^{2}-\left(x^{2}-4 x y+4 y^{2}\right)$ $=9 z^{2}-\left[x^{2}-2 \times x \times 2 y+(2 y)^{2}\right]$ $=(3 z)^{2}-(x-2 y)^{2}$ $=[3 z-(x-2 y)][3 z+(x-2 y)]$ $=(3 z-x+2 y)(3 z+x-2 y)$ $=(x-2 y+3 z)(-x+2 y+3 z)$...
Read More →The length of the diagonal of a square is 24 cm. Find its area.
Question: The length of the diagonal of a square is 24 cm. Find its area. Solution: The diagonal of a square forms the hypotenuse of an isosceles right triangle. The other two sides are the sides of the square of lengthacm. Using Pythagoras' theorem, we have: Diagonal $^{2}=a^{2}+a^{2}=2 a^{2}$ $\Rightarrow$ Diagonal $=\sqrt{2} a$ Diagonal of the square $=\sqrt{2} a$ $\Rightarrow 24=\sqrt{2} a$ $\Rightarrow a=\frac{24}{\sqrt{2}}$ Area of the square $=\operatorname{Side}^{2}=\left(\frac{24}{\sqrt...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:a2+ 2ab+b2 16 Solution: $a^{2}+2 a b+b^{2}-16$ $=a^{2}+2 \times a \times b+b^{2}-16$ $=(a+b)^{2}-4^{2}$ $=(a+b-4)(a+b+4)$...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:36a2+ 36a+ 9 Solution: $36 a^{2}+36 a+9$ $=9\left(4 \mathrm{a}^{2}+4 \mathrm{a}+1\right)$ $=9\left\{(2 \mathrm{a})^{2}+2 \times 2 \mathrm{a} \times 1+1^{2}\right\}$ $=9(2 \mathrm{a}+1)^{2}$ $=9(2 \mathrm{a}+1)(2 \mathrm{a}+1)$...
Read More →The longer side of a rectangular hall is 24 m and the length of its diagonal is 26 m.
Question: The longer side of a rectangular hall is 24 m and the length of its diagonal is 26 m. Find the area of the hall. Solution: Let the rectangleABCDrepresent the hall. Using the Pythagorean theorem in the right-angled triangle ABC, we have: Diagonal $^{2}=$ Length $^{2}+$ Breadth $^{2}$ $\Rightarrow$ Breadth $=\sqrt{\text { Diagonal }^{2}-\text { Length }^{2}}$ $=\sqrt{26^{2}-24^{2}}$ $=\sqrt{676-576}$ $=\sqrt{100}$ $=10 \mathrm{~m}$ $\therefore$ Area of the hall $=$ Length $\times$ Breadt...
Read More →The longer side of a rectangular hall is 24 m and the length of its diagonal is 26 m.
Question: The longer side of a rectangular hall is 24 m and the length of its diagonal is 26 m. Find the area of the hall. Solution: Let the rectangleABCDrepresent the hall. Using the Pythagorean theorem in the right-angled triangle ABC, we have: Diagonal $^{2}=$ Length $^{2}+$ Breadth $^{2}$ $\Rightarrow$ Breadth $=\sqrt{\text { Diagonal }^{2}-\text { Length }^{2}}$ $=\sqrt{26^{2}-24^{2}}$ $=\sqrt{676-576}$ $=\sqrt{100}$ $=10 \mathrm{~m}$ $\therefore$ Area of the hall $=$ Length $\times$ Breadt...
Read More →The value of the determinant
Question: The value of the determinant $\left|\begin{array}{ccc}x x+y x+2 y \\ x+2 y x x+y \\ x+y x+2 y x\end{array}\right|$ is (a) $9 x^{2}(x+y)$ (b) $9 y^{2}(x+y)$ (c) $3 y^{2}(x+y)$ (d) $7 x^{2}(x+y)$ Solution: $\left|\begin{array}{ccc}x x+y x+2 y \\ x+2 y x x+y \\ x+y x+2 y x\end{array}\right|$ $=\left|\begin{array}{ccc}-2 y y y \\ x+2 y x x+y \\ -y 2 y -y\end{array}\right|$ [Applying $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{3} \rightarrow R_{3}-R_{2}$ ] $=y^{2}\left|\begin{array}{ccc}-2 1 1 ...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:p2q2 6pqr+ 9r2 Solution: $p^{2} q^{2}-6 p q r+9 r^{2}$ $=(p q)^{2}-2 \times p q \times 3 r+(3 r)^{2}$ $=(p q-3 r)^{2}$ $=(p q-3 r)(p q-3 r)$...
Read More →Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.
Question: Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm. Solution: Area of an isosceles triangle: $=\frac{1}{4} b \sqrt{4 a^{2}-b^{2}}$ (Where $a$ is the length of the equal sides and $b$ is the base) $=\frac{1}{4} \times 24 \sqrt{4(13)^{2}-24^{2}}$ $=6 \sqrt{4 \times 169-576}$ $=6 \sqrt{676-576}$ $=6 \sqrt{100}$ $=6 \times 10$ $=60 \mathrm{~cm}^{2}$...
Read More →Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.
Question: Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm. Solution: Area of an isosceles triangle: $=\frac{1}{4} b \sqrt{4 a^{2}-b^{2}}$ (Where $a$ is the length of the equal sides and $b$ is the base) $=\frac{1}{4} \times 24 \sqrt{4(13)^{2}-24^{2}}$ $=6 \sqrt{4 \times 169-576}$ $=6 \sqrt{676-576}$ $=6 \sqrt{100}$ $=6 \times 10$ $=60 \mathrm{~cm}^{2}$...
Read More →O is a point in the interior of a square ABCD
Question: O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ΔOCD is an isosceles triangle. Solution: Given $O$ is a point in the interior of a square $A B C D$ such that $\triangle O A B$ is an equilateral triangle. Construction Join $O C$ and $O D$. To show $\triangle O C D$ is an isosceles triangle. Proof Since, $A O B$ is an equilateral triangle. $\therefore \quad \angle O A B=\angle O B A=60^{\circ}$ $\ldots(i)$ Also, $\angle D A B=\angle C B A...
Read More →The maximum value
Question: The maximum value of $\Delta=\left|\begin{array}{ccc}1 1 1 \\ 1 1+\sin \theta 1 \\ 1+\cos \theta 1 1\end{array}\right|$ is ( $\theta$ is real) (a) $\frac{1}{2}$ (b) $\frac{\sqrt{3}}{2}$ (c) $\sqrt{2}$ (d) $-\frac{\sqrt{3}}{2}$ Solution: $\Delta=\left|\begin{array}{ccc}1 1 1 \\ 1 1+\sin \theta 1 \\ 1+\cos \theta 1 1\end{array}\right|$ $=\left|\begin{array}{ccc}1 1 1 \\ 0 \sin \theta 0 \\ \cos \theta 0 0\end{array}\right|$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow ...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:9a2 24ab+ 16b2 Solution: $9 a^{2}-24 a b+16 b^{2}$ $=(3 a)^{2}-2 \times 3 a \times 4 b+(4 b)^{2}$ $=(3 a-4 b)^{2}$ $=(3 a-4 b)(3 a-4 b)$...
Read More →Find the area of an equilateral triangle having each side of length 10 cm.
Question: Find the area of an equilateral triangle having each side of length 10 cm. Solution: Given:Side of the equilateral triangle = 10 cmThus, we have: Area of the equilateral triangle $=\frac{\sqrt{3}}{4}$ side $^{2}$ $=\frac{\sqrt{3}}{4} \times 10 \times 10$ $=25 \times 1.732$ $=43.3 \mathrm{~cm}^{2}$...
Read More →Find the area of an equilateral triangle having each side of length 10 cm.
Question: Find the area of an equilateral triangle having each side of length 10 cm. Solution: Given:Side of the equilateral triangle = 10 cmThus, we have: Area of the equilateral triangle $=\frac{\sqrt{3}}{4}$ side $^{2}$ $=\frac{\sqrt{3}}{4} \times 10 \times 10$ $=25 \times 1.732$ $=43.3 \mathrm{~cm}^{2}$...
Read More →Factorize each of the following algebraic expression:
Question: Factorize each of the following algebraic expression:4x2+ 12xy+9y2 Solution: $4 x^{2}+12 x y+9 y^{2}$ $=(2 x)^{2}+2 \times 2 x \times 3 y+(3 y)^{2}$ $=(2 x+3 y)^{2}$ $=(2 x+3 y)(2 x+3 y)$...
Read More →The parallel sides of a trapezium are 9.7 cm and 6.3 cm, and the distance between them is 6.5 cm.
Question: The parallel sides of a trapezium are 9.7 cm and 6.3 cm, and the distance between them is 6.5 cm. The area of the trapezium is(a) 104 cm2(b) 78 cm2(c) 52 cm2(d) 65 cm2 Solution: (c) 52 cm2 Area of trapezium $=\frac{1}{2}$ (Sum of parallel sides) $\times$ Distance between them $=\frac{1}{2} \times(9.7+6.3) \times 6.5$ $=8 \times 6.5$ $=52.0 \mathrm{~cm}^{2}$...
Read More →If x, y ∈ R, then the determinant
Question: If $x, y \in \mathbb{R}$, then the determinant $\Delta=\left|\begin{array}{ccc}\cos x -\sin x 1 \\ \sin x \cos x 1 \\ \cos (x+y) -\sin (x+y) 0\end{array}\right|$ lies in the interval (a) $[-\sqrt{2}, \sqrt{2}]$ (b) $[-1,1]$ (c) $[-\sqrt{2}, 1]$ (d) $[-1,-\sqrt{2}]$ Solution: $\Delta=\left|\begin{array}{ccc}\cos x -\sin x 1 \\ \sin x \cos x 1 \\ \cos (x+y) -\sin (x+y) 0\end{array}\right|$ $=\left|\begin{array}{ccc}\cos x -\sin x 1 \\ \sin x \cos x 1 \\ 0 0 \sin y-\cos y\end{array}\right...
Read More →The sides of a triangle are in the ratio 12 : 14 : 25 and its perimeter is 25.5 cm.
Question: The sides of a triangle are in the ratio 12 : 14 : 25 and its perimeter is 25.5 cm. The largest side of the triangle is(a) 7 cm(b) 14 cm(c) 12.5 cm(d) 18 cm Solution: (c) 12.5 cmLet the sides of the triangle be 12xcm,14xcm and25xcm.Thus, we have: Perimeter $=12 x+14 x+25 x$ $\Rightarrow 25.5=51 x$ $\Rightarrow x=\frac{25.5}{51}=0.5$ $\therefore$ Greatest side of the triangle $=25 x=25 \times 0.5=12.5 \mathrm{~cm}$...
Read More →Factorize each of the following expression:
Question: Factorize each of the following expression:18a2x2 32 Solution: $18 a^{2} x^{2}-32$ $=2\left(9 a^{2} x^{2}-16\right)$ $=2\left[(3 a x)^{2}-4^{2}\right]$ $=2(3 a x-4)(3 a x+4)$...
Read More →In the given figure ABCD is a trapezium in which AB = 40 m,
Question: In the given figureABCDis a trapezium in whichAB= 40 m,BC= 15 m,CD= 28 m,AD= 9 m andCEAB.Area of trap.ABCDis (a) 306 m2(b) 316 m2(c) 296 m2(d) 284 m2 Solution: (a) 306 m2In the given figure,AECDis a rectangle.LengthAE= LengthCD= 28 m Now $B E=A B-A E=40-28=12 \mathrm{~m}$ Also,AD=CE= 9 m Area of trapezium $=\frac{1}{2} \times$ Sum of parallel sides $\times$ Distance between them $=\frac{1}{2} \times(D C+A B) \times C E$ $=\frac{1}{2} \times(28+40) \times 9$ $=\frac{1}{2} \times 68 \tim...
Read More →Factorize each of the following expression:
Question: Factorize each of the following expression:x3x Solution: $x^{3}-x$ $=x\left(x^{2}-1\right)$ $=x(x-1)(x+1)$...
Read More →In the given figure ABCD is a trapezium in which AB = 40 m,
Question: In the given figureABCDis a trapezium in whichAB= 40 m,BC= 15 m,CD= 28 m,AD= 9 m andCEAB.Area of trap.ABCDis (a) 306 m2(b) 316 m2(c) 296 m2(d) 284 m2 Solution: (a) 306 m2In the given figure,AECDis a rectangle.LengthAE= LengthCD= 28 m Now $B E=A B-A E=40-28=12 \mathrm{~m}$ Also,AD=CE= 9 m Area of trapezium $=\frac{1}{2} \times$ Sum of parallel sides $\times$ Distance between them $=\frac{1}{2} \times(D C+A B) \times C E$ $=\frac{1}{2} \times(28+40) \times 9$ $=\frac{1}{2} \times 68 \tim...
Read More →Factorize each of the following expression:
Question: Factorize each of the following expression:xy9yx9 Solution: $x y^{9}-y x^{9}$ $=x y\left(y^{8}-x^{8}\right)$ $=x y\left[\left(y^{4}\right)^{2}-\left(x^{4}\right)^{2}\right]$ $=x y\left(y^{4}+x^{4}\right)\left(y^{4}-x^{4}\right)$ $=x y\left(y^{4}+x^{4}\right)\left[\left(y^{2}\right)^{2}-\left(x^{2}\right)^{2}\right]$ $=x y\left(y^{4}+x^{4}\right)\left(y^{2}+x^{2}\right)\left(y^{2}-x^{2}\right)$ $=x y\left(y^{4}+x^{4}\right)\left(y^{2}+x^{2}\right)(y+x)(y-x)$...
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