ICSE-X-Mathematics
Previous Year Paper year:2020
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- #Section : A[40 Marks]
Answer
all questions from this Section.)
- #1
- #1-a [3]Solve the following quadratic equation:
``x^2 - 7x + 3 = 0 ``
Give your answer correct to two decimal places.Ans : Given ``x^2 - 7x + 3 = 0 ``
Comparing it with ``ax^2 + bx + c = 0 `` we get , ``a = 1, b = -7 `` and ``c = 3 ``
We know, ``x = `` ``\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ``
``x = `` ``\frac{7 \pm \sqrt{49-12}}{2} ``
``\Rightarrow x= `` ``\frac{7 \pm \sqrt{37}}{2} ``
``\Rightarrow x = `` ``\frac{7 \pm 6.08}{2} ``
Therefore ``x = `` ``\frac{7+6.08}{2} `` ``= 6.54 `` or
``x = `` ``\frac{7 - 6.08}{2} `` ``= 0.46 ``
Hence roots are ``6.54 `` and ``0.46 ``
- #1-b [3]Given ``A = \begin{bmatrix} x & 3 \\ y & 3 \end{bmatrix} ``
If ``A^2= 3 I ``, where ``I `` is the identity matrix of Order ``2 ``, find ``x `` and ``y ``.Ans : Given: ``A = \begin{bmatrix} x & 3 \\ y & 3 \end{bmatrix} `` and ``A^2 = 3I ``
``A^2 = \begin{bmatrix} x & 3 \\ y & 3 \end{bmatrix} \times \begin{bmatrix} x & 3 \\ y & 3 \end{bmatrix} ``
``\Rightarrow A^2 = \begin{bmatrix} x^2 + 3y & 3x+9 \\ xy+3y & 3y+9 \end{bmatrix} ``
``3I = 3 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} ``
Comparing we get
``\begin{bmatrix} x^2 + 3y & 3x+9 \\ xy+3y & 3y+9 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} ``
``\Rightarrow 3x + 9 = 0 \Rightarrow x = - 3 ``
Similarly, ``3y + 9 = 3 \Rightarrow y = - 2 ``
- #1-c [4]Using ruler and compass construct a triangle ``ABC `` where ``AB = 3 `` cm, ``BC = 4 `` cm and ``\angle ABC = 90^{\circ} ``. Hence construct a circle circumscribing the triangle ``ABC ``. Measure and write down the radius of the circle.Ans : ``BC = 4 `` cm and ``AB = 3 `` cm
Radius of the circle ``= 2.5 `` cm
- #2
- #2-a [3]Use factor theorem to factorize ``6x^3 + 17x^2 + 4x - 12 `` completely.Ans : Let ``f(x) = 6x^3 + 17x^2 + 4x - 12 ``
By trial and error, let ``x = -2 ``
Therefore ``f(-2) = 6(-8) + 17(4) + 4(-2) - 12 = -48 + 68 - 8 - 12 = 0 ``
Hence we can say that ``(x+2) `` is a factor of ``f(x) ``
Now,
``\begin{array}{r l l} x+2 ) & \overline{6x^3+ 17x^2 + 4x - 12} & (6x^2+5x-6 \\ (-) & 6x^3+12x^2 & \\ \hline & \hspace{1.0cm} 5x^2 + 4x & \\ (-) & \hspace{1.0cm} 5x^2 + 10x & \\ \hline & \hspace{2.0cm} -6x-12 & \\ (-) & \hspace{2.0cm} -6x - 12 & \\ \hline & \hspace{3.0cm} 0 & \end{array} ``
Now factorize
``6x^2+5x-6 = 6x^2 +9x - 4x - 6 = 3x( 2x+3) - 2 ( 2x+3) = (3x-2)(2x+3) ``
Therefore complete factorization is ``f(x) = (x+2)(3x-4)(2x+3) ``
- #2-b [3]Solve the following in-equation and represent the solution set on the number line.
``\frac{3x}{5} `` ``< x+4 \leq `` ``\frac{x}{5} `` ``, x \in R ``Ans : ``\frac{3x}{5} `` ``< x+4 \leq `` ``\frac{x}{5} `` ``, x \in R ``
First solve: ``\frac{3x}{5} `` ``+ 2 < x+4 ``
``\Rightarrow 3x + 10 < 5x + 20 ``
``\Rightarrow -10 < 2x ``
``\Rightarrow -5 < x ``
Also solve ``x+4 \leq `` ``\frac{x}{2} `` ``+ 5 ``
``\Rightarrow 2x + 8 \leq x + 10 ``
``\Rightarrow x \leq 2 ``
``\therefore -5 < x \leq 2, x \in R ``
Hence the solution set ``S = \{ x : x \in R, -5 < x \leq 2 \} ``
- #2-c [4]Draw a histogram for the given data using graph paper.
Weekly Wages (in Rs) No. of People 3000-4000 4 4000-5000 9 5000-6000 18 6000-7000 6 7000-8000 7 8000-9000 2 9000-10000 4 Ans :
- #3
- #3-aIn the figure given below, ``O `` is the center of the circle and ``AB `` is a diameter, if ``AC = BD `` and ``\angle ADC = 72^{\circ} ``, find
[3]
i) ``\angle ABC ``
ii) ``\angle BAD ``
iii) ``\angle ABD ``Ans : i) We know, The angle which an arc of a circle subtends at the center is double that which it subtends on any point of the part of the circumference.
Therefore ``\angle ABC = `` ``\frac{1}{2} `` ``\angle AOC = `` ``\frac{1}{2} `` ``(72) = 36^{\circ} ``
ii) Given ``AC = BD ``
The chord of the same length subtend the same angle
``\therefore \angle BAD = 36^{\circ} ``
iii) ``\angle ABD = 90^{\circ} `` (semi circular angles are ``90^{\circ} ``)
In ``\triangle ABD, \angle ABD = 180^{\circ} - 36^{\circ} - 90^{\circ} = 54^{\circ} ``
- #3-b [3]Prove that
``\frac{\sin A}{1 + \cot A} `` ``- `` ``\frac{\cos A}{1+\tan A} `` ``= \sin A - \cos A ``Ans : LHS ``= `` ``\frac{\sin A}{1 + \cot A} `` ``- `` ``\frac{\cos A}{1 + \tan A} ``
``= `` ``\frac{\sin^2 A}{\sin A + \cos A} `` ``- `` ``\frac{\cos^2 A}{\cos A + \sin A} ``
``= `` ``\frac{\sin^2 A - \cos^2 A}{\sin A + \cos A} ``
``= `` ``\frac{(\sin A - \cos A )(\sin A + \cos A )}{\sin A + \cos A } ``
``= \sin A - \cos A = `` RHS Hence proved.
- #3-c [4]In what ratio is the line joining ``P(5, 3) `` and ``Q( -5, 3) `` is divided by y-axis. Also find the coordinates of the point of intersection.Ans :
Let the point of intersection be ``(0, b) ``
Let the ratio of in which the line segment gets divided be ``k:1 ``
Using section formula
``x = `` ``\frac{mx_2+nx_1}{m+n} `` and ``y = `` ``\frac{my_2+ny_1}{m+n} ``
``\therefore 0 = `` ``\frac{-5k+5}{k+1} `` ``\Rightarrow -5k + 5 = 0 \Rightarrow k = 1 ``
Hence the ratio is ``1:1 ``
Therefore ``b = `` ``\frac{3+3}{1+1} `` ``= 3 ``
Therefore the coordinates of the point of intersection is ``(0, 3) ``
``\\ ``
- #4