11th Class Math Paper 2025 Federal Board
11th Class Math Paper 2025 Federal Board. Class 11th Math Paper 2025 Federal Board
Q2 :- Solve the following parts.
i):- Solve the quadratic equation
$$3z^2+2z+2=0\;\;,\;\;Z\in\mathbb{C}$$ by completing square method.
$$3z^2+2z+2=0\;\\\frac{3z^2+2z+2}3=\frac03$$
$$\frac{3z^2}3+\frac{2z}3+\frac23=0\\z^2+\frac{2z}3+\frac23=0$$
$$(z^2)\;+\;2\;(z)\;(\frac13)\;\\+{(\frac13)}^2-{(\frac13)}^2+\frac23=0$$
$${(z+\frac13)}^2-\frac19+\frac23=0\\{(z+\frac13)}^2=\frac19-\frac23$$
$$=\frac{1-6}9\\\sqrt{{(z+\frac13)}^2\;}=\pm\sqrt{\frac{-5}9}$$
$$z=\frac{-1}3\pm\frac{\sqrt{5i}}3\\=\frac{-1\pm\sqrt{5i}}3$$
i):- A triangular park has its three corners located at points A(2, 3, 1) , B(5, 7, 2) , C(1, 4, 3). Find area of the park using cross product of vector.
$$\overrightarrow{AB}\;=(3,\;4,\;1)\\\overrightarrow{AC}=(-1,\;1,\;2)\;$$
$$\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\widehat i&\widehat j&\widehat k\\3&4&1\\-1&1&2\end{vmatrix}$$
$$=\widehat i(8-1)-\widehat j(6+1)+\widehat k(3+4)\\=7\widehat i-7\widehat j+7\widehat k$$
$$\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|\\=\sqrt{{(7)}^2+{(-7)}^2+{(7)}^2}$$
$$=\sqrt{49+49+49}\\=\sqrt{147}\\=7\sqrt3$$
Area of triangle:-
$$=\frac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|\\=\frac12(7\sqrt3)\\=\frac{7\sqrt3}2square\;units$$
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ii):- Verify the Pascal’s Identity.
$$\begin{pmatrix}n\\r\end{pmatrix}+\begin{pmatrix}n\\r-1\end{pmatrix}=\;\begin{pmatrix}n+1\\r\end{pmatrix}$$
$${}^nC_r=\begin{pmatrix}n\\r\end{pmatrix}=\frac{n!}{r!\;(n-r)!}$$
$${}^nC_{r-1}=\begin{pmatrix}n\\r-1\end{pmatrix}=\frac{n!}{(r-1)!\;(n-(r-1))!}$$
$$=\frac{n!}{(r-1)!\;(n-r+1))!}$$
Lets Consider L.H.S
$$L.H.S=\frac{n!}{r!\;(n-r)!}+\frac{n!}{(r-1)!\;(n-r+1)\;)!}$$
$$L.H.S\;=\frac{n!}{r\;(r-1)!\;(n-r)!}\\+\frac{n!}{(r-1)!\;(n-r+1)\;)\;(n-r)!}$$
Lets Consider L.C.M
$$L.H.S\;=\;\frac{r\;(n!)\;+(n-r+1)\;n!}{(r-1)!\;(n-r)!\;r(n-r+1)}$$
$$=n!\;\left\{\frac{r+n-r+1}{(r-1)!\;(n-r)!\;r\;(n-r+1)}\right\}$$
$$=\frac{n!\;(n+1)}{r\;(r-1)!\;\;(n-r+1)\;(n-r)!\;}\\=\frac{(n+1)!}{r!\;\;(n+1-r)!}$$
$$=\;\begin{pmatrix}n+1\\r\end{pmatrix}\\=\;R.H.S$$
ii):- Verify that $$Sin\theta.Sin\;(\frac{\mathrm\pi}3-\theta).\\Sin\;(\frac{\mathrm\pi}3+\theta)=\frac14Sin\;3\theta$$
$$L.H.S=Sin\theta.Sin\;(\frac{\mathrm\pi}3-\theta).Sin\;(\frac{\mathrm\pi}3+\theta)$$
$$=Sin\theta\;(Sin\frac\pi3Cos\theta-Cos\frac{\mathrm\pi}3Sin\theta)\\(Sin\frac{\mathrm\pi}3Cos\theta+Cos\frac{\mathrm\pi}3Sin\theta)$$
$$=Sin\theta\;(\frac{\sqrt3}2Cos\theta-\frac12Sin\theta)\\(\frac{\sqrt3}2Cos\theta+\frac12Sin\theta)$$
$$=Sin\theta\;(\;(\frac{\sqrt3}2Cos^2\theta)-(\frac{\sqrt3}2Sin^2\theta)\;)\;$$
$$=Sin\theta\;(\;(\frac34Cos^2\theta-\frac14Sin^2\theta)$$
$$=Sin\theta\;(\frac{3\;Cos^2\theta-\;Sin^2\theta}4)\\=\frac{Sin\theta}4(\;3(1-Sin^2\theta)-Sin^2\theta\;)$$
$$=\frac{Sin\theta}4(\;3-3Sin^2\theta-Sin^2\theta\;)\\=\frac{Sin\theta}4(\;3-4Sin^2\theta\;)$$
$$=\frac14\;(\;3\;Sin\theta-4\;Sin\theta\;)\\=\frac14\;Sin\;3\theta\\=R.H.S$$
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iii) :- Find the value of P such that $$\widehat i+2\widehat{\;j}+p\widehat k\;,\;3\widehat i+p\widehat j+4\widehat k\\and\;2\widehat i+3\widehat j+4\widehat k$$
are coplaner.
If vector are coplanar, their scalar triple product is zero.
$$\overrightarrow a.\;(\;\widehat b\;\times\;\widehat c\;)=\begin{vmatrix}1&2&\rho\\3&\rho&4\\2&3&4\end{vmatrix}$$
$$=1\begin{vmatrix}\rho&4\\3&4\end{vmatrix}\;-2\begin{vmatrix}3&4\\2&4\end{vmatrix}\;+\rho\begin{vmatrix}3&\rho\\2&3\end{vmatrix}$$
$$=1(4\rho-12)-2(12-8)\;+\rho(9-2\rho)\\$$
$$=4\rho-12-2(4)\;+9\rho-2\rho^2\\=-2\rho^2+13\rho-20\\$$
$$-2\rho^2+13\rho-20=0\\2\rho^2-13\rho+20=0\\$$
$$\rho=\frac{-(-13)\pm\sqrt{{(13)}^2-4(2)(20)}}{2(2)}\\$$
$$\rho=\frac{13\pm\sqrt{169-160}}4\\=\frac{13\pm3}4\\$$
$$\rho=\frac{13+3}4=\frac{16}4=4\\\rho=\frac{13-3}4=\frac{10}4=\frac52\\$$
iii) :- Find the maximum and minimum values of the function
$$f\;(\theta)=\frac1{3+5\;Cos\;(2\theta+\pi)}\\=\frac1{3-5\;Cos\;2\theta}\\$$
$$f\;(\theta)=\frac1{3-5\;Cos\;2\theta}\\Cos\;2\theta\in\;\begin{bmatrix}-1,&1\end{bmatrix}\\$$
$$If\;Cos\;2\theta=\;-1\\$$
$$f(\theta)=\frac1{3-5(-1)}=\frac18\\=0.125\\$$
$$If\;Cos\;2\theta=1\\$$
$$f(\theta)=\frac1{3-5\;(1)}=\frac{-1}2\\=0.5\\$$
$$The\;denominator\;3-5\;Cos\;2\theta=0\\when\;Cos\;2\theta\;=\frac35and\;\frac35is\\between\;-1\;and\;1,\;such\;angle\;exist.\\$$
Therefore No finite maximum value and no finite minimum value will exist
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iv) :- If $$p\;(x)=x^4-6x^3+11x^2-6x\\$$
then,
a) :- Divide p (x) by (x-1)using synthetic division.
b) :- Solve the resulting depressed equation.
$$Q\;(x)=x^3-5x^2+6x\\x^3-5x^2+6x=0\\x\;(x^2-5x+6)=0\\\\$$
$$x\;(x^2-3x-2x+6)=0\\x(x\;(x-3)-2(x-3)\;)=0\\x\;(x-3)\;(x-2)\;=0\\\\$$
$$x=0,\;\\x-3=0\\\;x=3\\x-2=0\\x=2\\$$
$$Roots\;x\;=0\;,\;2,\;3,\;1\\$$
iv) :- A harmonic sequence has 2nd term 1/6 and 4th term 1/12. Find General term of the sequence.
In H.P the reciprocal of the terms form an A.P
2nd term = 1/6
4th term = 1/12
So, their reciprocal are
1/1/6 = 6 , 1/1/12 = 12
Hence in A.P
$$a_2=6\;,\;a_4=12\\a+d=6\;,\;a+3d=12\;eq…(ii)\\a=6-d\;eq…(i)\;,\;\\$$
Putting a = 6 – d in eq…(ii)
6 – d + 3d = 12
2d = 6
d = 3
Putting d = 3 in eq…(i)
a = 6 – 3
a = 3
$$an=a+(n-1)\;d\\=3+(n-1)\;3\\=3+3n-3=3n\\General\;term\;of\;H.P=\frac1{3n}\;\\$$
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v) :- Without drawing graph, find range, amplitude period and frequency of the function.
$$y=\;-4\;Cos(\;7x-\mathrm\pi\;)$$
$$Amplitude=\left|-4\right|=4\\Cos\theta\;\in\left[-1,\;1\right]\\Range=\left[-4,\;4\right]$$
$$-4\;Cos(7x-\mathrm\pi+2\mathrm\pi)\\-4\;\mathrm{Cos}(\;7\mathrm x+\mathrm\pi\;)\\$$
$$\mathrm y=\;\mathrm a\;\mathrm{Cos}(\mathrm{bx}+\mathrm c)\\\mathrm{Period}=\frac{2\mathrm\pi}{\left|\mathrm b\right|}\\$$
$$\mathrm{Hence}\;\mathrm{period}\;\mathrm{of},\;\\-4\;\mathrm{Cos}\;(\;7\mathrm x+\mathrm\pi\;)\\$$
$$\mathrm{Frequency}=\frac7{2\mathrm\pi}\\$$
$$\mathrm{Amplitude}=\left|\mathrm a\right|\\=\left|-4\right|\\=4\\$$
v) :- Find the term containing $$\mathrm x^3\\$$ in the expansion of $${(\mathrm x+2)}^6\\$$
$${\mathrm T}_{\mathrm r+1}={}^{\mathrm n}\mathrm C_{\mathrm r}\;,\;\mathrm a^{\mathrm n-\mathrm r}\;,\;\mathrm b^{\mathrm r}\\$$
$$\mathrm a\;=\;\mathrm x\;,\;\mathrm b\;=\;2\;,\;\mathrm n\;=\;6\\\mathrm n-\mathrm r\;=\;3\;,\;6-\mathrm r\;=\;3\\6-3\;=\;\mathrm r\;,\;\mathrm r\;=\;3\\$$
$${\mathrm T}_{\mathrm r+1}={}^6\mathrm C_3\;\mathrm a^3\;\mathrm b^3\\$$
$${\mathrm T}_4=\frac{6!}{3!\;(\;6-3\;)!}\mathrm x^3.2^3\\=\frac{6!}{3!\;3!}\mathrm x^3.8\\$$
$$=\frac{6\times5\times4\times3!}{3!\times3\times2}\mathrm x^3.8\\=160\;\mathrm x^3$$
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vi) :- In an arithmetic sequence, the 5th term is 15 and the 12th term is 50.
a) :- Find common difference
b) :- First term of the sequence
$$a_{5\;}=15\;,\;a_{12}=50\\a+4d=15\;,\;a+11d\;=50\\a=15-4d\;eq…\;(i)\;,\;a=50-11d\;eq…\;(ii)$$
Comparing eq… (i) and eq… (ii)
!5 – 4d = 50 – 11d
-4d + 11d = 50 – 15
7d = 35
d = 5
Putting d = 5 in eq… (i)
a = 15 – 4 (5)
= 15 – 20
a = -5
vi) :- Verify that
$$\frac{1+Sin\;2\theta+\;Cos\;2\theta}{1+\;Sin\;2\theta-\;Cos\;2\theta}=\;Cot\theta$$
$$L.H.S=\frac{1+Sin\;2\theta+\;Cos\;2\theta}{1+\;Sin\;2\theta-\;Cos\;2\theta}$$
$$=\frac{1+2\;Sin\theta\;Cos\theta+\;Cos^2\theta-Sin^2\theta}{1+\;2\;Sin\theta\;Cos\theta-(Cos^2\theta-Sin^2\theta)}$$
$$=\frac{Cos^2\theta+Sin^2\theta+2\;Sin\theta\;Cos\theta+\;Cos^2\theta-Sin^2\theta}{Cos^2\theta+Sin^2\theta+\;2\;Sin\theta\;Cos\theta-(Cos^2\theta-Sin^2\theta)}$$
$$=\frac{{(Cos\theta+Sin\theta)}^2+\;(Cos\theta+Sin\theta)\;(Cos\theta-Sin\theta)}{{(Cos\theta+Sin\theta)}^2-(Cos\theta+Sin\theta)\;(Cos\theta-Sin\theta)}$$
$$=\frac{(Cos\theta+Sin\theta)(Cos\theta+Sin\theta+Cos\theta-Sin\theta)}{(Cos\theta+Sin\theta)(Cos\theta+Sin\theta-Cos\theta+Sin\theta)}$$
$$=\frac{2\;Cos\theta}{2\;Sin\theta}\\=\;Cot\theta\\=R.H.S$$
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vii) :- Verify that
$$\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$$ =( b-a ) ( c-a ) ( c-b )
$$L.H.S=\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$$
$$=\begin{vmatrix}1&a&a^2\\0&b-a&b^2-a^22\\0&c-a&c^2-a^2\end{vmatrix}\;\\R_2-R_1\\R_3-R_1\\$$
$$Expanding\;through\;C_1\\=-1\;\begin{vmatrix}b-a&(b-a&(b+a)\\c-a&(c-a)&(c+a)\end{vmatrix}\\$$
$$=(b-a)\;\begin{vmatrix}1&b+a\\c-a&(c-a)(c+a)\end{vmatrix}\\\\$$
$$=(b-a)(c-a)\;\begin{vmatrix}1&b+a\\1&(c+a)\end{vmatrix}\\=(b-a)(c-a)(c+a-b-a)\\\\$$
$$=(b-a)(c-a)(c-b)\\=R.H.S\\\\$$
vii) :- If $$\tan\;\theta=\frac34with\;\mathrm\pi<\mathrm\theta<\frac{3\mathrm\pi}2\\$$. Find the exact value of,
a) :- $$Sin\;(\frac\theta2)\;and\;\\$$
b) :- $$Cos\mathit\;\mathit(\frac{\mathit\theta}{\mathit2}\mathit)\mathit\;without\mathit\;using\mathit\;calculater\\$$
Solution:-
$$2Cos^2\theta=1+Cos2\theta\\Cos^2\theta=\frac{1+Cos2\theta}2\\\sqrt{Cos^2\theta}=\;\pm\sqrt{\frac{1+Cos2\theta}2}\\Cos\theta\;=\;\pm\sqrt{\frac{1+Cos2\theta}2}\\\\$$
From this we conclude that
$$Cos\;(\frac\theta2)\;=\;\pm\sqrt{\frac{1+Cos\theta}2}\\\\$$
$$2Sin^2\theta=1-Cos2\theta\\Sin^2\theta=\frac{1-Cos2\theta}2\\\sqrt{Sin^2\theta}=\;\pm\sqrt{\frac{1-Cos2\theta}2}\\Sin\theta\;=\;\pm\sqrt{\frac{1-Cos2\theta}2}\\\\$$
From this we conclude that
$$Sin\;(\frac\theta2)\;=\;\pm\sqrt{\frac{1-Cos\theta}2}\\\\$$
Lets use Pythagorean identity
$$1+\tan^2\theta\;=\;Sec^2\theta\\1+{(\frac34)}^2\;=\;Sec^2\theta\\1+\frac9{16}=\;Sec^2\theta\\\frac{16+9}{16}=\;Sec^2\theta\\\frac{25}{16}=\;Sec^2\theta\\\sqrt{Sec^2\theta\;}=\pm\sqrt{\frac{25}{16}}\\Sec\;\theta\;=\;\pm\frac54\\But\;\theta\;lies\;in\;3rd\;Quadrant\\Hence\;\\Sec\;\theta\;=-\frac54\\So\\Cos\;\theta\;=\;-\frac45\\\\\\$$
We know that
$$Sin\;(\frac\theta2)\;=\;\pm\sqrt{\frac{1-Cos\theta}2}\\\\$$
But
$$\;\mathrm\pi<\mathrm\theta<\frac{3\mathrm\pi}2\\dividing\;by\;2\\\;\frac{\mathrm\pi}2<\frac{\mathrm\theta}2<\frac{3\mathrm\pi}4\\Its\;mean\;\frac{\mathrm\theta}2\;lies\;in\;2nd\;quadrant\\\\$$
$$In\;2nd\;Qudrant\;Sine\;is\;positive\;so\\\\Sin\;(\frac\theta2)\;=\;+\sqrt{\frac{1-Cos\theta}2}\\\\Sin\;(\frac\theta2)=+\sqrt{\frac{1-({\displaystyle\frac{-4}5})}2}\\=\;\sqrt{\frac{\displaystyle\frac{5+4}5}2}\\=\;\sqrt{\frac9{10}}\\=\frac3{\sqrt{10}}\\$$
We know that
$$Cos\;(\frac\theta2)\;=\;\pm\sqrt{\frac{1+Cos\theta}2}\\\\$$
$$In\;2nd\;Qudrant\;Co\sin e\;is\;negative\;so\\\\Cos\;(\frac\theta2)\;=\;-\sqrt{\frac{1+Cos\theta}2}\\\\Cos\;(\frac\theta2)=-\sqrt{\frac{1+({\displaystyle\frac{-4}5})}2}\\=-\;\sqrt{\frac{\displaystyle\frac{5-4}5}2}\\=-\;\sqrt{\frac1{10}}\\=-\frac1{\sqrt{10}}\\$$
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viii) :- A bouncing ball rebounds 80% of its previous height. If the ball dropped from a height of 50 m. Find the total distance travelled by the ball before coming to rest.
$$First\;term\;a_0=50\;m\\ar,\;(ar)r,\;(ar^2),\;(ar^3)r,\;…$$
$$ar,\;(ar)r,\;(ar^2),\;(ar^3)r,\;…\\2\;ar,\;2\;ar^2,\;2ar^3,\;2ar^4,\;…$$
$$Hence\;a_1=2\;ar\\Common\;ratio=r\\=\frac{2\;ar^2}{2ar}=r$$
$$S=\frac{a_1}{1-r}\\=\frac{2\;ar}{1-r}\\=\frac{2\;(50)\;80\%}{1-80\%}$$
$$=\frac{100\;(0.8)\;}{1-0.8}=400\\Total\mathit\;distance\;=50+400\\=\;450\;m\;$$
viii):- How many words can be formed by using the letters from the word “EDUCATION” such that all the vowels are never together.
Total letter = 9
To arrange 9 distinct
Letters = 9!
Vowels = E, U, A, I, O = 5 letters
Consonants = D, C, T, N = 4 letters
Vowels can be arranged in 5! way and consonant can be arranged in 4! ways.
If 5 vowels are together treat them like one block.
$$\underline E\;\underline D\;\underline U\;\underline C\;\underline A\;\underline T\;\underline I\;\underline O\;\underline N\\\underline{vowels}\;\underline D\;\underline C\;\underline T\;\underline N\\1!\;also\;4!$$
These five items can be arranged in 5! ways. Vowels can be arranged in 5! way. Number of arrangements with vowels all together
$$=\;5!\;\times\;5!$$
$$Requirement\;=\;9!-(5!\;\times\;5!)\\=\;348480$$
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ix) :- Use Method of Mathematical induction to prove that
$$1^3+2^3+3^3….n^3={(\frac{n(n+1)}2)}^2\\for\;all\;n\in\mathbb{N},\;n\geq1\\$$
$$Base\;case\;(n=1)\\{(1)}^3={(\frac{1(1+1)}2)}^2\;1={(\frac{1(2)}2)}^2$$
$$1=\;{(1)}^2\\1\;=\;1$$
Statement is true for n=1
Inductive step let the statement is true for n=k
$$1^3+2^3+3^3+…+k^3\\=\;{(\frac{k\;(k+1)}2)}^2$$
We’ll have to show that statement is true for n = k+1
$$1^3+2^3+3^3+…+k^3+{(k+1)}^3\\=\;{(\frac{(k+1)\;(k+2)}2)}^2$$
$$L.H.S=1^3+2^3+3^3+…+k^3+{(k+1)}^3\\=\;{(\frac{k\;(k+1)}2)}^2+{(k+1)}^3$$
$${(k+1)}^{2\;}(\frac{k^2}4+k+1)\\={(k+1)}^2(\frac{k^2+4k+4}4)$$
$$={(k+1)}^{2\;}\frac{{(k+2)}^2}{{(2)}^2}\\={(\frac{(k+1)(k+2)}2)}^2$$
$$R.H.S=>Identity\;holds\;for\;n\geq1\;$$
ix):- Find the rank of the matrix
$$\begin{bmatrix}1&1&0&-2\\2&0&2&2\\4&1&3&1\end{bmatrix}\\$$
$$\begin{bmatrix}1&1&0&-2\\0&-2&2&6\\0&-3&3&9\end{bmatrix}\\R_2-2R_1\\R_3-4R_1$$
$$\begin{bmatrix}1&1&0&-2\\0&1&-1&-3\\0&-3&3&9\end{bmatrix}\\-\frac12R_2$$
$$\begin{bmatrix}1&1&0&-2\\0&-1&-1&-3\\0&0&0&0\end{bmatrix}\\+R_3+3R_2$$
Number of nonzero rows in echelon form is called rank.
Hence Rank = 2
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x):- If
$$Z_1=\;10\;(Cos100+i\;Sin100)\\Z_2=\;5\;(Cos40\;+\;i\;Sin40\;$$
Then find the following in polar form
$$\;a):\;\;Z_1Z_2\;\;and\;\;b):\;\;\frac{Z_1}{Z_2}$$
$$Z_1Z_2=\left\{10\;(Cos\;100+i\;Sin\;100)\;\right\}\\\left\{5\;(Cos\;40+i\;Sin\;40)\;\right\}$$
$$=50\left\{(Cos(100+40)+iSin(100+40)\right\}\\=50\;(\;Cos\;140+i\;Sin\;140\;)\\=50\;Cis\;140$$
$$\frac{Z_1}{Z_2}=\frac{10\;(Cos\;100+i\;Sin\;100)}{5\;(Cos\;40+i\;Sin\;40)}\\=2\;(Cos\;(100-40)\\+i\;Sin\;(100-40)\;)$$
$$=\;2(\;Cos\;60-i\;Sin\;60)\\=2\;Cis\;60$$
x):-A solar panel’s Sunlight reception is modelled as
$$\;P(\theta)\;=\;100\;Cos(3\theta-90)\\with\;0\leq\theta\leq180$$
a):- Find angle for maximum sun light and the maximum sunlight received
b):- Find angle for minimum sun light and the minimum sun light received
$$p\;(\theta)=100\;Cos(30-90^\circ)\\Range\;of\;\cos ine\;is\;\left[-1,\;1\right]\\Cos\;(30-90^\circ)=1\\30-90^\circ=Cos^{-1}(1)$$
$$30-90^\circ=0^\circ\\30\;=\;90^\circ\\\theta\;=\;30^\circ\\p\;(\theta)=100(1)=100$$
$$Maximum\;angle\;=30^\circ\\Maximum\;sunlight=100\\p\;(\theta)=100\;Cos(30-90^\circ)$$
$$Cos(30-90)=-1\\30-90=Cos^{-1}(-1)\\30-90=180\\30=270\\\theta\;=\;90^\circ$$
$$p\;(\theta)=100\;(-1)\\=\;-100\\Minimum\;angle\;=90^\circ\\Minimum\;sunlight=-100$$
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xi):- Use binomial theorem to approximate the value of
$$\sqrt{101}\;\times\;\sqrt{99}$$
$$=\sqrt{101\times99}\\=\sqrt{9999}=\sqrt{10000-1}\\=\sqrt{10000(1-\frac1{10000})}$$
$$=\sqrt{10000}\sqrt{(1-\frac1{10000})}\\=100\;{(1-0.0001)}^\frac12\\(1-0.0001)\frac12=(1+{(-0.0001)\;)}^\frac12$$
$$=1+\frac12(-0.0001)+\frac{{\displaystyle\frac12}({\displaystyle\frac12}-1)}{2!}\\{(-0.0001)}^2+\frac{{\displaystyle\frac12}({\displaystyle\frac12}-1)({\displaystyle\frac12}-2)}{3!}\\{(-0.0001)}^3$$
$$=1-0.00005+(-0.125)(0.00000001)\\+(0.0625)(-0.000000000001)$$
=99.995
xi):- A force F=6i+8j+4k acts on a object. The object moves from A(1,2,3) to the point B(4,6,5). Find displacement vector d and work done by the force F.
$$\overrightarrow d=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$$
= ( 4, 6, 5 ) – ( 1, 2, 3 )
= ( 3, 4, 2 )
$$=\;3\widehat i+4\widehat j+2\widehat k\\w=\overrightarrow F.\overrightarrow d$$
= ( 6, 8, 4 ) . ( 3, 4, 2 )
=18 + 32 + 8
= 58
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xii):- If
$$p(x)\;=\;a\;x^3+\;b\;x^2\;+2x-1$$
leaves remainder 5 when divided by x+1. Find the values of a and b using remainder theorem.
x + 1 = 0 , x = -1
$$p\;(-1)\;=a{(-1)}^3+b{(-1)}^2+2(-1)-1$$
= – a + b – 2 – 1
= – a + b – 3
– a + b – 3 = 5
– a + b = 5 + 3
b – 8 = a
xii) :- Solve x + y + z ; y + z = 2 ; z = 2 using Cramer’s rule.
$$\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\;\begin{bmatrix}x\\y\\z\end{bmatrix}\;\begin{bmatrix}3\\2\\2\end{bmatrix}\\A\;X=B$$
$$\left|A\right|=\begin{vmatrix}1&1&1\\0&1&1\\0&0&1\end{vmatrix}$$
$$=1\begin{vmatrix}1&1\\0&1\end{vmatrix}=1(1-0)\\=\;1$$
$$\left|A_x\right|=\begin{vmatrix}3&1&1\\2&1&1\\2&0&1\end{vmatrix}\\=3(1-0)-1(2-2)+1(0-2)$$
$$\left|A_x\right|=3(1)-1(-2)\\=3-0-2=1\\\left|A_y\right|=\begin{vmatrix}1&3&1\\0&2&1\\0&2&1\end{vmatrix}$$
$$=\;1(2-2)=1\;(0)=0\\\left|A_z\right|=\begin{vmatrix}1&1&3\\0&1&2\\0&0&2\end{vmatrix}\\=1(2-0)=1\;(2)=2$$
$$x=\frac{\left|A_x\right|}{\left|A\right|}=\frac11=\;1\\y=\frac{\left|A_y\right|}{\left|A\right|}=\frac01=\;0\\z=\frac{\left|A_z\right|}{\left|A\right|}=\frac21=\;2$$
up to three decimal places
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SCCTION C
Q3 :- If
$$p\;(x)\;=\;2x^3+x^2-13x+6$$ , then
a) :- Verify that (x+3) is a factor of p (x)
b) :- Factorize p (x) using synthetic division.
c) :- Find all the roots of p (x).
d) :- Write complete factorized form of p (x) and confirm by expanding.
x + 3 = 0
x = – 3
Hence x + 3 is a factor of p (x)
$$2x^2-5x+2\\2x^2-4x-x+2\\2x(x-2)\;-1(x-2)\\(x-2)\;(2x-1)$$
$$x-2=0,\;x=2\\2x-1=0,\;x=\frac12\\Roots\mathit\;are\mathit\;\mathit-\mathit3\mathit,\mathit\;\mathit2\mathit,\mathit\;\frac{\mathit1}{\mathit2}$$
$$2x^3+x^2-13x+6\\=(x+3)\;(x-2)\;(2x-1)\\(x+3)\;(x-2)=x^2-2x+3x-6\\=x^2+x-6$$
$$(2x-1)(x^2+x-6)\\=2x^3+2x^2-12x-x^2-x+6\\=2x^3+x^2-13x+6\\=p\;(x)$$
Q3 :- If $$\widehat a=3\widehat i+2\widehat j+4\widehat k\\and\\\widehat b=\widehat i+2\widehat j+\widehat k\\\\$$ , then
a) :- Find $$\overrightarrow a\times\overrightarrow b$$
b) :- Find $$(\overrightarrow a\times\overrightarrow b).a\;and\;verify\;that\;\overrightarrow a\times\overrightarrow b\perp\overrightarrow a$$
c) :- Find $$(\overrightarrow a\times\overrightarrow b).b\;and\;verify\;that\;\overrightarrow a\times\overrightarrow b\perp\overrightarrow b$$
$$\overrightarrow a\times\overrightarrow b=\begin{vmatrix}i&j&k\\3&2&4\\1&2&-1\end{vmatrix}$$
$$=\begin{vmatrix}2&4\\2&-1\end{vmatrix}-j\begin{vmatrix}3&4\\1&-1\end{vmatrix}\\+k\begin{vmatrix}3&2\\1&2\end{vmatrix}$$
$$=i(-2-8)-i(-3-4)+k(6-2)\\\overrightarrow a\times\overrightarrow b=-10i+7j+4k$$
$$(\overrightarrow a\times\overrightarrow b).\overrightarrow b=(-10,\;7,\;4).(1,\;2,\;-1)\\=-10+14-4=0$$
$$Hence\;(\overrightarrow a\times\overrightarrow b)\perp\overrightarrow b$$
$$\;(\overrightarrow a\times\overrightarrow b).\overrightarrow a=(-10,\;7,\;4).(3,\;2,\;4)\\=-30+14+16=0\;$$
$$Hence\;\overrightarrow a\times\overrightarrow b\perp\overrightarrow a$$
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Q4:- If x is very small such that its square and higher powers can be neglected, then show that
$$\frac{{(16+4x)}^{\displaystyle\frac34}}{(4+x)\;\sqrt{9-6x}}=\frac23+\frac{13x}{72}$$
$$L.H.S=\frac{{(16+4x)}^{\displaystyle\frac34}}{(4+x)(\sqrt{9-6x})}\\=\frac{(4{(4+x)\;)}^{\displaystyle\frac34}}{(4+x)(\sqrt{9-6x})}$$
$$=\frac{{(4)}^{\displaystyle\frac34}\;(4+x)\;^{{\displaystyle\frac34}-1}}{\sqrt{9-6x}}\\$$
$$=\frac{{(2^2)}^{\displaystyle\frac34}{(4+x)}^{\displaystyle\frac{-1}4}}{{(9-6x)}^{\displaystyle\frac12}}$$
$$=\frac{2^{\displaystyle\frac{2\times3}4}{(4+x)}^{\displaystyle\frac{-1}4}}{{(9-6x)}^{\displaystyle\frac12}}$$
$$=\frac{2^{\displaystyle\frac32}(\;{(4)}^{\displaystyle\frac{-1}4}+({\displaystyle\frac{-1}4}){(4)}^{{\displaystyle\frac{-1}4}-1}(x)+{\displaystyle\frac{({\displaystyle\frac{-1}4})({\displaystyle\frac{-1}4}-1)}{21}}{(4)}^{{\displaystyle\frac{-1}4}-2}{(x)}^2+…)}{{(9)}^{\displaystyle\frac12}+{\displaystyle\frac12}{(9)}^{{\displaystyle\frac12}-1}(-6x)+{\displaystyle\frac12}({\displaystyle\frac12}-1){(9)}^{{\displaystyle\frac12}-2}{(6x)}^2+…}$$
$$=\frac{2^{\displaystyle\frac32}(\;{(2^2)}^{\displaystyle\frac{-1}4}+({\displaystyle\frac{-1}4}){(2^2)}^{\displaystyle\frac{-5}4}x+({\displaystyle\frac{-1}4})({\displaystyle\frac{-5}4}){(2^2)}^{\displaystyle\frac{-9}4}x^2+…}{{(3^2)}^{\displaystyle\frac12}+{\displaystyle\frac12}{(3^2)}^{\displaystyle\frac{-1}2}(-6x)+{\displaystyle\frac12}({\displaystyle\frac{-1}2}){(3^2)}^{\displaystyle\frac{-3}4}(36x^2)+…}$$
$$=\frac{2^{\displaystyle\frac32}(2^{2({\displaystyle\frac{-1}4})}+({\displaystyle\frac{-1}{2^2}}.2^{2({\displaystyle\frac{-5}4})}x)}{3+{\displaystyle\frac12}(3^{-1})(-6x)}$$
$$=\frac{2^{\displaystyle\frac32}(2^{\displaystyle\frac{-1}2}-2^{-2}.2^{\displaystyle\frac{-5}2}x)}{3-3^{-1}.3x}$$
$$=\frac{2^{\displaystyle\frac32}(2^{\displaystyle\frac{-1}2}-2^{\displaystyle\frac{-2-5}2}x)}{3-(3)^\circ x}$$
$$=\frac{2^{\displaystyle\frac32\;}.2^{\displaystyle\frac{-1}2}-2^{\displaystyle\frac32}.2x^{\displaystyle\frac{-9}2}}{3-x}$$
$$=\frac{{(2^{\displaystyle\frac32\;}}^{\displaystyle\frac{-1}2}-2x^{\displaystyle\frac32\frac{-9}2})}{3-x}$$
$$=\frac{2^1-2^{-3}x}{3-x}\\=\frac{2-{\displaystyle\frac x8}}{3-x}$$
$$=(2-\frac x8)(\;(3^{-1})+(-1)(3)(-x)\\+1(-1-1)(3^{-1-2}){(-x)}^2+…)\\=(2-\frac x8)(3^{-1}+3^{-2}x+3^{-3}x^2+…)$$
$$=2.3^{-1}+2.3^{-2}x+2.3^{-3}x^2.\frac x8+…\\=\frac23+\frac{2x}{3^2}-\frac x{3^1.8}$$
$$=\frac23+\frac x3\;(\frac23-\frac18)\\=\frac23+\frac x3\;(\frac{16-3}{24})$$
$$=\frac23+\frac{13x}{72}\\=\;R.H.S$$
Q4 Prove the Fundamental law of trigonometry
Lets consider a unit circle with centre o. Let a point A on the circumference on the circle such that OA is making an Acute central angle Alpha with the x-axis and a point B on the circumference on the circle such that OB is making an Acute central angle Beta with the x-axis. Angle AOB is equal to Alpha minus Beta. Consider a point D on x-axis and a point C on the circumference of the circle such that angle COD is equal to Alpha minus Bets.
Lets check the correspondence of triangle AOB and triangle COD
mOA= mOC because both are radii of the same circle
Angle AOB = Angle COD because both are equal to angle Alpha minus Beta
mOB= mOD because both are radii of the same circle
Hence triangle AOB is congruent to triangle COD
So
mAB=mCD because these are congruent sides of congruent triangle
Lets apply distance formula
$$\overline{AB}=\overline{CD}\\\sqrt{{(Cos\alpha-Cos\beta)}^2+{(Sin\alpha-Sin\beta)}^2}\\=\sqrt{{(Cos(\alpha-\beta)-1)}^2+{(Sin(\alpha-\beta)-0)}^2}\\or\\Cos^2\alpha+Cos^2\beta-2Cos\alpha Cos\beta+\\Sin^2\alpha+Sin^2\beta-2Sin\alpha Sin\beta\\=Cos^2(\alpha-\beta)+1-2Cos(\alpha-\beta)\\+Sin^2(\alpha-\beta)\\or\\1+1-2(Cos\alpha Cos\beta+Sin\alpha Sin\beta)\\=1+1-2Cos(\alpha-\beta)\\Hence\\Cos(\alpha-\beta)=Cos\alpha Cos\beta+Sin\alpha Sin\beta$$
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Q5:-
$$y=3Cos2x\;;\;\frac{-\mathrm\pi}2\leq x\leq\frac{\mathrm\pi}2$$
(i) make table values for given interval
(ii) Draw the graph of the function for given interval
| x | y=3Cos2x |
|---|---|
| -90 | -3 |
| -45 | 0 |
| 0 | 3 |
| 45 | 0 |
| 90 | -3 |
Q5 Solve x+2y-3z=4, 2x-3y+4z=5 and 3x+4y-5z=6 using Gaussian elimination method.
We will start with augumented matrix
$$\begin{bmatrix}1&2&-3&4\\2&-3&4&5\\3&4&-5&6\end{bmatrix}$$
Subtract two times of 1st row from 2nd row and three times of 1st row from 3rd row
$$\begin{bmatrix}1&2&-3&4\\0&-7&10&-3\\0&-2&4&-6\end{bmatrix}$$
Divide 2nd row by -7
$$\begin{bmatrix}1&2&-3&4\\0&1&\frac{-10}7&\frac37\\0&-2&4&-6\end{bmatrix}$$
Subtract two times of 2nd row from 1st row and add two times of 2nd row in 3rd row
$$\begin{bmatrix}1&0&\frac{-1}7&\frac{22}7\\0&1&\frac{-10}7&\frac37\\0&0&\frac87&\frac{-36}7\end{bmatrix}$$
Divide 3rd row by 8/7
$$\begin{bmatrix}1&0&\frac{-1}7&\frac{22}7\\0&1&\frac{-10}7&\frac37\\0&0&1&\frac{-9}2\end{bmatrix}$$
Add 1/7 of 3rd row in 1st row and 10/7 of 3rd row in 2nd row
$$\begin{bmatrix}1&0&0&\frac{85}{28}\\0&1&0&\frac{-57}{14}\\0&0&1&\frac{-9}2\end{bmatrix}$$
So x=85/28, y=-57/14 and z=-9/2
Verification:-
x+2y-3z=4
85/28+2(-57/14)-3(-9/2)=4
4=4
Verified
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Q6:- For an arithmetic geometric series
$$1+5(\frac12)+9{(\frac12)}^2+13{(\frac12)}^3+…..$$
a):- Find general term of the series
b):- Sum the series up to n-term
c):- Sum the series up to infinite terms
Solution:-
1,5,9,13,….
a=1, d=5-1=4
$$a_n=a-(n-1)\;d\\=1+(n-1)\;3\\=4n-3$$
$${(\frac12)}^0+{(\frac12)}^1+{(\frac12)}^2+{(\frac12)}^3+……\\a={(\frac12)}^0,\;r=\;\frac{{(\frac12)}^1}{{(\frac12)}^0}={\frac12\;}\\G_n=ar^{n-1}\\=(1)\;{(\frac12)}^{n-1}\\={(\frac12)}^{n-1}$$
Hence
$$General\;term={(4n-3)(\frac12)}^{n-1}$$
Formula for part b and c
$$Sum\;of\;n\;term\\=\frac a{1-r}+dr(\frac1{{(1-r)}^2})(1-r^n)-\frac{(a+nd)\;r^n}{1-r}\\Sum\;of\;infinite\;term\\=\frac a{1-r}+dr\;(\frac1{{(1-r)}^2})$$
Lets first calculate the sum of infinite terms
$$Sum\;of\;infinite\;term\\=\frac a{1-r}+dr\;(\frac1{{(1-r)}^2})\\=\frac1{1-{\displaystyle\frac12}}+(4)(\frac12)(\frac1{{(1-{\displaystyle\frac12})}^2})\\=2+2\times4\;=\;10$$
Lets check the sum of n terms
$$Sum\;of\;n\;term\\=\frac a{1-r}+dr(\frac1{{(1-r)}^2})(1-r^n)-\frac{(a+nd)\;r^n}{1-r}\\=2+8(1-{(\frac12)}^n)-\frac{(1+4n)\;{({\displaystyle\frac12})}^n}{1-{\displaystyle\frac12}}\\=10-\frac{10+8n}{2^n}\\=10-\frac{5+4n}{2^{n-1}}$$
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