{"id":19488,"date":"2020-01-13T12:44:26","date_gmt":"2020-01-13T07:14:26","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19488"},"modified":"2021-07-02T15:29:40","modified_gmt":"2021-07-02T09:59:40","slug":"2nd-puc-basic-maths-question-bank-chapter-4-ex-4-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-4-ex-4-2\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2"},"content":{"rendered":"

Students can Download Basic Maths Exercise 4.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Two Marks Questions and Answers<\/strong><\/p>\n

1. Find
\nQuestion (i).
\nThe 5th<\/sup> term in \\(\\left(\\frac{4 x}{5}-\\frac{5}{2 x}\\right)^{8}\\)
\nAnswer:
\n\\(\\left(\\frac{4 x}{5}-\\frac{5}{2 x}\\right)^{8}\\) compare with (x + a)n<\/sup>
\n\u21d2 x \u2192 \\(\\frac{4 x}{5}\\), a \u2192 \\(\\frac{4 x}{5}\\) n \u2192 8,
\nTo find th<\/sup> term put r = 4
\nTr+<\/sub> = n<\/sup>Cr<\/sub>.xn – r<\/sup>.ar<\/sup>
\n\"2nd
\nT5<\/sub> = 8<\/sup>C4<\/sub>.(22<\/sup>).2-4<\/sup>
\n8<\/sup>C4<\/sub>.28-4<\/sup> = 8<\/sup>C4<\/sub>.24<\/sup> = 1120<\/p>\n

Question (ii).
\nThe 8th <\/sup>term in \\(\\left(\\frac{a}{2}-\\frac{3}{b}\\right)^{10}\\)
\nAnswer:
\n\\(\\left(\\frac{a}{2}-\\frac{3}{b}\\right)^{10}\\) compare with (x+a)n<\/sup>
\n\u21d2 x \u2192 \\(\\frac{a}{5}\\), a \u2192 \\(\\frac{2}{b}\\) n = 10,
\nTo find 8th <\/sup>term put r = 7
\nTr+<\/sub> = n<\/sup>Cr<\/sub>.xn – r<\/sup>.ar<\/sup>
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question (iii).
\nThe 6th<\/sup> term in (\u221ax – \u221ay)17
\n<\/sup>Answer:
\nCompare (\u221ax – \u221ay)17<\/sup> with (x + a)n<\/sup>
\nx \u2192 \u221ax a \u2192 -\u221ay and n = 17
\nTo find 6th<\/sup>\u00a0term put r = 5
\n\"2nd<\/p>\n

Question (iv).
\nThe 7th<\/sup> term in (3x2<\/sup> – \\( \\frac{y}{3}\\) )9
\n<\/sup>Answer:
\nHere x \u2192 3x2 <\/sup>a \u2192 \\(-\\frac{y}{3}\\) nn = 9
\nPut r = 6
\nT6+1 <\/sub>=9<\/sup>C6<\/sub> (3x2<\/sup>)9-6<\/sup>. ( \\(-\\frac{y}{3}\\) )6<\/sup><\/p>\n

\"2nd<\/p>\n

Question (v).
\nthe 10th<\/sup> term in \\(\\left(\\frac{a}{b}-\\frac{2 b}{a^{2}}\\right)^{12}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question (vi).
\nthe 11th<\/sup> term in \\(\\left(x+\\frac{1}{\\sqrt{x}}\\right)^{14}\\)
\nAnswer:
\nHere x = x a = \\(\\frac{1}{\\sqrt{x}}\\) ,n = 14 and
\nput r = 10
\nT10+1<\/sub> = 14<\/sup>C10<\/sub> .x14 – 10<\/sup> . \\(\\left(\\frac{1}{\\sqrt{x}}\\right)^{10}\\) 14<\/sup>C4<\/sub>.x4<\/sup> \\(\\frac{1}{x^{5}}\\) = \\(\\frac{1001}{x}\\)<\/p>\n

\"KSEEB<\/p>\n

Part – B<\/span><\/p>\n

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Three Marks Questions and Answers<\/strong><\/p>\n

Question 2.
\nFind the middele term in the expansion of<\/p>\n

Question (i).
\n\\(\\left(x-\\frac{1}{2 y}\\right)^{10}\\)
\nAnswer:
\n\\(\\left(x-\\frac{1}{2 y}\\right)^{10}\\)
\nHere n = even i.e 10 \u2234 we have only one middle term\u00a0 \\(\\frac{\\mathrm{n}}{2}+1=\\frac{10}{2}+1\\) = 6th<\/sup> term
\n\"2nd<\/p>\n

Question (ii).
\n\\(\\left(\\frac{a}{x}+b x\\right)^{12}\\)
\nAnswer:
\n\\(\\left(\\frac{a}{x}+b x\\right)^{12}\\)
\nHere
\nn = 12 (even)
\n\u2234 We have only one middle term
\n\"2nd<\/p>\n

Question (iii).
\n\\(\\left(\\frac{2 a}{3}-\\frac{3}{3 a}\\right)^{6}\\)
\nAnswer:
\nHere n = 6 (even)
\n\"2nd<\/p>\n

Question (iv).
\n\\(\\left(3 x-\\frac{1}{6} x^{3}\\right)^{8}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question (v).
\n\\(\\left(\\frac{a}{3}+\\frac{b}{3}\\right)^{8}\\)
\nAnswer:
\nHere n = 8(even)
\n\u2234 middle term \\frac{\\mathrm{n}}{2} + 1 = 4 + 1 = 5th<\/sup>
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

3. Find the middle terms in the expansion of<\/p>\n

Question (i).
\nFind the middle term in the expansion of
\n\\(\\left(3 x-\\frac{2}{x^{2}}\\right)^{15}\\)
\nAnswer:
\nHere n = 15 and odd, so we have two middle terms i.e, \\(\\frac{n+1}{2}=\\frac{15+1}{2}=8^{t h}\\) and 8 + 1 = 9th<\/sup> terms to find 8 th<\/sup>term to find 8 th<\/sup> term pur r = 7
\n\"2nd<\/p>\n

Question (ii)
\n\\(\\left(\\frac{x}{2}+\\frac{3}{x^{2}}\\right)^{19}\\)
\nAnswer:
\nHere n = 19 (odd)\u2234 We have two middle terms \\(\\frac{n+1}{2}=\\frac{20}{2}\\) 10th<\/sup> and 10 + 1 =11th<\/sup> term To find 10th<\/sup> term put r = 9
\n\"2nd
\nQuestion (iii)
\n\\(\\left(2 x^{2}+\\frac{1}{\\sqrt{x}}\\right)^{11}\\)
\nAnswer:
\nHere n = 11 (odd)
\n\u2234 the two middle terms are \\(\\frac{13+1}{2}\\) = 7th<\/sup> and 6 + 1 = 7th<\/sup> terms
\nTo find 6th<\/sup> term put r = 5
\n\"2nd<\/p>\n

Question (iv).
\n\\(\\left(\\sqrt{x}-\\frac{4}{x^{2}}\\right)^{11}\\)
\nAnswer:
\nHere n = 11(odd) we have two middle term
\ni.e,\u00a0 \\( \\frac{1+1}{2}\\) = 6th<\/sup> and 6 + 1 = 7th<\/sup> terms
\nTo find 6th<\/sup> term put r = 5.
\n\"2nd
\n\"2nd<\/p>\n

Question (v).
\n\\(\\left(\\sqrt{x}-\\frac{3}{x^{2}}\\right)^{13}\\)
\nAnswer:
\n\\(\\left(\\sqrt{x}-\\frac{3}{x^{2}}\\right)^{13}\\)
\nHere n = 13 (odd)
\n\u2234 the two middle terms are \\(\\frac{13+1}{2}\\) = 7th<\/sup> and 7 + 1 = 8th<\/sup> terms
\nTo find 7th<\/sup> term put r = 6
\n\"2nd<\/p>\n

Part – C<\/span><\/p>\n

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Five Marks Questions and Answers<\/strong><\/p>\n

4.
\nQuestion (i).
\nFind the coefficent of xn<\/sup> in \\(\\left(x+\\frac{2}{x^{2}}\\right)^{17}\\)
\nAnswer:
\nHere x = x\u00a0 a\u00a0 = \\(\\frac{2}{x^{2}}\\) and\u00a0 n = 17
\nTr+1<\/sub> = n<\/sup>Cr<\/sub> . x n-r<\/sup>.ar<\/sup>
\n= 17<\/sup>Cr<\/sub> .x 17-r<\/sup>.( \\(\\frac{2}{x^{2}}\\) )r<\/sup> = 17<\/sup>Cr<\/sub> . 2r<\/sup>.x 17-r-2r<\/sup>
\n= 17<\/sup>Cr<\/sub> 2 r<\/sup>.x 17-3r<\/sup>
\nTo find coefficient of x11<\/sup> equate the power of x to 11
\n\u21d2 17 – 3r = 11 \u21d2 17 – 11 = 3r
\n\u21d2 3r = 6 \u21d2 r = 2
\nT2+1<\/sub> = 17<\/sup>C2<\/sub> . 22<\/sup>.x11<\/sup>
\n\u2234 Coefficient of x\u201d is 17<\/sup>C2<\/sub> .22<\/sup> = \\(\\frac{17 \\times 16 \\times 2}{2 \\times 1}\\) = 544<\/p>\n

Question (ii).
\nY3<\/sup> in \\(\\left(7 y^{2}-\\frac{2}{y}\\right)^{12}\\)
\nAnswer:
\nHere x =7y2<\/sup>, a = \\(-\\frac{2}{y}\\) and n = 12
\nTr+1<\/sub>\u00a0= 12Cr<\/sup>(7Y2<\/sup>)12-r<\/sup>.\\(\\left(\\frac{-2}{y}\\right)^{r}\\)
\n= 12<\/sup>Cr<\/sub>.712-r<\/sup> .y24-2r<\/sup>. y-r<\/sup>(-2)r<\/sup>
\nTr+1<\/sub>\u00a0= 12<\/sup>Cr<\/sub>.712-r<\/sup>.(-2)r<\/sup>.y24-3r<\/sup>
\nTo find the coefficient of y3<\/sup> equate the power of y ro 3
\ni.e., 24 – 3r = 3 \u21d2 21 = 3r \u21d2 r= 7
\n\u2234T7+1<\/sub>\u00a0= 12<\/sup>C7<\/sub> .712-7<\/sup>. (-2)7<\/sup> y3<\/sup>
\n= -12<\/sup>C7<\/sub> .75<\/sup> 27<\/sup> . y3<\/sup>
\n\u2234Coefficient of y3<\/sup> is -12<\/sup>C7<\/sub>.75<\/sup>.27<\/sup><\/p>\n

Question (iii).
\nx11<\/sup> in \\(\\left(\\sqrt{x}-\\frac{2}{x}\\right)^{17}\\)
\nAnswer:
\nlere, x \u2192 \u221ax, a \u2192\\(\\frac{-2}{x}\\) and n = 17
\n\"2nd
\nTo find the coefficient of x2 equate the power of x to -11
\n\u2234 \\(\\frac{17-3 r}{2}\\) = -11 \u21d2 17 – 3r = – 22
\n\u21d2 17+22 = 3r \u21d2 39 = 3.r \u21d2 r = 13
\nT13+1<\/sub> = 17<\/sup>C13<\/sub>(-2)13<\/sup>.x-11<\/sup>
\nCoefficient of x-11<\/sup> is -17<\/sup>C13<\/sub>.213<\/sup><\/p>\n

Question (iv).
\nX18<\/sup> in \\(\\left(x^{2}-\\frac{6}{x}\\right)^{15}\\)
\nlere, x \u2192 x2<\/sup>, a \u2192\\(\\frac{-6}{x}\\) and r = 15
\nTr+1<\/sub> = 15<\/sup>Cr<\/sub>.(x2<\/sup>)15-r<\/sup>\\(\\left(\\frac{-6}{x}\\right)^{r}\\)
\nTr+1<\/sub> = 15<\/sup>Cr<\/sub>.x30-2r<\/sup>.(-6)r<\/sup>.x-r<\/sup>
\n= 15<\/sup>Cr<\/sub>.(-6)r<\/sup>.x30-2r-r<\/sup>
\n= 15<\/sup>Cr<\/sub>(-6)r<\/sup>.x30-3r<\/sup>.
\nTo find the coefficient of x18<\/sup>,equate the power of x to 18
\n\u2234 30 – 3r = 18 \u21d2 30 – 18 = 3r \u21d2 3r = 12 \u21d2 r = 4
\nT4+1<\/sub> = 15<\/sup>C4<\/sub>(-6) 4<\/sup>x18<\/sup>
\n\u2234 Coefficient of x18<\/sup> is 15<\/sup>C4<\/sub>. (6)4<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question (v).
\nX-2<\/sup> in \\(\\left(x+\\frac{1}{x^{2}}\\right)^{17}\\)
\nAnswer:
\nHere, x \u2192 x, a \u2192 \\(\\frac{1}{x^{2}}\\) and n = 17
\n\u2234 Tr+1<\/sub> = 17<\/sup>Cr<\/sub>.x17-r<\/sup>\\(\\left(\\frac{1}{x^{2}}\\right)^{r}\\)
\nTr+1<\/sub> = 17<\/sup>Cr<\/sub>.x17-r-2r<\/sup>
\n= 17<\/sup>Cr<\/sub>x17-3r<\/sup>
\nTo find the coefficient of x-2<\/sup>,equate the power of x to -2
\n17 – 3r = -2 \u21d2 17 + 2 = 3r \u21d2 3r \u21d2 r = \\(\\frac{19}{3}\\)
\nSince r is a fraction the coefficient of x-2<\/sup> is 0.<\/p>\n

Question (vi).
\nX5<\/sup> in \\(\\left(x+\\frac{1}{x^{2}}\\right)^{17}\\)
\nAnswer:
\nHere, x \u2192 x, a \u2192 \\(\\frac{1}{x^{2}}\\) and n = 17
\nTr+1<\/sub> = 17<\/sup>Cr<\/sub>.x17-r<\/sup>\\(\\left(\\frac{1}{x^{2}}\\right)^{r}\\)
\n= 17<\/sup>Cr<\/sub>x17-3r<\/sup>
\nTo find the coefficient of x5<\/sup>, equate the power of x to 5
\n\u2234 17 – 3r = 5
\n12 = 23 \u21d2 r = 4
\nT4+1<\/sub> = T5<\/sub> = 17<\/sup>C4<\/sub>.x5<\/sup>
\n\u2234 the coefficient of x5<\/sup> is 17<\/sup>C4<\/sub><\/p>\n

Question (vii).
\nX18<\/sup> in \\(\\left(x^{2}+\\frac{3 a}{x}\\right)^{15}\\)
\nAnswer:
\nHere x \u2192 x2<\/sup> a \u2192 \\(\\frac{3 \\mathrm{a}}{\\mathrm{x}}\\) , n=15
\n\u2234 Tr+1<\/sub> = 15<\/sup>Cr<\/sub>.(x2<\/sup>)15-r<\/sup> \\(\\left(\\frac{3 a}{x}\\right)^{r}\\)
\n= 15<\/sup>Cr<\/sub>x30-2r<\/sup>.(3a)r<\/sup>xr<\/sup>
\n=15<\/sup>Cr<\/sub> .3r<\/sup>.ar<\/sup>.x30-3r<\/sup>
\nTo find the coefficient of x18<\/sup>, we get 30 – 3r = 18
\n12 = 3r \u21d2 r =4
\n\u2234 T4+1<\/sub> = 15<\/sup>C4<\/sub>.34<\/sup>.a4<\/sup>.x18<\/sup>
\n\u2234 coefficient of x18<\/sup> is 15<\/sup>C4<\/sub>.(3a)4<\/sup><\/p>\n

5. Find the term independent of x in<\/p>\n

Question (i).
\n\\(\\left(\\frac{4 x^{2}}{3}+\\frac{3}{2 x}\\right)^{9}\\)
\nHere x \u2192 \\(\\frac{4 x^{2}}{3}\\) , a = \\(\\frac{3}{2 x}\\) and n = 9
\n\"2nd
\nTo find the term independent of x, equate the power of x to zero.
\ne., 18 – 3r = 0 \u21d2 r = 6
\n\"2nd<\/p>\n

Question (ii).
\n\\(\\left(x^{3}-\\frac{3}{x^{2}}\\right)^{15}\\)
\nAnswer:
\nHere x \u2192 x3<\/sup>, a = \\(\\frac{-3}{x^{2}}\\) and n = 15
\nTr+1<\/sub> = 15<\/sup>Cr<\/sub>.(x3<\/sup>)15-r<\/sup>.\\(\\left(\\frac{-3}{x^{2}}\\right)^{r}\\)
\n= 15<\/sup>Cr<\/sub>.x45-r<\/sup>.x-2r<\/sup> (-3)r<\/sup>
\n= 15<\/sup>Cr<\/sub>.(-3)r<\/sup>.x45-5r<\/sup>
\nTo find the term independent of x we have 45 – 5r = O
\n:. 45 = 5r \u21d2 r = 9
\nT9+1<\/sub> = 15<\/sup>C9<\/sub>.(-3)9<\/sup>.x0<\/sup>
\nT10<\/sub> = -15<\/sup>C9<\/sub>.(3)9<\/sup> is the term independent of x.<\/p>\n

\"KSEEB<\/p>\n

Question (iii).
\n\\(\\left(\\sqrt{x}+\\frac{1}{3 x^{2}}\\right)^{10}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question (iv).
\n\\(\\left(\\frac{3 x^{2}}{2}-\\frac{1}{3 x}\\right)^{9}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question (v).
\n\\(\\left(3 x-\\frac{2}{x^{2}}\\right)^{15}\\)
\nAnswer:
\nHere x \u2192 3x a \u2192 \\(\\frac{-2}{x^{2}}\\) and n =15
\nTr+1<\/sub> = 15<\/sup>Cr<\/sub>.(3x)15-r<\/sup>.\\(\\left(\\frac{-2}{x^{2}}\\right)^{r}\\)
\n= 15<\/sup>Cr<\/sub>.315-r<\/sup>.(-2)r<\/sup> . x15-r<\/sup>
\nWe have x15-3r<\/sup>. = x0<\/sup> \u21d2 15 = 3r \u21d2 r = 5
\nT = 15<\/sup>C5<\/sub>.315-5<\/sup>.(-2)5<\/sup> = -15<\/sup>C5<\/sub> .310<\/sup> 25<\/sup>
\n\u2234 The term independent of x is -15<\/sup>C5<\/sub>.310<\/sup>.25<\/sup><\/p>\n

Question (vi).
\n\\(\\left(x^{2}-\\frac{2}{x^{3}}\\right)^{5}\\)
\nAnswer:
\nHere x \u2192 x2<\/sup> a \u2192 \\(\\frac{-2}{x^{3}}\\) = 5r<\/sub>(x2<\/sup>)5 – r<\/sup> . \\(\\left(\\frac{-2}{x^{3}}\\right)^{r}\\)= 5r<\/sub>.x10-2r<\/sup>.(-2)r<\/sup>
\n= 5Cr<\/sub>(-2)r<\/sup>.x10-5r<\/sup>
\nWe have 10 – 5r = 0 \u21d2 r = 2
\nT2+1<\/sub> = T3<\/sub> = 5<\/sup>C2<\/sub>(-2)2<\/sup>.x0<\/sup> = 4. \\(\\frac { 5.4}{ 2.1 }\\) = 40
\n\u2234\u00a0 The term independent of x is 40<\/p>\n

Question (vii).
\n\\(\\left(x-\\frac{1}{x^{2}}\\right)^{21}\\)
\nAnswer:
\nHere x \u2192x and a \u2192 \\(\\frac{-1}{x^{2}}\\) and n= 21
\nTr+1<\/sub> = 21<\/sup>Cr<\/sub>.(x)21-r<\/sup>. \\(\\left(\\frac{-1}{x^{2}}\\right)^{r}\\)
\n= 21<\/sup>Cr<\/sub>.x21-3r<\/sup>.(-1)r<\/sup>
\nWe have x0<\/sup>= x21-3r<\/sup> \u21d221 = 3r \u21d2 r = 7
\n\u2234 T7+1<\/sub> = T8<\/sub> = 21<\/sup>C7<\/sub> (-1)7<\/sup>.x0<\/sup> = -21<\/sup>C7<\/sub>
\n\u2234 The term independent of x is -21C7<\/sub><\/p>\n

Question (viii).
\n\\(\\left( \\sqrt { 2 } \\frac { 2 }{ { x }^{ 2 } } \\right)\\)<sup>20<\/sup>
\nAnswer:
\n\"2nd<\/p>\n

Part – D<\/span><\/p>\n

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Four Marks Questions and Answers<\/strong><\/p>\n

6. Use binomial theorem to evaluate upt 4 decimals place<\/p>\n

Questoin (i)
\n(102)6<\/sup>
\n(102)6<\/sup> = (100 + 2)6<\/sup>
\n= (100)6<\/sup> + 6<\/sup>C1<\/sub>(1 0O)5<\/sup>. 2 + 6<\/sup>C2<\/sub>(100)4<\/sup>.22 <\/sup>+ 6<\/sup>C3<\/sub>(100)3<\/sup>.23<\/sup> +\u00a06<\/sup>C4<\/sub>(100)2<\/sup> 24<\/sup> + 6<\/sup>C5<\/sub>.100.25<\/sup> + 6<\/sup>C6<\/sub>26<\/sup>
\n= 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64
\n= 1,126,162,419,264<\/p>\n

Question (ii).
\n(98)4<\/sup>
\nAnswer:
\n(98)4<\/sup> = (100 – 2 )4<\/sup>
\n= (100) 4<\/sup>– 4<\/sup>C1<\/sub> (100)3<\/sup>.2 + 4<\/sup>C2<\/sub>(100)2<\/sup>.22 <\/sup>– 4<\/sup>C3<\/sub>(100).23<\/sup> + 4<\/sup>C4<\/sub>.24<\/sup>
\n= 100000000 – 8000000 + 240000 – 3200 + 16
\n= 92236816<\/p>\n

\"KSEEB<\/p>\n

Question (iii).
\n(1.0005)4<\/sup>
\nAnswer:
\n(1.0005)4<\/sup> = (1 + 0.0005)4<\/sup>
\n= 14<\/sup> + 4<\/sup>C1<\/sub>(O.0005) + 4<\/sup>C2<\/sub>(0.0005)2 <\/sup>+ 4<\/sup>C3<\/sub>(0.0005)3<\/sup> + 4<\/sup>C4<\/sub>(0.0005)4<\/sup>
\n1 + 0.002 + 0.0000015 + …………….
\n= 1.00200150 \u2248 1.0020<\/p>\n

Question (iv).
\n(0.99)4<\/sup>
\nAnswer:
\n(0.99)4<\/sup> = (1- 0.01)4<\/sup> = 4<\/sup>C0<\/sub>(0.01) – 4<\/sup>C1<\/sub> (0.01) + 4<\/sup>C2<\/sub>(0.O1)2<\/sup> – 4<\/sup>C3<\/sub>(0.01)3<\/sup> + 4<\/sup>C4<\/sub>(0.01)4<\/sup>
\n= 1 – 0.04 + 0.0006 – 0.000004 + 0.00000001
\n= 0.96059601 \u2248 0.9606<\/p>\n

Question 7.
\nThe first three terms in (1 + ax)n<\/sup> where n is a positive integer are 1, 6x, 16x2<\/sup>. Find the vaIue
\nAnswer:
\nGiven . T1<\/sub> = lin (1 + ax)n<\/sup>; T2<\/sub> = 6x
\nn<\/sup>C1 .<\/sub>ax = 6x
\nnax = 6x
\n\u21d2 na = 6 \u21d2 a = \\(\\frac{6}{n}\\)
\nand T3<\/sub> = 16x2<\/sup>
\n\\(\\frac{n(n-1)}{2}\\) a2<\/sup>x2<\/sup> = 16x2<\/sup><\/p>\n

Question 8.
\nIn the expansion of (3 + kx)9<\/sup> the x2<\/sup> and x3<\/sup> are equal. Find k.
\nAnswer:
\nGiven , Tr+1<\/sub>= 9<\/sup>Cr<\/sub>.39-r<\/sup>.(kx) r<\/sup>
\n= 9<\/sup>Cr<\/sub>.39-r<\/sup>.kr<\/sup>xr<\/sup>
\nCoefficient of x2<\/sup> \u21d2 x2<\/sup> \u21d2 r = 2
\n\"2nd
\n36x 37<\/sup>.k2<\/sup> x x2<\/sup>
\nCoefficient of x3<\/sup> \u21d2 x3<\/sup> = xr<\/sup>\u21d2 r = 3
\n\"2nd
\nT3+1<\/sub> = T4<\/sub> = 9<\/sup>3<\/sub>.39-3<\/sup>.k3<\/sub>.x3<\/sub> = 9<\/sup>C3<\/sub>.36<\/sup>
\nk3<\/sub>.x3<\/sub> = 84.36<\/sup>k3<\/sub>.x3<\/sub>
\n84 x 36<\/sup>k2<\/sub>= 36 x 37<\/sub> x k2<\/sup>
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nFind the ratio of the coefficient of x4<\/sup> in the two expansions (1+x)7<\/sup> and (1+x)10<\/sup>?
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 4.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 Part – A 2nd PUC …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19488"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=19488"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19488\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19488"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19488"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19488"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}