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Ncert Solutions Maths class 11th

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Ncert Solutions Maths class 11th Limits and Derivatives

48.

Kindly consider the following

$\frac{1}{a x^{2} + b x + c}$

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alok kumar singh

Contributor-Level 10

48. Given, f (x) = $\frac{1}{a x^{2} + b x + c}$

So, f? (x) = $\frac{(a x^{2} + b x + c) \frac{d}{d x} (1) - 1 \cdot \frac{d}{d x} (a x^{2} + b x + c)}{{(a x^{2} + b x + c)}^{2}}$

$= \frac{(a x^{2} + b x + c) (0) - (2 a x + b)}{{(a x^{2} + b x + c)}^{2}}$

$= - \frac{(2 a x + b)}{{(a x^{2} + b x + c)}^{2}}$

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New answer posted

4 months ago

Ncert Solutions Maths class 11th Limits and Derivatives

49.

Kindly consider the following

$\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$

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alok kumar singh

Contributor-Level 10

49. Given, f (x) = $\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} = \frac{\frac{x + 1}{x}}{\frac{x - 1}{x}} = \frac{x + 1}{x - 1}$

So, f (x) = $\frac{(x - 1) \frac{d}{d x} (x + 1) - (x + 1) \frac{d}{d x} (x - 1)}{{(x - 1)}^{2}}$

= $\frac{(x - 1) - (x + 1)}{{(x - 1)}^{2}}$

$= \frac{x - 1 - x - 1}{{(x - 1)}^{2}} = \frac{- 2}{{(x - 1)}^{2}}$

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New answer posted

4 months ago

Ncert Solutions Maths class 11th Limits and Derivatives

48.

Kindly consider the following

$\frac{a x + b}{c x + d}$

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alok kumar singh

Contributor-Level 10

48. Given, f (x) = $\frac{a x + b}{c x + d}$

So, f (x) = $\frac{(c x + d) \frac{d}{d x} (a x + b) - (a x + b) \frac{d}{d x} (c x + d)}{({(x + d)}^{2}}$

$= \frac{(c x + d) a - (a x + b) c}{{(c x + d)}^{2}}$

$= \frac{a c x + a d - a c x - c b}{{(c x + d)}^{2}}$

$= \frac{a d - b c}{{(c x + d)}^{2}}$

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New answer posted

4 months ago

Ncert Solutions Maths class 11th Sequence and Series

106. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

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Payal Gupta

Contributor-Level 10

106. Let 'x' be the no of days in which 150 workers took to finish the job.

If 150 workers worked for x days then number of workers for x days =150 x.

But given that number of works dropped 4 on 2^nd day, then 4 on 3^rd day and so on taking 8 more

days to finish the work. i.e., x + 8 days we can express as.

150 x = 150 + (150 $-$ 4) + (150 $-$ 4 $-$ 4)+……+ (x + 8) days.

150 x = 150 + 146 + 142 +……… (x+8) days which

R.H.S. from as A.P. of

a = 150

d = -4 and n = x +8

So, S_n = 150 x

$\Rightarrow \frac{n}{2} [2 * 150 + (n - 1) (- 4)] = 150 x$

n [ 150 + (n - 1) (-2)] = 150 (n - 8) [ $?$ n = x +8 x $-$ 8 $-$ x]

150n 2n (n - 1) 150n 1200

2n² + 2n 1200 =0

n² - n - 6

...more

106. Let 'x' be the no of days in which 150 workers took to finish the job.

If 150 workers worked for x days then number of workers for x days =150 x.

But given that number of works dropped 4 on 2^nd day, then 4 on 3^rd day and so on taking 8 more

days to finish the work. i.e., x + 8 days we can express as.

150 x = 150 + (150 $-$ 4) + (150 $-$ 4 $-$ 4)+……+ (x + 8) days.

150 x = 150 + 146 + 142 +……… (x+8) days which

R.H.S. from as A.P. of

a = 150

d = -4 and n = x +8

So, S_n = 150 x

$\Rightarrow \frac{n}{2} [2 * 150 + (n - 1) (- 4)] = 150 x$

n [ 150 + (n - 1) (-2)] = 150 (n - 8) [ $?$ n = x +8 x $-$ 8 $-$ x]

150n 2n (n - 1) 150n 1200

2n² + 2n 1200 =0

n² - n - 600 = 0 [ dividing by -2 throughout]

n² - 25 n + 24 n 600 = 0

n (n - 25) + 24 (n - 25) = 0

(n - 25) (n + 24) = 0

n = 25 or n = -24 (which is not possible)

So, n = 25 days

i.e. x + 8 = 25 days

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New answer posted

4 months ago

Ncert Solutions Maths class 11th Sequence and Series

105. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

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Payal Gupta

Contributor-Level 10

105. Given,

Cost of machine =? 15625

depreciation rate = 20 % each year.

We have,

Depreciated value after 1st year =? 15625 - 20 % of 15625

=? $15625 - \frac{20}{100} *$ ?15625

=? $15625 (1 - \frac{20}{100})$

=? $15625 * (1 - \frac{1}{5})$

=? $15625 * \frac{4}{5}$

Similarly,

Depreciated value after 2nd year =? 15625 $* {(\frac{4}{5})}^{2}$ and so on.

This, Depreciated value at end of 5 years

=? 15625 $* {(\frac{4}{5})}^{5}$

=? 15625 $* \frac{1024}{3125}$

=? 5120

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New answer posted

4 months ago

Ncert Solutions Maths class 11th Sequence and Series

104. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

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Payal Gupta

Contributor-Level 10

104. Given,

Principal amount =? 10000

Amount at end of 1 year =? $(10000 + \frac{10000 + 5 * 1}{100})$

=? (10000 + 500)

=? 10500 {Amount paid = principal + S.I. in a year}

$S . I = \frac{Principal*rate *time}{100}$

Amount at end of 2nd year

=? $(10000 + \frac{10000 * 5 * 2}{100}) {\frac{Principal* rate*time}{100}}$

=? (10000 + 1000)10500 + 500

=? 11000

Amount at end of 3rd year

=? $(10000 + \frac{10000 * 5 * 3}{100})$

=? (1000 + 1500)

=? 11500

So, amount at end of 1^st, 2^nd, 3^rd ………, n^th year forms as A.P.

i.e.? 10500? 11000? 115000, ………. with

a = 10500

d = 11000 - 10500 = 500

Now, Amount in 15th year = Amount at end of 14th year

=? 10500 + ( 14 - 1) * 500

=? 10500 + 13 * 500

=? 10500 + 6500

=? 17000

Similarly amount after 20th year, =? 10500+ (20 - 1) * 500

=? 10500 + 19 * 500

=? 10500 + 9500

...more

104. Given,

Principal amount =? 10000

Amount at end of 1 year =? $(10000 + \frac{10000 + 5 * 1}{100})$

=? (10000 + 500)

=? 10500 {Amount paid = principal + S.I. in a year}

$S . I = \frac{Principal*rate *time}{100}$

Amount at end of 2nd year

=? $(10000 + \frac{10000 * 5 * 2}{100}) {\frac{Principal* rate*time}{100}}$

=? (10000 + 1000)10500 + 500

=? 11000

Amount at end of 3rd year

=? $(10000 + \frac{10000 * 5 * 3}{100})$

=? (1000 + 1500)

=? 11500

So, amount at end of 1^st, 2^nd, 3^rd ………, n^th year forms as A.P.

i.e.? 10500? 11000? 115000, ………. with

a = 10500

d = 11000 - 10500 = 500

Now, Amount in 15th year = Amount at end of 14th year

=? 10500 + ( 14 - 1) * 500

=? 10500 + 13 * 500

=? 10500 + 6500

=? 17000

Similarly amount after 20th year, =? 10500+ (20 - 1) * 500

=? 10500 + 19 * 500

=? 10500 + 9500

=? 20,000

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New answer posted

4 months ago

Ncert Solutions Maths class 11th Sequence and Series

103. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

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Payal Gupta

Contributor-Level 10

103. As the number of letters mailed forms a G.P. i.e., 4, 4², ……., 4⁸

We have,

a = 4

$r = \frac{4^{2}}{4} = 4$

and n = 8

So, Total numbers of letters = 4 + 4³ + ……. + 4⁸

$= \frac{4 (4^{8} - 1)}{4 - 1}$

$= \frac{4 * (65536 - 1)}{3}$

$= \frac{4}{3} * 65535$

$= \frac{4}{8} * 21, 845$

= 87380

As amount spent on one postage = 50 paise $=$ ? $\frac{50}{100}$

So, for reqd. postage =? $\frac{50}{100} * 87380$

=? 43690

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New answer posted

4 months ago

Ncert Solutions Maths class 11th Sequence and Series

102. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

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Payal Gupta

Contributor-Level 10

102. Given, cost of scooter = ?22,000

Amount paid =? 4000

Amount unpaid =? 22,000 -? 4000 =? 18,000

Now, Number of installments =Amount unpaid/Amount of each instalment

As he per 10% interest on up unpaid amount and ?1000 each installment

Amount of 1st instalment =? 1000 +? $\frac{18000 * 10 * 1}{100}$ = ?1000 + ?1800 =? 2800

Amount of 2nd installment =? 1000 +? $\frac{(18000 - 1000) * 10 * 1}{100}$ =? 1000 +? 1700 =? 2700

Similarly,

Amount of 3rd installment =? 1000 +? $\frac{(17000 - 1000) * 10 * 1}{100}$ =? 1000 +? 1600 =? 2600

So, Total installment paid =? 2800 +? 1600 =? 2600 + …… upto 18 installment

$= \frac{18}{2} [2 * 2800 + (18 - 1) (- 100)]$

= 9 [5600 - 1700]

= 9 * 3900

=? 35,000

Total cost of scooter = Amount + Total instal

...more

102. Given, cost of scooter = ?22,000

Amount paid =? 4000

Amount unpaid =? 22,000 -? 4000 =? 18,000

Now, Number of installments =Amount unpaid/Amount of each instalment

As he per 10% interest on up unpaid amount and ?1000 each installment

Amount of 1st instalment =? 1000 +? $\frac{18000 * 10 * 1}{100}$ = ?1000 + ?1800 =? 2800

Amount of 2nd installment =? 1000 +? $\frac{(18000 - 1000) * 10 * 1}{100}$ =? 1000 +? 1700 =? 2700

Similarly,

Amount of 3rd installment =? 1000 +? $\frac{(17000 - 1000) * 10 * 1}{100}$ =? 1000 +? 1600 =? 2600

So, Total installment paid =? 2800 +? 1600 =? 2600 + …… upto 18 installment

$= \frac{18}{2} [2 * 2800 + (18 - 1) (- 100)]$

= 9 [5600 - 1700]

= 9 * 3900

=? 35,000

Total cost of scooter = Amount + Total installment paid

=? 4000 +? 35,100

=? 39100

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New answer posted

4 months ago

Ncert Solutions Maths class 11th Sequence and Series

101. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

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Payal Gupta

Contributor-Level 10

101. Given,

cost of tractor =? 12,000

Amount paid =? 62,000

Amount unpaid =? 12000 -? 6000 =? 6000

So, number of instalments =

$=$ ? $\frac{6000}{500}$

= 12 = n

Now, interest on 1st installment = interest on unpaid amount ( i.e.? 6000) for 1 Year.

=? $\frac{6000 * 12 * 1}{100}$ = ?720

Similarly,

Interest on 2nd installment = interest on (? 6000 -? 500) for the next 1 year

=? $\frac{5500 * 12 * 1}{100}$ = ?660

And,

Interest on 2nd installment =? $\frac{(5500 - 500) * 12 * 1}{100}$

=? 600

Here, Total (interest per) installment paid =? (720 + 660 + 600 + ……. upto 12 terms)

$= \frac{12}{2} [2 (720) + (12 - 1) (- 60)]$ { $?$ 720 + 660 + 600 …….is an A.P. of a = 720, d = 660 - 720 = - 60}

= 6 [ 1440 -

...more

101. Given,

cost of tractor =? 12,000

Amount paid =? 62,000

Amount unpaid =? 12000 -? 6000 =? 6000

So, number of instalments =

$=$ ? $\frac{6000}{500}$

= 12 = n

Now, interest on 1st installment = interest on unpaid amount ( i.e.? 6000) for 1 Year.

=? $\frac{6000 * 12 * 1}{100}$ = ?720

Similarly,

Interest on 2nd installment = interest on (? 6000 -? 500) for the next 1 year

=? $\frac{5500 * 12 * 1}{100}$ = ?660

And,

Interest on 2nd installment =? $\frac{(5500 - 500) * 12 * 1}{100}$

=? 600

Here, Total (interest per) installment paid =? (720 + 660 + 600 + ……. upto 12 terms)

$= \frac{12}{2} [2 (720) + (12 - 1) (- 60)]$ { $?$ 720 + 660 + 600 …….is an A.P. of a = 720, d = 660 - 720 = - 60}

= 6 [ 1440 - 660]

= 6 * 780

=? 4680

So, the total cost of tractor = cost price + installments

=? 12,000 +? 4680

=? 16680

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New answer posted

4 months ago

Ncert Solutions Maths class 11th Sequence and Series

100. Show that $\frac{1 * 2^{2} + 2 * 3^{2} + ? + n {(n + 1)}^{2}}{1^{2} * 2 + 2^{2} + 3 + ? + n^{2} (n + 1)} = \frac{3 n + 5}{3 n + 1}$

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Payal Gupta

Contributor-Level 10

100. Given, $\frac{1 * 2^{2} + 2 * 3^{2} + ? + n {(n + 1)}^{2}}{1^{2} * 2 + 2^{2} + 3 + ? + n^{2} (n + 1)}$

For numerator,

a (n^th term) = n (n + 1)² = n (n + 1)² = n (n² + 2n +1) = n³ + 2n² + n

So, $S_{n} = \sum_{}^{} a_{n} = \sum_{}^{} n^{3} + 2 \sum_{}^{} n^{2} + \sum_{}^{} n$

$= \frac{n^{2} (n + 1)^{2}}{4} + \frac{2 \cdot (n + 1) (2 n + 1) n}{6} + \frac{n (n + 1)}{2}$

$= \frac{n (n + 1)}{2} [\frac{n (n + 1)}{2} + \frac{2 (2 n + 1)}{3} + 1]$

$= \frac{n (n + 1)}{2} [\frac{3 n (n + 1) + 2 * 2 (2 n + 1) + 6}{6}]$

$= \frac{n (n + 1)}{12} [3 n^{2} + 3 n + 8 n + 4 n + 6]$

$= \frac{n (n + 1)}{12} [3 n^{2} + 11 n + 10]$

$= \frac{n (n + 1)}{12} [3 n^{2} + 6 n + 5 n + 10]$

$= \frac{n (n + 1)}{12} [(33 n (n + 2) + 5 (n + 2)]$

$= \frac{n (n + 1) (n + 2) (3 n + 5)}{12}$

For denominator,

a_n (n^th term) = n² (n + 1) = n³ + n²

So,

$S_{n} = \sum_{}^{} a_{n} = \sum_{}^{} n^{3} + \sum_{}^{} n^{2}$

$= \frac{n^{2} (n + 1)^{2}}{4} + \frac{n (n + 1) (2 n + 1)}{6}$

$= \frac{n (n + 1)}{2} [\frac{n (n + 1)}{2} + \frac{(2 n + 1)}{3}]$

$= \frac{n (n + 1)}{2} [\frac{3 n (n + 1) + 2 (2 n + 1)}{6}]$

$= \frac{n (n + 1)}{2} [\frac{3 n^{2} + 3 n + 4 n + 2}{6}]$

$= \frac{n (n + 1)}{2} [\frac{3 n^{2} + 6 n + 1 n + 2}{6}]$

$= \frac{n (n + 1)}{2} [\frac{3 n (n + 2) + (n + 2)}{6}]$

$= \frac{n (n + 1) (n + 2) (3 n + 1)}{12}]$

So, $\frac{1 * 2^{2} + 2 * 3^{2} + ? + n {(n + 1)}^{2}}{1^{2} * 2 + 2^{2} * 3 + ? + n^{2} (n + 1)} = \frac{\frac{n (n + 1) (n + 2) (3 n + 5)}{12}}{\frac{n (n + 1) (n + 2) (3 n + 1)}{12}}$

$= \frac{3 n + 5}{3 n + 1}$

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