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 Multiple Choice QuestionsMultiple Choice Questions

1.

If a1, a2, … , an are in H.P., then the expression a1a2 + a2a3 + … + an−1an is equal to

  • n(a1 − an)

  • (n − 1) (a1 − an)

  • na1an

  • na1an


D.

na1an

1 over straight a subscript 2 space minus 1 over straight a subscript 1 space equals 1 over straight a subscript 3 minus 1 over straight a subscript 2 space equals space........ space equals 1 over straight a subscript straight n minus 1 over straight a subscript straight n minus 1 end subscript space equals straight d
Then space straight a subscript 1 straight a subscript 2 space equals space fraction numerator straight a subscript 1 minus straight a subscript 2 over denominator straight d end fraction comma space straight a subscript 2 straight a subscript 3 space equals space fraction numerator straight a subscript 2 minus straight a subscript 3 over denominator straight d end fraction comma........ comma straight a subscript straight n minus 1 end subscript straight a subscript straight n space equals space fraction numerator straight a subscript straight n minus 1 end subscript minus straight a subscript straight n over denominator straight d end fraction
therefore space straight a subscript 1 straight a subscript 2 space plus space straight a subscript 2 straight a subscript 3 space plus......... space plus straight a subscript straight n minus 1 end subscript straight a subscript straight n space equals space fraction numerator straight a subscript 1 minus straight a subscript straight n over denominator straight d end fraction space Also comma space 1 over straight a subscript straight n space equals 1 over straight a subscript 1 space plus space left parenthesis straight n minus 1 right parenthesis straight d
rightwards double arrow space fraction numerator straight a subscript 1 minus straight a subscript straight n over denominator straight d end fraction space equals space left parenthesis straight n minus 1 right parenthesis straight a subscript 1 straight a subscript straight n
230 Views

2.

The value of integral subscript 1 superscript straight a left square bracket straight x right square bracket straight f apostrophe left parenthesis straight x right parenthesis space dx comma space straight a space greater than 1,where [x] denotes the greatest integer not exceeding x is

  • af(a) − {f(1) + f(2) + … + f([a])}

  • [a] f(a) − {f(1) + f(2) + … + f([a])}

  • [a] f([a]) − {f(1) + f(2) + … + f(a)}

  • [a] f([a]) − {f(1) + f(2) + … + f(a)}


B.

[a] f(a) − {f(1) + f(2) + … + f([a])}

Let a = k + h, where [a] = k and 0 ≤ h < 1

integral subscript 1 superscript straight a left square bracket straight x right square bracket straight f apostrophe left parenthesis straight x right parenthesis space dx space integral subscript 1 superscript 2 1 straight f apostrophe left parenthesis straight x right parenthesis space dx space plus integral subscript 2 superscript 3 2 straight f apostrophe space left parenthesis straight x right parenthesis dx space plus
space....... integral subscript straight k minus 1 end subscript superscript straight k left parenthesis straight k minus 1 right parenthesis space dx space plus integral subscript straight k superscript straight k plus straight h end superscript kf apostrophe left parenthesis straight x right parenthesis space dx

{f(2) − f(1)} + 2{f(3) − f(2)} + 3{f(4) − f(3)}+…….+ (k−1) – {f(k) − f(k − 1)} + k{f(k + h) − f(k)}

= − f(1) − f(2) − f(3)……. − f(k) + k f(k + h)
= [a] f(a) − {f(1) + f(2) + f(3) + …. + f([a])}

117 Views

3.

If xm.yn = (x+y)m+n, then dy/dx is

  • y/x

  • x+y/xy

  • xy

  • xy


A.

y/x

straight x to the power of straight m. straight y to the power of straight n space equals space left parenthesis straight x plus straight y right parenthesis to the power of straight m plus straight n end exponent
rightwards double arrow space mln space straight x space plus straight n space lny space equals space left parenthesis straight m plus straight n right parenthesis ln left parenthesis straight x plus straight y right parenthesis
therefore space straight m over straight x space plus straight n over straight y dy over dx space equals fraction numerator straight m plus straight n over denominator straight x plus straight y end fraction open parentheses 1 plus dy over dx close parentheses
rightwards double arrow space open parentheses straight m over straight x minus fraction numerator straight m plus straight n over denominator straight x plus straight y end fraction close parentheses space equals space open parentheses fraction numerator my minus nx over denominator straight y left parenthesis straight x plus straight y right parenthesis end fraction close parentheses dy over dx
rightwards double arrow space dy over dx space equals straight y over straight x
229 Views

4.

The larger of 9950 + 10050 and 10150 is

  • 9950 + 10050

  • both are equal

  • 10150

  • None of the above


C.

10150

We have,

     10150 = (100 + 1)50 = 10050 + 50 . 1004950 . 492 . 110048 + ...   ...(i)

and 9950 = (100 - 1)50 = 10050 - 50 . 10049 50 . 492 . 110048 + ...     ...(ii)

On subtracting Eq. (ii) from Eq. (i), we get

10150 - 9950 = 10050250 . 49 . 481 . 2 . 310047 + ... > 10050

Hence, 10150 - 9950 > 10050

 10150 > 10050 + 9950


5.

In the binomial expansion of (a - b)n, n ≥ 5, the sum of 5th and 6th terms is zero, then
a/b equals

  • 5/n −4

  • 6 /n −5

  • n -5 /6

  • n -5 /6


D.

n -5 /6

straight C presuperscript straight n subscript 4 space straight a to the power of straight n minus 4 end exponent space left parenthesis negative straight b right parenthesis to the power of 4 space plus straight C presuperscript straight n subscript 5 straight a to the power of straight n minus 5 end exponent space left parenthesis negative straight b right parenthesis to the power of 5 space equals space 0
rightwards double arrow space open parentheses straight a over straight b close parentheses space equals space fraction numerator straight n minus 5 plus 1 over denominator 5 end fraction
159 Views

6.

The sum of the series 20C020C1 + 20C220C3 + …… - ….. + 20C10 is-

  • 20C10

  • 1 half straight C presuperscript 20 subscript 10
  • 0

  • 0


B.

1 half straight C presuperscript 20 subscript 10

(1 + x)20 = 20C0 + 20C1x + … + 20C10x10 + …+ 20C20x20
put x = − 1,
0 = 20C0 − 20C1 + … − 20C9 + 20C1020C11 + … + 20C20
0 = 2 (20C020C1 + … − 20C9) + 20C10
20C020C1 + … + 20C10 =1 half straight C presuperscript 20 subscript 10

1417 Views

7. If space straight S subscript straight n space equals space sum from straight r equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight r end fraction space and space straight t subscript straight n space equals space sum from straight r equals space 0 to straight n of space straight r over straight C subscript straight r space then comma straight t subscript straight n over straight S subscript straight n space is
  • 1 half straight n
  • 1 half straight n minus 1
  • n-1

  • n-1


A.

1 half straight n
straight S subscript straight n space equals space sum from straight r space equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction
straight S subscript straight n space equals space sum from straight r space equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction space left parenthesis because space to the power of straight n straight C subscript straight r space equals straight C presuperscript straight n subscript straight n minus straight r end subscript right parenthesis
ns subscript straight n space equals space sum from straight r space equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction space
ns subscript straight n space equals space sum from straight r space equals 0 to straight n of open square brackets fraction numerator straight n minus straight r over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction plus fraction numerator straight r over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction close square brackets
ns subscript straight n space equals space sum from straight r space equals 0 to straight n of space fraction numerator straight n minus straight r over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction space plus space sum from straight r space equals 0 to straight n of space fraction numerator straight r over denominator straight C presuperscript straight n subscript straight r end fraction
ns subscript straight n space equals space straight t subscript straight n plus straight t subscript straight n
ns subscript straight n space equals space 2 straight t subscript straight n
straight t subscript straight n over straight s subscript straight n space equals straight n over 2
161 Views

8.

For natural numbers m, n if (1 − y)m (1 + y)n = 1 + a1y + a2y2 + … , and a1 = a2 = 10, then (m, n) is

  • (20, 45)

  • (35, 20)

  • (45, 35)

  • (45, 35)


D.

(45, 35)

left parenthesis 1 minus straight y right parenthesis to the power of straight m space left parenthesis 1 plus straight y right parenthesis to the power of straight n space equals space left square bracket 1 minus to the power of straight m straight C subscript 1 straight y space plus to the power of straight m straight C subscript 2 straight y squared space minus........ right square bracket left square bracket 1 plus to the power of straight n straight C subscript 1 straight y space plus to the power of straight n straight C subscript 2 straight y squared plus... right square bracket
space equals space 1 space plus space left parenthesis straight n minus straight m right parenthesis space plus open curly brackets fraction numerator straight m left parenthesis straight m minus 1 right parenthesis over denominator 2 end fraction plus fraction numerator straight n left parenthesis straight n minus 1 right parenthesis 2 over denominator 2 end fraction minus mn close curly brackets straight y squared space plus....
therefore space straight a subscript 1 space equals straight n minus straight m space equals space 10 space and space straight a subscript 2 space equals space fraction numerator straight m squared plus straight n squared space minus straight m minus straight n minus 2 mn over denominator 2 end fraction space equals space 10
So comma space straight n minus straight m equals 10 space and space left parenthesis straight m minus straight n right parenthesis squared space minus left parenthesis straight m plus straight n right parenthesis space equals space 20
rightwards double arrow straight m plus straight n space equals space 80
therefore comma space straight m space equals space 35 comma space straight n space equals space 45
180 Views

9.

If the expansion in powers of x of the function fraction numerator 1 over denominator left parenthesis 1 minus ax right parenthesis left parenthesis 1 minus bx right parenthesis end fraction is a0 + a1x + a2x2 + a3x3 + … , then an is

  • fraction numerator straight b to the power of straight n minus straight a to the power of straight n over denominator straight b minus straight a end fraction
  • fraction numerator straight a to the power of straight n minus straight b to the power of straight n over denominator straight b minus straight a end fraction
  • fraction numerator straight a to the power of straight n plus 1 end exponent minus straight b to the power of straight n plus 1 end exponent over denominator straight b minus straight a end fraction
  • fraction numerator straight a to the power of straight n plus 1 end exponent minus straight b to the power of straight n plus 1 end exponent over denominator straight b minus straight a end fraction

D.

fraction numerator straight a to the power of straight n plus 1 end exponent minus straight b to the power of straight n plus 1 end exponent over denominator straight b minus straight a end fraction

(1-ax)-1(1-bx)-1 = (1+ax+a2x2+.....)(1+bx+b2x2+....)
therefore coefficient of xn = bn +abn-1 +a2bn-2 +.....+an-1b +an =fraction numerator straight b to the power of straight n plus 1 end exponent minus straight a to the power of straight n plus 1 end exponent over denominator straight b minus straight a end fraction

182 Views

10.

The coefficient of the middle term in the binomial expansion in powers of x of (1 +αx)4  and of (1−αx )6  is the same if α equals

  • -5/3

  • 3/5

  • -3/10

  • -3/10


C.

-3/10

The coefficient of x in the middle term of expansion of (1+ αx)4= 4C22
The coefficient of x in middle term of the expansion of 3(1− αx) = 6C3 (−α)3
According to question
straight C presuperscript 4 subscript 2 straight alpha squared space equals space to the power of 6 straight C subscript 3 space left parenthesis negative straight alpha right parenthesis cubed
fraction numerator 4 factorial over denominator 2 factorial 2 factorial end fraction straight alpha squared space equals space minus space fraction numerator 6 factorial over denominator 3 factorial 3 factorial end fraction straight alpha cubed
6 straight alpha squared space equals space minus space 20 straight alpha cubed
straight alpha space equals space minus space 6 over 20
straight alpha space equals space minus 3 over 10

219 Views