Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

If the coefficients of rth, (r+ 1)th and (r + 2)th terms in the binomial expansion of (1 + y)m are in A.P., then m and r satisfy the equation

  • m2 – m(4r – 1) + 4r2 – 2 = 0

  • m2 – m(4r+1) + 4r2 + 2 = 0

  • m2 – m(4r + 1) + 4r2 – 2 = 0

  • m2 – m(4r – 1) + 4r2 + 2 = 0


C.

m2 – m(4r + 1) + 4r2 – 2 = 0

Given space to the power of straight m straight C subscript straight r minus 1 end subscript space comma space to the power of straight m straight C subscript straight r space plus straight C presuperscript straight m subscript straight r plus 1 end subscript space are space in space straight A. straight P.
2 to the power of space straight m end exponent straight C subscript 2 space equals space fraction numerator blank to the power of straight m straight C subscript straight r minus 1 end subscript over denominator straight C presuperscript straight m subscript straight r end fraction space plus fraction numerator straight C presuperscript straight m subscript straight r plus 1 end subscript over denominator straight C presuperscript straight m subscript straight r end fraction
space equals space fraction numerator straight r over denominator straight m minus straight r plus 1 end fraction space plus fraction numerator straight m minus straight r over denominator straight r plus 1 end fraction
rightwards double arrow space straight m squared minus straight m space left parenthesis 4 straight r plus 1 right parenthesis space plus 4 straight r squared space minus 2 space equals space 0 space
197 Views

2.

Let P be the point (1, 0) and Q a point on the locus y2 = 8x. The locus of mid point of PQ is

  • y2 – 4x + 2 = 0

  • y2 + 4x + 2 = 0

  • x2 + 4y + 2 =

  • x2 – 4y + 2 = 0


A.

y2 – 4x + 2 = 0

P = (1, 0) Q = (h, k) such that
k2 = 8h
Let (α, β) be the midpoint of PQ

straight alpha space equals space fraction numerator straight h plus 1 over denominator 2 end fraction comma space straight beta space equals space fraction numerator straight k plus 0 over denominator 2 end fraction
2 straight alpha space minus 1 space equals straight h comma
2 straight beta space equals straight k
left parenthesis 2 straight beta right parenthesis squared space equals space 8 space left parenthesis 2 straight alpha minus 1 right parenthesis
rightwards double arrow straight beta squared space equals space 4 straight alpha space minus 2
rightwards double arrow space straight y squared minus 4 straight x space plus 2 space equals space 0

442 Views

3.

The value of α for which the sum of the squares of the roots of the equation x2 – (a – 2)x – a – 1 = 0 assume the least value is

  • 1

  • 0

  • 3

  • 2


A.

1

x2 – (a – 2)x – a – 1 = 0
⇒ α + β = a – 2
α β = –(a + 1)
α2 + β2 = (α + β)2 - 2αβ
= a2 – 2a + 6 = (a – 1)2 + 5
⇒ a = 1

152 Views

4.

If roots of the equation x2 – bx + c = 0 be two consectutive integers, then b2 – 4c equals

  • – 2

  • 3

  • 2

  • 1


D.

1

Let α, α + 1 be roots
α + α + 1 = b
α(α + 1) = c
∴ b2 – 4c = (2α + 1)2 - 4α(α + 1) = 1.

170 Views

5.

In a triangle PQR, ∠R =π/2. If (P/2) and tan (Q/2) are the roots of ax2 +bx+ c = 0, a ≠ 0 then 

  • a = b + c

  • c = a + b

  • b = c

  • b = a + c


B.

c = a + b

tan space open parentheses straight P over 2 close parentheses comma space tan space open parentheses straight Q over 2 close parentheses space are space the space roots space of space ax squared space plus bx plus space straight c space equals space 0
tan space open parentheses straight P over 2 close parentheses comma space tan space open parentheses straight Q over 2 close parentheses space equals space minus space straight b over straight a
tan space open parentheses straight P over 2 close parentheses comma space tan space open parentheses straight Q over 2 close parentheses space space equals straight c over straight a

fraction numerator tan space open parentheses straight P over 2 close parentheses comma space tan space open parentheses straight Q over 2 close parentheses space over denominator 1 minus space tan space open parentheses straight P over 2 close parentheses comma space tan space open parentheses straight Q over 2 close parentheses space end fraction space equals space tan space open parentheses straight P over 2 plus straight Q over 2 close parentheses space equals space 1
rightwards double arrow space fraction numerator negative begin display style straight b over straight a end style over denominator 1 minus begin display style straight c over straight a end style end fraction space equals space 1
rightwards double arrow negative straight b over straight a equals space straight a over straight a minus straight c over straight a rightwards double arrow space minus straight b space equals straight a minus straight c
straight c space equals straight a plus straight b
239 Views

6.

The system of equations 
αx + y + z = α - 1, 
x + αy + z = α - 1, 
x + y + αz = α - 1 

has no solution, if α is

  • -2

  • either-2 or 1

  • not -2

  • 1


A.

-2

αx + y + z = α - 1, 
x + αy + z = α - 1, 
x + y + αz = α - 1 

increment space equals space open vertical bar table row straight alpha 1 1 row 1 straight alpha 1 row 1 1 straight alpha end table close vertical bar

= α(α2 – 1) – 1(α - 1) + 1(1 - α)
= α (α - 1) (α + 1) – 1(α - 1) – 1(α - 1)
⇒ (α - 1)[α2 + α - 1 – 1] = 0
⇒ (α - 1)[α2 + α - 2]= 0 [α2 + 2α - α - 2] = 0
(α - 1) [α(α + 2) – 1(α + 2)]= 0
(α - 1) = 0,
α + 2 = 0 ⇒ α = –2, 1; but α ≠ 1

275 Views

7.

If the cube roots of unity are 1, ω, ω2 then the roots of the equation (x – 1)+ 8 = 0, are

  • -1 , - 1 + 2ω, - 1 - 2ω2

  • -1 , -1, - 1

  • -1 , 1 - 2ω, 1 - 2ω2

  • -1 , 1 + 2ω, 1 + 2ω2


C.

-1 , 1 - 2ω, 1 - 2ω2

(x – 1)3 + 8 = 0
⇒ (x – 1) = (-2) (1)1/3
⇒ x – 1 = -2 or -2ω or -2ω2 or
n = -1 or 1 – 2ω or 1 – 2ω2 .

446 Views

8. limit as straight n rightwards arrow infinity of space open square brackets 1 over straight n squared sec squared space 1 over straight n squared plus 2 over straight n squared space plus 2 over straight n squared sec squared 4 over straight n squared plus.....1 over straight n squared sec squared 1 close square brackets equal
  • 1 half sec space 1
  • 1 half cosec space 1
  • tan 1

  • 1 half tan space 1

D.

1 half tan space 1
limit as straight n rightwards arrow infinity of space open square brackets 1 over straight n squared sec squared 1 over straight n squared plus 2 over straight n squared sec squared 4 over straight n squared plus 3 over straight n squared sec squared 9 over straight n squared plus..... plus 1 over straight n sec squared 1 close square brackets is equals
limit as straight n rightwards arrow infinity of space straight r over straight n squared space sec squared space straight r squared over straight n squared space equals space limit as straight n rightwards arrow infinity of 1 over straight n. straight r over straight n space sec squared straight r squared over straight n squared
rightwards double arrow space Given space limit space is space equal space to space value space of space integral space integral subscript 0 superscript 1 straight x space sec squared space straight x squared space dx
or space 1 half integral subscript 0 superscript 1 space 2 straight x space sec space straight x squared space dx space equals space 1 half integral subscript 0 superscript 1 space sec squared space tdt
space equals space 1 half left parenthesis tan space straight t right parenthesis subscript 0 superscript 1 space equals space 1 half space tan space 1
107 Views

9.

Area of the greatest rectangle that can be inscribed in the ellipse straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1

  • 2ab

  • ab

  • square root of ab
  • a/b


A.

2ab



Area of rectangle ABCD = (2acosθ)
(2bsinθ) = 2absin2θ
⇒ Area of greatest rectangle is equal to 2ab when sin2θ = 1.
178 Views

10.

If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

  • 22.0

  • 20.5

  • 25.5

  • 24.0


D.

24.0

Mode + 2Mean = 3 Median
⇒ Mode = 3 × 22 – 2 × 21= 66 – 42= 24.

154 Views