Subject

Mathematics

Class

JEE Class 12

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JEE Mathematics 2014 Exam Questions

Multiple Choice Questions

21.

If g is the inverse of a function f and f'(x) = fraction numerator 1 over denominator 1 plus straight x to the power of 5 end fraction comma then g'(x) is equal to 

  • 1+ x6

  • 5x4

  • fraction numerator 1 over denominator 1 plus space left curly bracket straight g left parenthesis straight x right parenthesis right square bracket to the power of 5 end fraction
  • 1+{g(x)}5


D.

1+{g(x)}5

Here 'g' is the inverse of f(x)
⇒ fog (x) =x
On differentiating w.r.t x, we get
f'{g(x)} x g'(x) =1
space straight g apostrophe left parenthesis straight x right parenthesis space equals space fraction numerator 1 over denominator straight f apostrophe left parenthesis straight g left parenthesis straight x right parenthesis right parenthesis end fraction space equals space fraction numerator 1 over denominator begin display style fraction numerator 1 over denominator 1 plus left curly bracket straight g left parenthesis straight x right parenthesis to the power of 5 right curly bracket end fraction end style end fraction
open square brackets because space straight f apostrophe left parenthesis straight x right parenthesis space equals space fraction numerator 1 over denominator 1 plus straight x to the power of 5 end fraction close square brackets
straight g apostrophe left parenthesis straight x right parenthesis space equals space 1 space plus space left curly bracket straight g space left parenthesis straight x right parenthesis right curly bracket to the power of 5

178 Views

22.

Let A and B be two events such that straight P left parenthesis space stack straight A space union straight B right parenthesis with bar on top space equals space 1 over 6 comma space straight P space left parenthesis straight A space intersection straight B right parenthesis space equals space 1 fourth space and space space straight P space left parenthesis straight A with bar on top right parenthesis space equals space 1 fourth where straight A with bar on top. Then, the events A and B are

  • independent but not equally likely

  • independent and equally likely

  • mutually exclusive and independent

  • equally likely but not independent


A.

independent but not equally likely

158 Views

23.

The Integral integral subscript 0 superscript straight pi square root of 1 plus 4 space sin squared straight x over 2 minus 4 sin straight x over 2 dx end root is equal to

  • π-4

  • fraction numerator 2 space straight pi over denominator 3 end fraction minus 4 minus 4 square root of 3
  • 4 square root of 3 minus 4
  • 4 square root of 3 minus 4 minus 4 square root of 3

D.

4 square root of 3 minus 4 minus 4 square root of 3

By using the formula, vertical line straight x minus straight a vertical line space equals open curly brackets table attributes columnalign left end attributes row cell straight x minus straight a comma space space space space space space space space space straight x greater or equal than straight a end cell row cell negative left parenthesis straight x minus straight a right parenthesis comma space space space straight x less than straight a end cell end table close
It breaks given integral in two parts and then integrates separately.

integral subscript 0 superscript straight x square root of open parentheses 1 minus 2 sin straight x over 2 close parentheses squared end root dx space equals space integral subscript 0 superscript straight pi vertical line 1 minus 2 space sin straight x over 2 vertical line dx
equals integral subscript 0 superscript straight pi over 3 end superscript open parentheses 1 minus 2 space sin space straight x over 2 close parentheses dx space minus integral subscript straight pi over 3 end subscript superscript straight pi open parentheses 1 minus 2 space sin space straight x over 2 close parentheses dx
equals space open parentheses straight x plus 4 space cos space straight x over 2 close parentheses subscript 0 superscript straight pi over 3 end superscript space minus space open parentheses straight x plus 4 space cos space straight x over 2 close parentheses subscript straight pi over 3 end subscript superscript straight pi
equals space 4 square root of 3 space minus 4 minus straight pi over 3

120 Views

24.

The integral integral open parentheses 1 plus straight x space minus 1 over straight x close parentheses straight e to the power of straight x plus 1 over straight x end exponent dx is equal to 

  • left parenthesis straight x minus 1 right parenthesis straight e to the power of straight x plus 1 over straight x space plus straight C end exponent
  • xe to the power of straight x plus 1 over straight x end exponent plus straight C
  • left parenthesis straight x plus 1 right parenthesis straight e to the power of straight x plus 1 over straight x space plus straight C end exponent
  • negative xe to the power of straight x plus 1 over straight x end exponent plus straight C

B.

xe to the power of straight x plus 1 over straight x end exponent plus straight C
integral open parentheses 1 plus straight x minus 1 over straight x close parentheses straight e to the power of straight x plus 1 over straight x dx end exponent
space equals space integral straight e to the power of straight x plus 1 half end exponent dx space plus integral straight x open parentheses 1 minus 1 over straight x squared close parentheses straight e to the power of straight x space plus 1 over straight x end exponent dx
equals space integral straight e to the power of straight x plus 1 over straight x end exponent dx space plus xe to the power of straight x plus 1 over straight x end exponent minus integral straight e to the power of straight x plus 1 over straight x end exponent dx
space equals space straight e to the power of straight x plus 1 over straight x dx end exponent space plus xe to the power of straight x plus 1 over straight x end exponent minus integral straight e to the power of straight x plus 1 over straight x end exponent dx
open square brackets because space integral straight e to the power of straight x plus 1 over straight x end exponent dx space plus xe to the power of straight x plus 1 over straight x end exponent minus integral straight e to the power of straight x plus 1 over straight x end exponent dx space equals space straight e to the power of straight x plus 1 over straight x end exponent close square brackets
equals space straight e to the power of straight x plus 1 over straight x end exponent dx space plus xe to the power of straight x plus 1 over straight x end exponent minus integral ex to the power of straight x plus 1 over straight x end exponent dx
space equals space xe to the power of straight x plus 1 over straight x end exponent space plus straight C
131 Views

25.

If =-1 and x =2 are extreme points of f(x) =α log|x| + βx2 +x, then

  • α = -6, β = 1/2

  • α = -6, β = -1/2

  • α = 2, β = -1/2

  • α = 2, β = 1/2


C.

α = 2, β = -1/2

Here, x =-1 and x = 2 are extreme points of f(x) = α log|x| +βx2 +x then,
f'(x) = α/x +2βx + 1
f'(-1) = -α -2β +1 = 0  .... (i)
[At extreme point f'(x) = 0]
f'(2) = α/x +4βx + 1 = 0 .. (ii)
On solving Eqs (i) and (ii), we get
α = 2 and β = -1/2

307 Views

26.

The area of the region described by A = {(x,y): x2 +y2 ≤ 1 and y2 ≤1-x} is

  • straight pi over 2 plus 4 over 3
  • straight pi over 2 minus 2 over 3
  • straight pi over 2 minus 2 over 3
  • straight pi over 2 space plus 2 over 3

A.

straight pi over 2 plus 4 over 3

Given, {(x,y): x2 +y2 ≤ 1 and y2 ≤1-x}
Required area = 1 half πr squared space plus space 2 space integral subscript 0 superscript 1 left parenthesis 1 minus straight y squared right parenthesis dy
equals space 1 half straight pi left parenthesis 1 right parenthesis squared space plus space 2 space open parentheses straight y minus straight y cubed over 3 close parentheses subscript 0 superscript 1
equals space straight pi over 2 plus 4 over 3

145 Views

27.

If fk(x) = 1/k (sink x + cosk x), where x ε R and k ≥1, then f4 (x)-fo (x) equal to 

  • 1/6

  • 1/3

  • 1/4

  • 1/12


D.

1/12

Given comma space straight f subscript straight x left parenthesis straight x right parenthesis space equals space 1 over straight k space left parenthesis sin to the power of 4 space straight x space plus space cos to the power of straight k space straight x right parenthesis
where space straight x element of space straight R space and space straight k greater than 1
straight f subscript 4 left parenthesis straight x right parenthesis space minus straight f subscript 6 left parenthesis straight x right parenthesis space equals space 1 fourth space left parenthesis sin to the power of 4 space straight x space plus space cos to the power of 4 space straight x right parenthesis minus 1 over 6 space left parenthesis sin to the power of 6 straight x space plus space cos to the power of 6 space straight x right parenthesis
equals space 1 fourth left parenthesis 1 minus 2 sin squared straight x. cos squared straight x right parenthesis
minus 1 over 6 left parenthesis 1 minus 3 sin squared straight x. cos squared straight x right parenthesis space equals space 1 fourth minus 1 over 6 space equals space 1 over 12
222 Views

28.

If f and ga re differentiable  functions in (0,1) satisfying f(0) =2= g(1), g(0) = 0 and f(1) = 6, then for some c ε] 0,1[

  • 2f'(c) = g'(c)

  • 2f'(c) = 3g'(c)

  • f'(c) = g'(c)

  • f'(c) = 2g'(c)


D.

f'(c) = 2g'(c)

Given, f(0) = 2 = g(1), g(0) and f(1) = 6
f and g are differentiable in (0,1)
Let h(x) = f(x)-2g(x)  .... (i)
h(0) = f(0)-2g(0)
h(0) = 2-0
h(0) = 2
and h(1) = f(1)-2g(1) = 6-2(2)
h(1) = 2, h(0) = h(1) = 2
Hence, using rolle's theorem 
h'(c) = 0, such that cε (0,1)
Differentiating Eq. (i) at c, we get
f'(c) -2g'(c) = 0
f'(c) = 2g'(c)

149 Views

29.

Let the population of rabbits surviving at a time t be governed by the differential equation.fraction numerator dp left parenthesis straight t right parenthesis over denominator dt end fraction space equals space 1 half straight p left parenthesis straight t right parenthesis space minus 200 If p(0) = 100 then p(t) is equal to 

  • 400 minus 300 straight e to the power of straight t over 2 end exponent
  • 300 minus 200 straight e to the power of negative straight t over 2 end exponent
  • 600 minus 500 straight e to the power of straight t over 2 end exponent
  • 400 minus 300 straight e to the power of negative straight t over 2 end exponent

A.

400 minus 300 straight e to the power of straight t over 2 end exponent

Given differential equation dp over dt minus 1 half straight p left parenthesis straight t right parenthesis space equals space minus 200 is a linear differential equation
Here, p(t) = fraction numerator negative 1 over denominator 2 end fraction comma space straight Q space left parenthesis straight t right parenthesis space equals space minus space 200
 If space equals space straight e to the power of integral negative open parentheses 1 half close parentheses dt end exponent space equals space straight e to the power of negative 1 half end exponent
Hence, solution is 
p(t), IF = ∫Q(t)IF dt
straight p left parenthesis straight t right parenthesis. straight e to the power of negative straight t over 2 end exponent space equals space integral negative 200. straight e to the power of negative straight t over 2 end exponent dt
straight p left parenthesis straight t right parenthesis. straight e to the power of negative straight t over 2 end exponent space equals integral negative 200 space straight e to the power of negative straight t over 2 end exponent. dt
straight p left parenthesis straight t right parenthesis. straight e to the power of negative straight t over 2 end exponent space equals space 400 space straight e to the power of negative straight t over 2 end exponent space plus space straight K
rightwards double arrow space straight p left parenthesis straight t right parenthesis space equals space 400 space plus ke to the power of negative 1 half end exponent
If space straight p left parenthesis 0 right parenthesis space equals space 100 space comma then space straight k space equals negative 300
rightwards double arrow space straight p left parenthesis straight t 0 space equals space 400 space minus 300 to the power of straight t over 2 end exponent

147 Views

30.

If A is a 3x3 non- singular matrix such that AAT = ATA, then BBT is equal to

  • l +B
  • l
  • B-1

  • (B-1)T


B.

l

If A is non - singular matrix then |A| ≠0
AAT = ATA and B = A-1AT
BBT = (A-1AT)(A-1AT)T
= A-1ATA(A-1)T       [∵ (AB)T= BTAT]
=A-1AAT(A-1)T        [∵ AAT = ATA]
=AT(A-1)T              [ ∵A-1A = l]
=A-1A)T                 [∵ (AB)T = BTAT]
lTl

284 Views

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