If the integral   then an equal to  from Mathematics JEE Yea

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

21.

The population p(t) at time t of a certain mouse species satisfies the differential equation fraction numerator dp space left parenthesis straight t right parenthesis over denominator dt end fraction space equals space 0.5 space left parenthesis straight t right parenthesis space minus 450. if p (0) = 850, then the  time at which the population becomes zero is

  • 2 log 18

  • log 9

  • 1 half space log space 18
  • 1 half space log space 18
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22.

Let a, b ∈ R be such that the function f given by f(x) = ln |x| + bx
2+ ax, x ≠ 0 has extreme values at x = –1 and x = 2.
Statement 1: f has local maximum at x = –1 and at x = 2.
Statement 2: straight a space equals space 1 half space and space straight b space equals space fraction numerator negative 1 over denominator 4 end fraction

  • Statement 1 is false, statement 2 is true

  • Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

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23.

If: R →R is a function defined by straight f space left parenthesis straight x right parenthesis space equals space left square bracket straight x right square bracket space cos space open parentheses fraction numerator 2 straight x minus 1 over denominator 2 end fraction close parentheses straight pi where [x] denotes the greatest integer function, then f is

  • continuous for every real x

  • discontinous only at x = 0

  • discontinuous only at non-zero integral values of x

  • discontinuous only at non-zero integral values of x

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24.

Let P and Q be 3 × 3 matrices with P ≠ Q. If P3= Q3 and P2Q = Q2P, then determinant of(P2+ Q2) is equal to

  • -2

  • 1

  • 0

  • 0

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25.

Consider the function f(x) = |x – 2| + |x – 5|, x ∈ R.
Statement 1: f′(4) = 0
Statement 2: f is continuous in [2, 5], differentiable in (2, 5) and f(2) = f(5).

  • Statement 1 is false, statement 2 is true

  • Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

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26.

Let straight a with hat on top space and space straight b with hat on top be  two unit vectors. If the vectors straight c equals space straight a with hat on top space plus 2 straight b with hat on top and straight d space equals space 5 straight a with hat on top space minus 4 straight b with hat on top are perpendicular to each other, then the angle between straight a with hat on top space and space straight b with hat on top is 

  • Ï€/6

  • Ï€/2

  • Ï€/3

  • Ï€/3

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27.

If the integral integral fraction numerator 5 space tan space straight x over denominator tan space straight x minus 2 end fraction straight d space straight x space equals space straight x
space plus space straight a space log space vertical line space sin space straight x minus space 2 space cos space straight x vertical line space plus space straight k comma  then an equal to 

  • -1

  • -2

  • 1

  • 1


D.

1

integral fraction numerator 5 space tan space straight x over denominator tan space straight x minus 2 end fraction straight d space straight x space equals space straight x space plus space straight a space log space vertical line space sin space straight x minus space 2 space cos space straight x vertical line space plus space straight k comma space... space left parenthesis straight i right parenthesis
Now, let us assume that I,

straight I space equals integral fraction numerator 5 space tan space straight x over denominator tan space straight x minus 2 end fraction straight d space straight x
Multiplying by cos x in numerator and denominator, we get
straight I space equals space integral fraction numerator 5 space sin space straight x over denominator sin space straight x minus 2 space cos space straight x end fraction straight d space straight x
This special integration requires special substitution of type
N' = A (D') +straight B space open parentheses fraction numerator dD apostrophe over denominator dx end fraction close parentheses
⇒ Let 5 sin x = (A + 2B) sin x + (B- 2A) cos x
Comapring the coefficients of sin x and cosx, 
we get
A +2B =5 and B- 2A = 0
Solving the above two equations in A and B
we get
A = 1 an B= 2
⇒ 5 sin x = ( sin x - 2 cos x)+ 2 (cos x + 2 sin x)
rightwards double arrow space straight I space equals space integral fraction numerator 5 space sin space straight x over denominator sin space straight x minus space 2 space cos space straight x end fraction dx
space equals space integral fraction numerator sin space straight x minus space 2 space cos space straight x right parenthesis space plus space 2 space left parenthesis cos space straight x space plus 2 space sin space straight x right parenthesis space dx over denominator left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis end fraction
rightwards double arrow space straight I space equals space integral fraction numerator left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis plus space 2 space left parenthesis cos space straight x space plus 2 space sin space straight x right parenthesis over denominator sin space straight x minus space 2 cos space straight x end fraction space dx
rightwards double arrow space straight I space equals space integral fraction numerator sin space straight x minus space 2 space cos space straight x over denominator sin space straight x minus space 2 space cos space straight x end fraction dx space plus space 2 space integral fraction numerator left parenthesis cos space straight x space plus 2 space sin space straight x right parenthesis over denominator left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis end fraction dx
rightwards double arrow space straight I space equals integral space 1 space dx space plus space 2 space integral fraction numerator straight d left parenthesis sin space straight x minus 2 space cos space straight x right parenthesis over denominator left parenthesis sin space straight x minus space 2 space cos space straight x right parenthesis space end fraction
rightwards double arrow space straight I space space equals space straight x space plus 2 space log space vertical line sin space straight x minus space 2 space cos space straight x right parenthesis vertical line space plus straight k space... space left parenthesis ii right parenthesis

where k is the constant of integration. Now, by comparing the value of l in eq. (i) and (ii) we get a = 2
⇒
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28.

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is

  • 880

  • 629

  • 630

  • 630

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29.

If g(x) = integral subscript 0 superscript straight x cos space 4 straight t space dt comma then g(x +π) equals

  • g(x)/g(Ï€)

  • g(x) +g(Ï€)

  • g (x) - g(Ï€)

  • g (x) - g(Ï€)

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30.

Let ABCD be a parallelogram such that AB with rightwards arrow on top space equals space straight q with rightwards arrow on top comma space AD with rightwards arrow on top space equals space straight p with rightwards arrow on top and ∠BAD be an acute angle. If straight r with rightwards arrow on top  is the vector that coincides with the altitude directed from the vertex B the side AD, then straight r with rightwards arrow on top is given byLet ABCD be a parallelogram such that AB = q,AD = p and ∠BAD be an acute angle. If r is the vector that coincides with the altitude directed from the vertex B to the side AD, then r is given by (1)

  • straight r with rightwards arrow on top space equals space 3 straight q with rightwards arrow on top space minus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
  • straight r with rightwards arrow on top space equals negative space straight q with rightwards arrow on top space plus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
  • straight r with rightwards arrow on top space equals space straight q with rightwards arrow on top space minus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
  • straight r with rightwards arrow on top space equals space straight q with rightwards arrow on top space minus fraction numerator open parentheses straight p with rightwards arrow on top stack. straight q with rightwards arrow on top close parentheses over denominator straight p with rightwards arrow on top straight p with rightwards arrow on top end fraction straight p with rightwards arrow on top
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