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Integrals

Multiple Choice Questions

291.

The number of points having both coordinates as integers that lie in the interior of the triangle with vertices (0,0), (0,41) and (41,0) is

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292.

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

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

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293.

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

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294.

At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P with respect to the additional number of workers x is given by $dP over dx space equals space 100 minus 12 square root of straight x$ . If the firm employs 25 more workers, then
the new level of production of items is

2500
3000
3500
3500

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295.

If ∫f x dx =Ψ (x),then ∫x⁵ f(x³)dx is equal to

$1 third space open square brackets straight x cubed straight psi left parenthesis straight x cubed right parenthesis minus integral straight x squared space straight psi left parenthesis straight x cubed right parenthesis dx close square brackets space plus straight C$
$1 third space straight x cubed straight psi left parenthesis straight x cubed right parenthesis minus 3 integral straight x squared space straight psi left parenthesis straight x cubed right parenthesis dx space plus straight C$
$1 third straight x cubed straight psi left parenthesis straight x cubed right parenthesis minus integral straight x squared space straight psi left parenthesis straight x cubed right parenthesis dx space plus space straight C$
$1 third straight x cubed straight psi left parenthesis straight x cubed right parenthesis minus integral straight x squared space straight psi left parenthesis straight x cubed right parenthesis dx space plus space straight C$

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296.

Statement I - The value of the integral $integral subscript straight pi divided by 6 end subscript superscript straight pi divided by 3 end superscript fraction numerator dx over denominator 1 plus space square root of tan space straight x end fraction$ is equal to π/6.
Statement II- $integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript straight a superscript straight b straight f left parenthesis straight a plus straight b minus straight x right parenthesis space dx$

Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I
Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I
Statement -I is True; Statement -II is False.
Statement -I is True; Statement -II is False.

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297.
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

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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298.

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(π)
g(x) +g(π)
g (x) - g(π)
g (x) - g(π)

146 Views

299.

The value of $integral subscript 0 superscript 1 fraction numerator 18 space log space left parenthesis 1 plus straight x right parenthesis over denominator 1 plus straight x squared end fraction space dx$ is

$straight pi over 8 space log space 2$
$straight pi over 2 space log space 2$
log 2
log 2

91 Views

300.

The integral $integral subscript straight pi over 4 end subscript superscript fraction numerator 3 straight pi over denominator 4 end fraction end superscript fraction numerator dx over denominator 1 plus space cos space straight x end fraction$ is equal to

-1
-2
2
2

194 Views

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

297. If the integral then an equal to -1 -2 1 1