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Question

$not stretchy integral fraction numerator e to the power of x end exponent d x over denominator cos invisible function application h x plus sin invisible function application h x end fraction$ is

$\frac{1}{2} e^{2 x} + c$
$x + c$
$l o g (c o t h x) + c$
$l o g (t a n h x) + c$

The correct answer is: $x plus c$

Related Questions to study

$integral open parentheses x minus blank to the power of 10 end exponent C subscript 1 end subscript x to the power of 2 end exponent plus blank to the power of 10 end exponent C subscript 2 end subscript x to the power of 3 end exponent minus blank to the power of 10 end exponent C subscript 3 end subscript x to the power of 4 end exponent horizontal ellipsis horizontal ellipsis plus blank to the power of 10 end exponent C subscript 10 end subscript x to the power of 11 end exponent close parentheses d x$ dx equals:

$integral open parentheses x minus blank to the power of 10 end exponent C subscript 1 end subscript x to the power of 2 end exponent plus blank to the power of 10 end exponent C subscript 2 end subscript x to the power of 3 end exponent minus blank to the power of 10 end exponent C subscript 3 end subscript x to the power of 4 end exponent horizontal ellipsis horizontal ellipsis plus blank to the power of 10 end exponent C subscript 10 end subscript x to the power of 11 end exponent close parentheses d x$ dx equals:

If I = $not stretchy integral fraction numerator d x over denominator x square root of 1 minus x to the power of 3 end exponent end root end fraction$ , then I equals:

If I = $not stretchy integral fraction numerator d x over denominator x square root of 1 minus x to the power of 3 end exponent end root end fraction$ , then I equals:

The indefinite integral of $left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent$ is, for any arbitrary constant -

The indefinite integral of $left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent$ is, for any arbitrary constant -

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If f $open parentheses fraction numerator 3 x minus 4 over denominator 3 x plus 4 end fraction close parentheses$ = x + 2 then $not stretchy integral f left parenthesis x right parenthesis d x$ is equal to

If f $open parentheses fraction numerator 3 x minus 4 over denominator 3 x plus 4 end fraction close parentheses$ = x + 2 then $not stretchy integral f left parenthesis x right parenthesis d x$ is equal to

Let $x to the power of 2 end exponent not equal to n pi minus 1 comma n element of N$ , then $integral x square root of fraction numerator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses minus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses over denominator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses plus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction end root d x$ is equal to:

Let $x to the power of 2 end exponent not equal to n pi minus 1 comma n element of N$ , then $integral x square root of fraction numerator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses minus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses over denominator 2 s i n invisible function application open parentheses x to the power of 2 end exponent plus 1 close parentheses plus s i n invisible function application 2 open parentheses x to the power of 2 end exponent plus 1 close parentheses end fraction end root d x$ is equal to:

Statement I : y = f(x) = $fraction numerator x to the power of 2 end exponent minus 2 x plus 4 over denominator x to the power of 2 end exponent minus 2 x plus 5 end fraction$ , x $element of$ R Range of f(x) is [3/4, 1)
Statement II : $left parenthesis x minus 1 right parenthesis to the power of 2 end exponent equals fraction numerator 4 y minus 3 over denominator 1 minus y end fraction$ .

Statement I : y = f(x) = $fraction numerator x to the power of 2 end exponent minus 2 x plus 4 over denominator x to the power of 2 end exponent minus 2 x plus 5 end fraction$ , x $element of$ R Range of f(x) is [3/4, 1)
Statement II : $left parenthesis x minus 1 right parenthesis to the power of 2 end exponent equals fraction numerator 4 y minus 3 over denominator 1 minus y end fraction$ .

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Statement I : Function f(x) = sinx + {x} is periodic with period $2 pi$
Statement II : sin x and {x} are both periodic with period $2 pi$ and 1 respectively.

Statement I : Function f(x) = sinx + {x} is periodic with period $2 pi$
Statement II : sin x and {x} are both periodic with period $2 pi$ and 1 respectively.

Statement 1 : f : R $rightwards arrow$ R and $f left parenthesis x right parenthesis equals e to the power of x end exponent plus e to the power of negative x end exponent$ is bijective.
Statement 2 : $f colon R rightwards arrow R comma space f left parenthesis x right parenthesis equals e to the power of x minus e to the power of negative x end exponent$ is bijective.

Statement 1 : f : R $rightwards arrow$ R and $f left parenthesis x right parenthesis equals e to the power of x end exponent plus e to the power of negative x end exponent$ is bijective.
Statement 2 : $f colon R rightwards arrow R comma space f left parenthesis x right parenthesis equals e to the power of x minus e to the power of negative x end exponent$ is bijective.

Statement I : Graph of y = tan x is symmetrical about origin
Statement II : Graph of $y equals sec squared invisible function application x$ is symmetrical about y-axis

Statement I : Graph of y = tan x is symmetrical about origin
Statement II : Graph of $y equals sec squared invisible function application x$ is symmetrical about y-axis

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Statement- $1 colon$ If $f left parenthesis x right parenthesis equals vertical line x minus 1 vertical line plus vertical line x minus 2 vertical line plus vertical line x minus 3 vertical line$ Where $2 less than x less than 3$ is an identity function.
Statement- $2 colon f colon A rightwards arrow$ R defined by $f left parenthesis x right parenthesis equals x$ is an identity function.

Statement- $1 colon$ If $f left parenthesis x right parenthesis equals vertical line x minus 1 vertical line plus vertical line x minus 2 vertical line plus vertical line x minus 3 vertical line$ Where $2 less than x less than 3$ is an identity function.
Statement- $2 colon f colon A rightwards arrow$ R defined by $f left parenthesis x right parenthesis equals x$ is an identity function.

Assertion (A) : Graph of $open curly brackets left parenthesis x comma y right parenthesis divided by y equals 2 to the power of negative x end exponent text and end text x comma y element of R close curly brackets$
Reason (R) : In the expression a^m/n, where a, m, n × J⁺, m represents the power to which a is be raised, whereas n determines the root to be taken; these two processes may be administered in either order with the same result.

Assertion (A) : Graph of $open curly brackets left parenthesis x comma y right parenthesis divided by y equals 2 to the power of negative x end exponent text and end text x comma y element of R close curly brackets$
Reason (R) : In the expression a^m/n, where a, m, n × J⁺, m represents the power to which a is be raised, whereas n determines the root to be taken; these two processes may be administered in either order with the same result.

Assertion: The period of $f left parenthesis x right parenthesis equals s i n invisible function application 2 x c o s invisible function application left square bracket 2 x right square bracket minus c o s invisible function application 2 x s i n invisible function application left square bracket 2 x right square bracket$ is 1/2.
Reason: The period of x – [x] is 1.

Assertion: The period of $f left parenthesis x right parenthesis equals s i n invisible function application 2 x c o s invisible function application left square bracket 2 x right square bracket minus c o s invisible function application 2 x s i n invisible function application left square bracket 2 x right square bracket$ is 1/2.
Reason: The period of x – [x] is 1.

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Assertion : Fundamental period of $c o s invisible function application x plus c o t invisible function application x text is end text 2 pi$ .
Reason : If the period of f(x) is $T subscript 1 end subscript$ and the period of g(x) is $T subscript 2 end subscript$ , then the fundamental period of f(x) + g(x) is the L.C.M. of $T subscript 1 end subscript$ and T

Assertion : Fundamental period of $c o s invisible function application x plus c o t invisible function application x text is end text 2 pi$ .
Reason : If the period of f(x) is $T subscript 1 end subscript$ and the period of g(x) is $T subscript 2 end subscript$ , then the fundamental period of f(x) + g(x) is the L.C.M. of $T subscript 1 end subscript$ and T

Assertion: The function defined by $f left parenthesis x right parenthesis equals x to the power of 3 end exponent plus a x to the power of 2 end exponent plus b x plus c$ is invertible if and only if $a to the power of 2 end exponent less or equal than 3 b$ .
Reason: A function is invertible if and only if it is one-to-one and onto function.

Assertion: The function defined by $f left parenthesis x right parenthesis equals x to the power of 3 end exponent plus a x to the power of 2 end exponent plus b x plus c$ is invertible if and only if $a to the power of 2 end exponent less or equal than 3 b$ .
Reason: A function is invertible if and only if it is one-to-one and onto function.

Assertion : $f left parenthesis x right parenthesis equals s g n left parenthesis x minus vertical line x vertical line right parenthesis$ can never become positive.
Reason : f(x) = sgn x is always a positive function.

Assertion : $f left parenthesis x right parenthesis equals s g n left parenthesis x minus vertical line x vertical line right parenthesis$ can never become positive.
Reason : f(x) = sgn x is always a positive function.

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