Please help me with this 6 MCQs Calculus. The tangent line to the curve y = tan

Question

Please help me with this 6 MCQs Calculus.

The tangent line to the curve y = tan x at the point (x/4, 1) is given by the equation: 2y - 4x = pi -2 4x - 2y = 2 - pi 4x - 2y = pi -2 (y x/4) = 2(x 1) Evaluate the definite integral integral_ln8^ln0 8e^x dx. The correct answer is 8e^ln6 8c^lna 24 -24 Evaluate the indefinite integral integral cos(ln x)/x dx. The correct answer is cos(ln z) + c sin(ln x) + c -cos(ln x)+ c -sin(ln x) + c Evaluate the integral integral x10^x^2 dx. The correct answer is (ln 10/2)10^x2 + e 2.10^x2/ln 10 + c - 10^x^2/2ln 10 + e 10^x^2/2ln 10 + e Let f(x) = Squareroot x + 1, x elementof [3, 8]. The area between the graph of f and the x-axis is given by: 38/3 3/38 5/38 38/5 Let f(x) = 8-x^2 and g(X) = x^2. The area of the region bounded by the curves f and g is given by: 3/64 64/3 68/3 3/68

Explanation / Answer

9)

given curve y=tanx

for slope of tangent , differentiate with respect to x

dy/dx=sec2x

at (/4,1)

dy/dx=sec2(/4)

dy/dx=2

slope of tangent =2

now,equation of tangent with slope 2 at point (/4,1) is of form y-1=2(x-(/4))

=> y-1=(2x-(/2))

=> y-1=(4x-)/2

=> 2y-2=(4x-)

=> 4x-2y=-2

option (3)

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10)limts are not clearly visible

ln3ln0 8exdx

= 8ex ln3|ln0

=8eln0-8eln3

=(8*0)-(8*3)

=0-24

=-24

option(4)

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11) cos(lnx)/x dx

substitute lnx =p , differentiate =>(1/x)dx =dp

=>cos(lnx)/x dx

=cos(p) dp

=sin(p) +c

=sin(lnx) +c

option(2)

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12)

x10x^2dx

substitute x2=u , differentiate =>2x dx =du => x dx =(1/2)du

=>x10x^2dx

=10u(1/2)du

=(1/2)10u(1/ln(10)) +c

=(10u/2ln(10)) +c

=(10x^2/2ln(10)) +c

option(4)

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