1
AP EAPCET 2025 - 23rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the lengths of the tangent, subtangent, normal and subnormal for the curve $y=x^2+x-1$ at the point $(1,1)$ are $a, b, c$ and $d$ respectively, then their increasing order is

A

$b, d, a, c$

B

b, a, c, d

C

$a, b, c, d$

D

$b, a, d, c$

2
AP EAPCET 2025 - 23rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{x+1}{x^3-1} d x= $$

A

$\frac{1}{3} \log \left(\frac{x+1}{x^2+x+1}\right)+C$

B

$\frac{1}{3} \log \left(\frac{(x-1)^2}{x^2+x+1}\right)+C$

C

$\frac{1}{3} \log \left(\frac{x-1}{x^2+x+1}\right)+C$

D

$\frac{1}{3} \log \left(\frac{(x+1)^2}{x^2-x+1}\right)+C$

3
AP EAPCET 2025 - 23rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{x^4-16 x^2+2 x+8}{x^3-4 x^2+2} d x= $$

A

$\frac{x^2+8 x+C}{2}$

B

$x^2+8 x+C$

C

$x^3-4 x+C$

D

$\frac{x^2-8 x+C}{2}$

4
AP EAPCET 2025 - 23rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{\sec ^2 x}{(\sec x+\tan x)^{\frac{5}{2}}} d x= $$

A

$-\frac{(\sec x+\tan x)^{\frac{5}{2}}}{5}-\frac{(\sec x+\tan x)^{\frac{7}{2}}}{7}+C$

B

$-\frac{(\sec x-\tan x)^{\frac{5}{2}}}{5}-\frac{(\sec x-\tan x)^{\frac{7}{2}}}{7}+C$

C

$-\frac{(\sec x+\tan x)^{\frac{3}{2}}}{3}-\frac{(\sec x+\tan x)^{\frac{7}{2}}}{7}+C$

D

$-\frac{(\sec x-\tan x)^{\frac{3}{2}}}{3}-\frac{(\sec x-\tan x)^{\frac{7}{2}}}{7}+C$