1
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}, x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

A
$\frac{1}{x^3 y}$
B
$\frac{1}{x y^3}$
C
$-\frac{1}{x y^3}$
D
$-\frac{1}{x^3 y}$
2
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $y$ is a function of $x$ and $\log (x+y)=2 x y$, then the value of $y^{\prime}(0)$ is

A
1
B
$-$1
C
2
D
0
3
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $x^2 y^2=\sin ^{-1} x+\cos ^{-1} x$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ and $y=2$ is

A
$\frac{1}{2}$
B
$2$
C
$-\frac{1}{2}$
D
$-2$
4
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\frac{\mathrm{d}}{\mathrm{d} x} \mathrm{f}(x)=4 x^3-\frac{3}{x^4}$ such that $\mathrm{f}(2)=0$, then $\mathrm{f}(x)$ is equal to

A
$x^4+\frac{1}{x^3}+\frac{129}{8}$
B
$x^4+\frac{1}{x^3}-\frac{129}{8}$
C
$x^3+\frac{1}{x^4}+\frac{129}{8}$
D
  $x^3+\frac{1}{x^4}-\frac{129}{8}$
MHT CET Subjects
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