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

Let $\mathrm{f}(x)=\frac{x}{\sqrt{\mathrm{a}^2+x^2}}-\frac{\mathrm{d}-x}{\sqrt{\mathrm{~b}^2+(\mathrm{d}-x)^2}}, x \in \mathbb{R}$ where $\mathrm{a}, \mathrm{b}, \mathrm{d}$ are non-zero real constants. Then

A
$\mathrm{f}^{\prime}$ is not a continuous function of $x$.
B
f is neither increasing nor decreasing function of $x$.
C
f is an increasing function of $x$.
D
f is a decreasing function of $x$.
2
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $y=(\sin x)^{\tan x}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to

A
$(\sin x)^{\tan x}\left(1+\sec ^2 x \log (\sin x)\right)$
B
$\tan x(\sin x)^{\tan x-1} \cos x$
C
$(\sin x)^{\tan x} \sec ^2 x \log \sin x$
D
$\tan x(\sin x)^{\tan x-1}$
3
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{f}(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)$, then $f^{\prime}(1)=$

A
60
B
240
C
80
D
120
4
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
$-\frac{1}{x^3 y}$
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