1
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{f}(x)=\frac{x+x^2+x^3+\ldots \ldots \ldots \ldots+x^{\mathrm{n}}-\mathrm{n}}{x-1}$, for $x \neq 1$ is continuous at $x=1$, then $\mathrm{f}(1)=$

A
$\frac{\mathrm{n}(\mathrm{n}+1)(4 \mathrm{n}-1)}{6}$
B
$\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
C
$\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$
D
$\frac{\mathrm{n}(2 \mathrm{n}+1)}{4}$
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\lim _\limits{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=7$, then $a+b$ is equal to

A
$-$1
B
1
C
$-$11
D
11
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $f(x)=\left(\sin ^4 x+\cos ^4 x\right), 0< x<\frac{\pi}{2}$, then the function has minimum value at $x=$

A
$0.7934, \frac{\pi}{9}$
B
$\frac{1}{2}, \frac{\pi}{4}$
C
$\frac{5}{8}, \frac{\pi}{3}$
D
$0.75, \frac{\pi}{8}$
4
MHT CET 2024 4th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

$\lim _\limits{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ has the value

A
2
B
$\frac{1}{2}$
C
4
D
3
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