1
VITEEE 2022
MCQ (Single Correct Answer)
+1
-0

The value of $$f(x) = \mathop {\lim }\limits_{x \to 2} {{{x^3} - 3{x^2} + 4} \over {{x^4} - 7x - 2}}$$

A
0
B
3
C
1/4
D
5
2
VITEEE 2022
MCQ (Single Correct Answer)
+1
-0

The solution of the equation $$\frac{d y}{d x}+x(x+y)=x^3(x+y)^3-1$$ is

A
$$\frac{1}{(x+y)^2}=x^2+1+c e^x$$
B
$$\frac{1}{(x+y)^2}=x^2+1+c e^x$$
C
$$\frac{1}{(x+y)^2}=x^2+1+c e^{x^2}$$
D
$$\frac{1}{x+y}=x+1+c e^{x^2}$$
3
VITEEE 2022
MCQ (Single Correct Answer)
+1
-0

If $$\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$$, where $$|x|< \frac{1}{\sqrt{3}}$$ then value of $$y$$ is

A
$$\frac{3 x-x^3}{1-3 x^2}$$
B
$$\frac{3 x+x^3}{1-3 x^2}$$
C
$$\frac{3 x-x^3}{1+3 x^2}$$
D
$$\frac{3 x+x^3}{1+3 x^2}$$
4
VITEEE 2022
MCQ (Single Correct Answer)
+1
-0

If $$p$$ and $$q$$ are order and degree of the question $$\left(\frac{d^2 y}{d x^2}\right)^4+4 \frac{\left(\frac{d^2 y}{d x^2}\right)^2}{\left(\frac{d^3 y}{d x^3}\right)^3}+\frac{d^3 y}{d x^3}=x^2-1$$, then

A
$$p=3, q=3$$
B
$$p=3, r=2$$
C
$$p=4, r=3$$
D
$$p=3, r=4$$
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