AP EAMCET · Maths · Basic of Mathematics
If \(\frac{x+1}{(x-1)^2\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C x+D}{x^2+1}\), then \(\sqrt{3 A^2+4 D^2+5 C^2+B^2}=\)
- A \(\frac{3}{2}\)
- B \(\frac{1}{2}\)
- C 1
- D 2
Answer & Solution
Correct Answer
(D) 2
Step-by-step Solution
Detailed explanation
\(x+1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2\) \(x=1 \Rightarrow 1+1 = A(0) + B(1^2+1) + (C(1)+D)(0)^2 \Rightarrow 2=2B \Rightarrow B=1\) \(x+1 = A(x^3-x^2+x-1) + (x^2+1) + (Cx+D)(x^2-2x+1)\) Coefficient of \(x^3: 0 = A+C \Rightarrow C=-A\) Constant term:…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- The probability distribution of a random variable X is as follows. Then the mean of X is
\(\mathrm{X} = \mathrm{x}_{\mathrm{i}}\) \(-2\) \(-1\) \(0\) \(1\) \(2\) \(\mathrm{P}\left(\mathrm{X} = \mathrm{x}_{\mathrm{i}}\right)\) \(\mathrm{k}^2 / 3\) \(\mathrm{k}^2\) \(2 \mathrm{k}^2 / 3\) \(k / 2\) \(k / 2\) AP EAMCET 2025 Medium - Thecoefficientof \(x^3\) in the expansion of \((1-2 x)^{\frac{1}{2}}(1+3 x)^{\frac{1}{3}}\) isAP EAMCET 2023 Hard
- \[
\int \frac{1}{(\sin x+\cos x+\sqrt{2} \sqrt{\sin 2 x})^2} d x=
\]AP EAMCET 2022 Hard - The equation of a plane passing through \((-1,2,3)\) and whose normal makes equal angles with the coordinate axes isAP EAMCET 2023 Easy
- The circle \(4 x^2+4 y^2-12 x-12 y+9=0\)AP EAMCET 2013 Medium
- If \(x^\alpha \frac{d y}{d x}=y^\beta(\gamma \log x+\delta \log y+1)\) is a homogeneous differential equation, thenAP EAMCET 2023 Medium
More PYQs from AP EAMCET
- The greatest integer \(\mathrm{r}\) such that \(30^{\mathrm{r}}\) divides 30 ! isAP EAMCET 2023 Easy
- An unbiased coin is tossed to get 2 points for turning up a head and one point for the tail. If three unbiased coins are tossed simultaneously, then the probability of getting a total of odd number of points isAP EAMCET 2004 Medium
- \(A\) and \(B\) each select one number at random from the distinct numbers \(1,2,3, \ldots, n\) and the probability that the number selected by \(A\) is less than the number selected by \(B\) is \(\frac{1009}{2019}\). Now, the probability that the number selected by \(B\) is the number immediately next to the number selected by \(A\) isAP EAMCET 2019 Hard
- \(\begin{aligned} & \text { If } \mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{b}=2 \mathbf{i}-\mathbf{j}+3 \mathbf{k} \text { and } \mathbf{c}=\mathbf{i}-\mathbf{j} \\ & \text { and if } 6 \mathbf{i}+2 \mathbf{j}+3 \mathbf{k}=\lambda_1(\mathbf{a} \times \mathbf{b}) \\ & +\lambda_2(\mathbf{b} \times \mathbf{c})+\lambda_3(\mathbf{c} \times \mathbf{a}) \text {, then }\left(\lambda_1, \lambda_2, \lambda_3\right)=\end{aligned}\)AP EAMCET 2018 Easy
- When three NAND logic gates are connected as shown in the figure, then the logic gate equivalent to the circuit is
AP EAMCET 2025 Medium - \(\lim _{n \rightarrow \infty}\left[\begin{array}{l}\frac{1}{n}+\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(n+3)^3} \\ +\ldots+\frac{1}{125 n}\end{array}\right]=\)AP EAMCET 2019 Hard