NIMCET 2020 Previous Year Questions (PYQs) – Page 10 of 12
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Roots of equation are $ax^2-2bx+c=0$ are n and m ,
then the value of $\frac{b}{an^2+c}+\frac{b}{am^2+c}$ is
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Solution
Given quadratic equation is $ax^2 - 2bx + c = 0$.
Since $n$ and $m$ are the roots of this equation:
$an^2 - 2bn + c = 0 \implies an^2 + c = 2bn$
$am^2 - 2bm + c = 0 \implies am^2 + c = 2bm$
Now, the given expression is:
$\frac{b}{an^2+c} + \frac{b}{am^2+c}$
Substitute $an^2 + c = 2bn$ and $am^2 + c = 2bm$:
$= \frac{b}{2bn} + \frac{b}{2bm}$
Canceling $b$ from numerator and denominator:
$= \frac{1}{2n} + \frac{1}{2m}$
$= \frac{1}{2} \left( \frac{n+m}{nm} \right)$
From Vieta's formulas:
Sum of roots, $n + m = \frac{-(-2b)}{a} = \frac{2b}{a}$
Product of roots, $nm = \frac{c}{a}$
Substituting these values:
$= \frac{1}{2} \left( \frac{\frac{2b}{a}}{\frac{c}{a}} \right)$
$= \frac{1}{2} \left( \frac{2b}{c} \right)$
$= \frac{b}{c}$
Therefore, the value is $\frac{b}{c}$.
The number of values of $k$ for which the linear
equations
4x + ky + 2z = 0
kx + 4y + z = 0
2x + 2y + z = 0
posses a non-zero solution is
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Solution
A homogeneous system has a non-trivial solution $\iff$ the determinant of its coefficient matrix is $0$.
Coefficient matrix $A=\begin{bmatrix}4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1\end{bmatrix}$. Hence, $$ \det(A)= \begin{vmatrix} 4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1 \end{vmatrix} =-(k-4)(k-2). $$ Setting $\det(A)=0 \Rightarrow -(k-4)(k-2)=0 \Rightarrow k=2 \text{ or } k=4.$
Therefore, the number of values of $k$ is $\boxed{2}$.
Coefficient matrix $A=\begin{bmatrix}4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1\end{bmatrix}$. Hence, $$ \det(A)= \begin{vmatrix} 4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1 \end{vmatrix} =-(k-4)(k-2). $$ Setting $\det(A)=0 \Rightarrow -(k-4)(k-2)=0 \Rightarrow k=2 \text{ or } k=4.$
Therefore, the number of values of $k$ is $\boxed{2}$.
Let A = (aij) and B = (bij) be two square matricesof order n and det(A) denotes the determinant of A.
Then, which of the following is not correct.
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Solution
The tangent to an ellipse x2 + 16y2 = 16 and making angel 60° with X-axis is:
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Solution
Find the number of point(s) of intersection of the
ellipse $\dfrac{x^2}{4}+\dfrac{(y-1)^2}{9}=1$ and the circle x2 + y2 = 4
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Solution
An arithmetic progression has 3 as its first term.
Also, the sum of the first 8 terms is twice the sum of
the first 5 terms. Then what is the common
difference?
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Solution
Step 1: Write sum formula
$S_n = \dfrac{n}{2}[2a + (n-1)d]$
$S_8 = \dfrac{8}{2}[6 + 7d] $
$S_n = \dfrac{n}{2}[2a + (n-1)d]$
$S_8 = \dfrac{8}{2}[6 + 7d] $
$= 4(6+7d) $
$= 24 + 28d$
$S_5 = \dfrac{5}{2}[6 + 4d] $
$S_5 = \dfrac{5}{2}[6 + 4d] $
$= \dfrac{5}{2}(6+4d) $
$= 15 + 10d$
Step 2: Apply condition $S_8 = 2S_5$
$24 + 28d = 2(15 + 10d)$
$24 + 28d = 30 + 20d$
$8d = 6$
$d = \dfrac{3}{4}$
Answer: $d = \boxed{\dfrac{3}{4}}$
Step 2: Apply condition $S_8 = 2S_5$
$24 + 28d = 2(15 + 10d)$
$24 + 28d = 30 + 20d$
$8d = 6$
$d = \dfrac{3}{4}$
Answer: $d = \boxed{\dfrac{3}{4}}$
If a + b + c = 0, then the value of $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}$
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Solution
Just Let a=1, b=1 and c=-2 and put in given question
$\frac{1}{-2}+\frac{1}{-2}+\frac{4}{1}$
$\frac{1}{-2}+\frac{1}{-2}+\frac{4}{1}$
$=-1+4=3$
If $f(x)=\begin{cases}{{x}^2} & {,\leq0} \\ {2\sin x} & {,0}\end{cases}$, then x = 0 is
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Solution
If 
is a continuous function at x = 0, then the value of k is
is a continuous function at x = 0, then the value of k is1
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Solution
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