NIMCET 2027 Quadratic Equations Previous Year Questions (PYQs) – Page 1 of 3
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Showing 29 questions
If $x,y$ are real numbers such that
$2^{x+\frac{1}{2}}\times 4^{y-\frac{5}{6}}=3^{x-\frac{1}{2}}\times 9^{y-\frac{1}{3}}$ then which of the following is true?
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Solution
$2^{x+\frac{1}{2}}\times 4^{y-\frac{5}{6}}=3^{x-\frac{1}{2}}\times 9^{y-\frac{1}{3}}$
Convert powers:
$4=2^2,\quad 9=3^2$
$2^{x+\frac{1}{2}}\times 2^{2y-\frac{5}{3}}=3^{x-\frac{1}{2}}\times 3^{2y-\frac{2}{3}}$
$2^{x+2y-\frac{7}{6}}=3^{x+2y-\frac{7}{6}}$
Since bases are different,
$x+2y-\frac{7}{6}=0$
$6x+12y-7=0$
If the equation $|x^2 – 6x + 8| = a$ has four real solution then find the value of $a$?
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Solution
If
$2x^4 + x^3 - 11x^2 + x + 2 = 0$
then the values of
$x + \dfrac{1}{x}$
are:
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Solution
If x is real, then the minimum value of $\dfrac{x^2-x+1}{x^2+x+1}$is:
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Solution
Roots of $x^{2} - 2x + 4 = 0$ are $\alpha, \beta$.
Compute $ \alpha^{6} + \beta^{6} $.
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Solution
Roots:
$ \alpha = 1 + i\sqrt{3},\ \beta = 1 - i\sqrt{3} = 2(\cos 60^\circ \pm i\sin 60^\circ) $
So:
$ \alpha = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}) $
$ \Rightarrow \alpha^{6} = 2^{6} (\cos 2\pi + i\sin 2\pi) = 64 $
Same for $\beta$.
So
$ \alpha^{6} + \beta^{6} = 64 + 64 = 128 $
Let C denote the set of all tuples (x,y) which satisfy $x^2 -2^y=0$ where x and y are natural numbers. What is the cardinality of C?
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Solution
If $\alpha , \beta$ are the roots of $x^2-x-1=0$ and $A_n=\alpha^n+\beta^n$, the Arithmetic mean of $A_{n-1}$ and $A_n$ is
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Solution
If the roots of the quadratic equation $x^2+px+q=0$ are tan 30° and tan 15°
respectively, then the value of 2 + p - q is
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Solution
If $a,b$ are the roots of $x^2+px+1=0$ and $c,d$ are the roots of $x^2+qx+1=0$, the value of
$E=(a-c)(b-c)(a+d)(b+d)$ is
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Solution
Using symmetric sums and simplification,
$E=q^2-p^2$
Answer: $\boxed{q^2-p^2}$
Let $\alpha$ and $\beta$ be the roots of $x^2+x+1=0$. The equation whose roots are $\alpha^{19}$ and $\beta^{19}$ is
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Solution
$\alpha,\beta$ are complex cube roots of unity.
$\alpha^{19}=\alpha,\ \beta^{19}=\beta$
Same equation remains.
Answer: $\boxed{x^2+x+1=0}$
NIMCET 2027
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and More.
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