NIMCET Inequation Previous Year Question Paper and Solution with PDF

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NIMCET Previous Year Questions (PYQs)

NIMCET Inequation PYQ

Qus : 1 NIMCET PYQ

Find least integer $k$ such that $(k-2)x^2 + k + 8x + 4 > 0$ for all $x\in\mathbb{R}$.

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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

For quadratic $ax^2+bx+c>0$ for all $x$: $a>0$ → $k-2>0$ → $k>2$ Discriminant $<0$ $D=b^2-4ac=8^2-4(k-2)(k+4)$ Compute: $D=64-4(k^2+2k-8)=64-4k^2-8k+32$ $D=96-4k^2-8k<0$ Divide by $-4$: $k^2+2k-24>0$ $(k+?)(k+?)$ → roots $4$ and $-6$ So $k>4$ or $k<-6$ Combine with $k>2$ ⇒ $k>4$ Least integer = $5$

Qus : 2 NIMCET PYQ

If $x < -1$ and $2^{|x+1|} - 2^x = |2x - 1| + 1$ then the value of $x$ is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2009 PYQ

Solution

Since $x < -1$, then

$|x+1| = -(x+1)$ and $|2x - 1| = -(2x - 1)$ because $2x - 1 < -3 < 0$.

Equation: $2^{-(x+1)} - 2^x = -(2x - 1) + 1$

Right side: $-2x + 1 + 1 = -2x + 2$

Left side: $2^{-(x+1)} = \dfrac{1}{2^{x+1}}$

Try $x = -2$:

LHS: $2^{ -(-2+1) } - 2^{-2} = 2^{1} - \dfrac{1}{4} = 2 - 0.25 = 1.75$

RHS: $-2(-2) + 2 = 4 + 2 = 6$ → not equal Try $x = -2$ again carefully: Wait — check systematically. Better approach: substitute $x=-2$ in original equation: Left side: $2^{| -2 + 1 |} - 2^{-2} = 2^{1} - \dfrac{1}{4} = \dfrac{7}{4}$ Right side: $|2(-2) - 1| + 1 = | -5 | + 1 = 6$ Not equal → reject. Try $x = -1.5$ type pattern? Better substitute only options (valid since only one option has $x< -1$): Only option with $x < -1$ is (1) −2. But we tested −2 and it doesn't satisfy? Check original equation carefully: Original: $2^{|x+1|} - 2^x = |2x - 1| + 1$ Try $x=-2$ again: Left: $2^{|-2+1|} - 2^{-2} = 2^{1} - \dfrac{1}{4} = \dfrac{7}{4}$ Right: $|2(-2)-1| + 1 = |-5| +1 = 6$