NIMCET 2008 Previous Year Questions (PYQs) – Page 3 of 12
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If $f:\mathbb R\to\mathbb R$ and $g:\mathbb R\to\mathbb R$ are continuous functions, then evaluate
$\displaystyle \int_{-\pi/2}^{\pi/2}[f(x)+f(-x)][g(x)-g(-x)],dx$
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Solution
$f(x)+f(-x)$ is an even function
$g(x)-g(-x)$ is an odd function
Product of even and odd function is odd
Integral of odd function over symmetric limits is $0$
The maximum value of
$(\cos\alpha_1)(\cos\alpha_2)\cdots(\cos\alpha_n)$
where $0\le \alpha_1,\alpha_2,\ldots,\alpha_n\le\pi$ and
$(\cot\alpha_1)(\cot\alpha_2)\cdots(\cot\alpha_n)=1$ is
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Solution
By AM–GM, maximum occurs when
$\alpha_1=\alpha_2=\cdots=\alpha_n=\frac{\pi}{4}$
Then
$\cos\alpha_i=\frac{1}{\sqrt2}$
Product $=\left(\frac{1}{\sqrt2}\right)^n=\frac{1}{2^{n/2}}$
Let $M$ be a point inside the triangle $ABC$. Then which one of the following is true?
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Solution
By triangle inequality in $\triangle ABM$ and $\triangle ACM$,
$AB>MB-AM,\ AC>MC-AM$
Adding,
$AB+AC>MB+MC$
Answer: $\boxed{AB+AC>MB+MC}$
A line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line has intercepts $p$ and $q$. Which of the following is true?
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Solution
Intercept form of line remains invariant under rotation in terms of reciprocal squares.
Answer: $\boxed{\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{1}{p^2}+\dfrac{1}{q^2}}$
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: $\boxed{q^2-p^2}$
If $f(x)+f(1-x)=2$, then the value of
$f\left(\dfrac{1}{2001}\right)+f\left(\dfrac{2}{2001}\right)+\cdots+f\left(\dfrac{2000}{2001}\right)$ is
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Solution
Terms pair as
$f\left(\frac{k}{2001}\right)+f\left(1-\frac{k}{2001}\right)=2$
There are $1000$ such pairs.
Sum $=1000\times2=2000$
Answer: $\boxed{2000}$
Suppose $a,b,c$ are in A.P. with common difference $d$. Then $e^{1/c},e^{1/b},e^{1/a}$ are
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Solution
$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in H.P.
Exponentials of H.P. form G.P.
Answer: $\boxed{\text{G.P.}}$
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: $\boxed{x^2+x+1=0}$
In the expression
$(x+1)(x+4)(x+9)(x+16)\cdots(x+400)$
the coefficient of $x^{19}$ is
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Solution
Coefficient of $x^{19}$ equals sum of constants:
$1+4+9+\cdots+400$
This is sum of squares from $1^2$ to $20^2$:
$\frac{20(21)(41)}{6}=2870$
Answer: $\boxed{2870}$
The value of
$y=0.36\log_{0.25}\left(\dfrac13+\dfrac1{3^2}+\cdots\right)$
is
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Solution
The series is a G.P. with first term $\dfrac13$ and ratio $\dfrac13$.
Sum $=\dfrac{\frac13}{1-\frac13}=\dfrac12$
So
$y=0.36\log_{0.25}\left(\dfrac12\right)$
$\log_{0.25}\left(\dfrac12\right)=\dfrac{\log(1/2)}{\log(1/4)}=\dfrac{-1}{-2}=\dfrac12$
Hence
$y=0.36\times\dfrac12=0.18$
Answer: $\boxed{0.18}$
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