NIMCET 2021 Previous Year Questions (PYQs) – Page 6 of 12
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In a triangle, if the sum of two sides is x and their product is y such that (x+z)(x-z)=y, where z is the third side of the triangle , then triangle is
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Solution
If $H_1,H_2,\ldots,H_n$ are n harmonic means between a and b $(b\ne a)$;,then $\frac{{{H}}_n+a}{{{H}}_n-a}+\frac{{{H}}_n+b}{{{H}}_n-b}$
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Solution
Step 1: Convert HP to AP
Since $a, H_1, H_2, \ldots, H_n, b$ is in HP,
$\dfrac{1}{a},\ \dfrac{1}{H_1},\ \ldots,\ \dfrac{1}{H_n},\ \dfrac{1}{b}$ is in AP with $n+2$ terms.
Step 2: Find common difference $d$
$d = \dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{n+1} = \dfrac{a-b}{ab(n+1)}$
Step 3: Find $H_1$ and $H_n$
$\dfrac{1}{H_1} = \dfrac{1}{a} + d = \dfrac{a+bn}{ab(n+1)}$ $\Rightarrow H_1 = \dfrac{ab(n+1)}{a+bn}$
$\dfrac{1}{H_n} = \dfrac{1}{b} - d = \dfrac{an+b}{ab(n+1)}$ $\Rightarrow H_n = \dfrac{ab(n+1)}{an+b}$
Step 4: Evaluate $\dfrac{H_1+a}{H_1-a}$ using Componendo-Dividendo
$\dfrac{H_1}{a} = \dfrac{b(n+1)}{a+bn}$
Applying componendo-dividendo:
$\dfrac{H_1+a}{H_1-a} = \dfrac{b(n+1)+(a+bn)}{b(n+1)-(a+bn)} $
Since $a, H_1, H_2, \ldots, H_n, b$ is in HP,
$\dfrac{1}{a},\ \dfrac{1}{H_1},\ \ldots,\ \dfrac{1}{H_n},\ \dfrac{1}{b}$ is in AP with $n+2$ terms.
Step 2: Find common difference $d$
$d = \dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{n+1} = \dfrac{a-b}{ab(n+1)}$
Step 3: Find $H_1$ and $H_n$
$\dfrac{1}{H_1} = \dfrac{1}{a} + d = \dfrac{a+bn}{ab(n+1)}$ $\Rightarrow H_1 = \dfrac{ab(n+1)}{a+bn}$
$\dfrac{1}{H_n} = \dfrac{1}{b} - d = \dfrac{an+b}{ab(n+1)}$ $\Rightarrow H_n = \dfrac{ab(n+1)}{an+b}$
Step 4: Evaluate $\dfrac{H_1+a}{H_1-a}$ using Componendo-Dividendo
$\dfrac{H_1}{a} = \dfrac{b(n+1)}{a+bn}$
Applying componendo-dividendo:
$\dfrac{H_1+a}{H_1-a} = \dfrac{b(n+1)+(a+bn)}{b(n+1)-(a+bn)} $
$= \dfrac{a+b(2n+1)}{b-a} \quad \cdots(1)$
Step 5: Evaluate $\dfrac{H_n+b}{H_n-b}$ using Componendo-Dividendo
$\dfrac{H_n}{b} = \dfrac{a(n+1)}{an+b}$
Applying componendo-dividendo:
$\dfrac{H_n+b}{H_n-b} = \dfrac{a(n+1)+(an+b)}{a(n+1)-(an+b)} $
Step 5: Evaluate $\dfrac{H_n+b}{H_n-b}$ using Componendo-Dividendo
$\dfrac{H_n}{b} = \dfrac{a(n+1)}{an+b}$
Applying componendo-dividendo:
$\dfrac{H_n+b}{H_n-b} = \dfrac{a(n+1)+(an+b)}{a(n+1)-(an+b)} $
$= \dfrac{b+a(2n+1)}{a-b} \quad \cdots(2)$
Step 6: Add (1) and (2)
$\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} $
Step 6: Add (1) and (2)
$\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} $
$= \dfrac{a+b(2n+1)}{b-a} + \dfrac{b+a(2n+1)}{a-b}$
$= \dfrac{a+b(2n+1) - b - a(2n+1)}{b-a}$
$= \dfrac{(a-b) + (2n+1)(b-a)}{b-a}$
$= \dfrac{(b-a)(2n+1)-(b-a)}{b-a}$
$= (2n+1) - 1$
$= 2n$
Answer: $\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} = \boxed{2n}$
$= \dfrac{a+b(2n+1) - b - a(2n+1)}{b-a}$
$= \dfrac{(a-b) + (2n+1)(b-a)}{b-a}$
$= \dfrac{(b-a)(2n+1)-(b-a)}{b-a}$
$= (2n+1) - 1$
$= 2n$
Answer: $\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} = \boxed{2n}$
If $y=\sin ^{-1}(\frac{{x}^2+1}{\sqrt[]{1+3{x}^2+{x}^4}}),\, (x>0),$ then $\frac{dy}{dx}$=
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Solution
If $32\, \tan ^8\theta=2\cos ^2\alpha-3\cos \alpha$ and $3\, \cos \, 2\theta=1$, then the general value of $\alpha$ =
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Solution
In a triangle ABC, $a\cos ^2\frac{C}{2}+\, c\, \, {\cos }^2\frac{A}{2}=\frac{3b}{2}$ then the sides of the triangle are in
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Solution
If |k|=5 and 0° ≤ θ ≤ 360°, then the number of distinct solutions of 3cosθ + 4sinθ = k is
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Solution
If θ is acute angle between the pair of lines $x^2-7xy+12y^2=0$, then $\frac{2\cos \theta+3\sin \theta}{4\sin \theta+5\cos \theta}=$
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Solution
The four geometric means between 2 and 64 are
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Solution
Step 1: Set up the GP
The sequence is: $2,\ G_1,\ G_2,\ G_3,\ G_4,\ 64$
Total terms $= 6$, so $n = 6$
Step 2: Find common ratio $r$
$a = 2,\ a_6 = 64$
$a_6 = a \cdot r^{n-1}$
$64 = 2 \cdot r^5$
$r^5 = 32$
$r^5 = 2^5$
$r = 2$
Step 3: Find the four geometric means
$G_1 = ar = 2 \times 2 = 4$
$G_2 = ar^2 = 2 \times 4 = 8$
$G_3 = ar^3 = 2 \times 8 = 16$
$G_4 = ar^4 = 2 \times 16 = 32$
Answer: The four geometric means are $\boxed{4, 8, 16, 32}$
The sequence is: $2,\ G_1,\ G_2,\ G_3,\ G_4,\ 64$
Total terms $= 6$, so $n = 6$
Step 2: Find common ratio $r$
$a = 2,\ a_6 = 64$
$a_6 = a \cdot r^{n-1}$
$64 = 2 \cdot r^5$
$r^5 = 32$
$r^5 = 2^5$
$r = 2$
Step 3: Find the four geometric means
$G_1 = ar = 2 \times 2 = 4$
$G_2 = ar^2 = 2 \times 4 = 8$
$G_3 = ar^3 = 2 \times 8 = 16$
$G_4 = ar^4 = 2 \times 16 = 32$
Answer: The four geometric means are $\boxed{4, 8, 16, 32}$
The lines $px+qy=1$ and $qx+py=1$ are respectively the sides AB, AC of the triangle ABC and the base BC is bisected at $(p,q)$. Equation of the median of the triangle through the vertex A is
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Solution
If the vectors $a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k},\hat{i}+\hat{j}+c\hat{k}$ , $(a,b,c\ne1)$ are coplanar, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$
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Solution
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