MathematicsQ1–2 of 2 questions
Page 1 of 1 · Gujarati
1
MathematicsMediumMCQIIT JEE · 1960
$\frac{18^3 + 7^3 + 3 \times 18 \times 7 \times 25}{3^6 + 6 \times 243 \times 2 + 15 \times 81 \times 4 + 20 \times 27 \times 8 + 15 \times 9 \times 16 + 6 \times 3 \times 32 + 64}$ ની કિંમત શોધો.
A
$1$
B
$5$
C
$25$
D
$100$
Solution
(A) અંશ $a^3 + b^3 + 3ab(a + b) = (a + b)^3$ સ્વરૂપમાં છે,જ્યાં $a = 18$ અને $b = 7$ છે.
$a + b = 18 + 7 = 25$ હોવાથી,અંશ $25^3$ થાય.
છેદ $(x + y)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} y^k$ સ્વરૂપમાં છે.
અહીં,$x = 3$ અને $y = 2$ છે,તેથી છેદ $(3 + 2)^6 = 5^6$ થાય.
$5^6 = (5^2)^3 = 25^3$ હોવાથી,છેદ $25^3$ થાય.
તેથી,પદાવલિની કિંમત $\frac{25^3}{25^3} = 1$ છે.
$a + b = 18 + 7 = 25$ હોવાથી,અંશ $25^3$ થાય.
છેદ $(x + y)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} y^k$ સ્વરૂપમાં છે.
અહીં,$x = 3$ અને $y = 2$ છે,તેથી છેદ $(3 + 2)^6 = 5^6$ થાય.
$5^6 = (5^2)^3 = 25^3$ હોવાથી,છેદ $25^3$ થાય.
તેથી,પદાવલિની કિંમત $\frac{25^3}{25^3} = 1$ છે.
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2
MathematicsEasyMCQIIT JEE · 1960
જો $A(6, 3)$,$B(-3, 5)$,$C(4, -2)$ અને $D(x, 3x)$ ચાર બિંદુઓ હોય. જો $\Delta DBC$ અને $\Delta ABC$ ના ક્ષેત્રફળનો ગુણોત્તર $1 : 2$ હોય,તો $x$ ની કિંમત શોધો.
A
$\frac{11}{8}$
B
$\frac{8}{11}$
C
$3$
D
$\text{આપેલ પૈકી કોઈ નહીં}$
Solution
(A) ત્રિકોણનું ક્ષેત્રફળ $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ સૂત્ર દ્વારા મળે છે.
$\Delta ABC$ માટે: $\text{ક્ષેત્રફળ} = \frac{49}{2}$.
$\Delta DBC$ માટે: $\text{ક્ષેત્રફળ} = |14x - 7|$.
આપેલ છે કે $\frac{\text{Area}(\Delta DBC)}{\text{Area}(\Delta ABC)} = \frac{1}{2}$,તેથી $2 \times |14x - 7| = \frac{49}{2}$.
$|14x - 7| = \frac{49}{4}$.
ઉકેલતા,$x = \frac{11}{8}$ અથવા $x = -\frac{3}{8}$.
તેથી,સાચો વિકલ્પ $\frac{11}{8}$ છે.
$\Delta ABC$ માટે: $\text{ક્ષેત્રફળ} = \frac{49}{2}$.
$\Delta DBC$ માટે: $\text{ક્ષેત્રફળ} = |14x - 7|$.
આપેલ છે કે $\frac{\text{Area}(\Delta DBC)}{\text{Area}(\Delta ABC)} = \frac{1}{2}$,તેથી $2 \times |14x - 7| = \frac{49}{2}$.
$|14x - 7| = \frac{49}{4}$.
ઉકેલતા,$x = \frac{11}{8}$ અથવા $x = -\frac{3}{8}$.
તેથી,સાચો વિકલ્પ $\frac{11}{8}$ છે.
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