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IIT JEE Main Maths -Unit 2 -Quadratic equations- Exam Style Questions- New Syllabus

IIT JEE Main Maths - Exam Style Practice Questions - All Chapters
Question 
The product of all solutions of the equation \(e^{5\ln x} + 3 = 8e^x, \ x>0\) is:
\((1)\ e^{8/5}\)
\((2)\ e^{6/5}\)
\((3)\ e^{2}\)
\((4)\ e\)
▶️ Answer/Explanation
\(e^{5\ln x} + 3 = 8e^x \ \Rightarrow \ x^5 + 3 = 8x \ \Rightarrow \ x^5 – 8x + 3 = 0\). Let \(t = \ln x\), then \(5t^2 – 8t + 3 = 0 \ \Rightarrow\ t_1 + t_2 = \dfrac{8}{5}\). Hence \(x_1 x_2 = e^{8/5}\).
The correct answer is \((1)\ e^{8/5}\).

Question 
Let $\alpha_\theta$ and $\beta_\theta$ be the distinct roots of $2x^2 + (\cos \theta)x – 1 = 0$, $\theta \in (0, 2\pi)$. If $m$ and $M$ are the minimum and the maximum values of $\alpha_\theta^4 + \beta_\theta^4$, then $16(M + m)$ equals :
(1) 24
(2) 25
(3) 27
(4) 17
▶ Answer/Explanation
Ans. (2)
Sol. $(\alpha^2 + \beta^2)^2 – 2\alpha^2\beta^2$
$[(\alpha + \beta)^2 – 2\alpha\beta]^2 – 2(\alpha\beta)^2$
$\frac{(\cos \theta)^2 + 1}{4} – \frac{1}{2}$
$\frac{(\cos \theta)^2 + 1 – 2}{4} = \frac{\cos^2 \theta – 1}{4}$
$M = \frac{25}{16} – 1 = \frac{9}{16}$
$m = \frac{1}{2}$, $16(M + m) = 25$
Question 
If the set of all values of \(a\), for which the equation \(5x^3 – 15x – a = 0\) has three distinct real roots, is the interval \((\alpha, \beta)\), then \(\beta – 2\alpha\) is equal to ________
▶️ Answer/Explanation

Ans. (30)

Sol. \(5x^3 – 15x – a = 0\)
\(f(x) = 5x^3 – 15x\)
\(f'(x) = 15x^2 – 15 = 15(x-1)(x + 1)\)


\(a \in (-10, 10)\)
\(\alpha = -10, \beta = 10\)
\(\beta – 2\alpha = 10 + 20 = 30\)

Question 
If the equation \(a(b – c)x^2 + b(c – a)x + c(a – b) = 0\) has equal roots, where \(a + c = 15\) and \(\frac{b}{36} = \frac{5}{5}\), then \(a^2 + c^2\) is equal to ________
▶️ Answer/Explanation

Ans. (117)

Sol. \(a(b – c) x^2 + b (c – a) x + c(a – b) = 0\)
\(x = 1\) is root \(\therefore\) other root is 1
\(\alpha + \beta = -\frac{b(c-a)}{2a(b-c)}\)
\(\Rightarrow -bc + ab = 2ab – 2ac\)
\(\Rightarrow 2ac = b(a + c)\)
\(\Rightarrow 2ac = 15b\)
\(\Rightarrow 2ac = 108\)
\(\Rightarrow ac = 54\)
\(a + c = 15\)
\(a^2 + c^2 + 2ac = 225\)
\(a^2 + c^2 = 225 – 108 = 117\)

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